Proof of “every convex function is continuous”












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A real valued function $f$ defined in $(a,b)$ is said to be convex if
$$f(lambda x+(1-lambda)y)le lambda f(x)+(1-lambda)f(y)$$
whenever $a < x < b,; a < y < b,; 0< lambda <1$.

Prove that every convex function is continuous.




Usually it uses the fact:

If $a < s < t < u < b$ then $$frac{f(t)-f(s)}{t-s}le frac{f(u)-f(s)}{u-s}lefrac{f(u)-f(t)}{u-t}.$$



I wonder whether any other version of this proof exists or not?










share|cite|improve this question











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  • 2




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    All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
    $endgroup$
    – copper.hat
    Dec 14 '12 at 7:39








  • 14




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    Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:33








  • 3




    $begingroup$
    Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:37






  • 2




    $begingroup$
    Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:44






  • 2




    $begingroup$
    What is the "usual proof" that uses that fact?
    $endgroup$
    – MCT
    Mar 8 '15 at 22:01
















57












$begingroup$



A real valued function $f$ defined in $(a,b)$ is said to be convex if
$$f(lambda x+(1-lambda)y)le lambda f(x)+(1-lambda)f(y)$$
whenever $a < x < b,; a < y < b,; 0< lambda <1$.

Prove that every convex function is continuous.




Usually it uses the fact:

If $a < s < t < u < b$ then $$frac{f(t)-f(s)}{t-s}le frac{f(u)-f(s)}{u-s}lefrac{f(u)-f(t)}{u-t}.$$



I wonder whether any other version of this proof exists or not?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
    $endgroup$
    – copper.hat
    Dec 14 '12 at 7:39








  • 14




    $begingroup$
    Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:33








  • 3




    $begingroup$
    Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:37






  • 2




    $begingroup$
    Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:44






  • 2




    $begingroup$
    What is the "usual proof" that uses that fact?
    $endgroup$
    – MCT
    Mar 8 '15 at 22:01














57












57








57


49



$begingroup$



A real valued function $f$ defined in $(a,b)$ is said to be convex if
$$f(lambda x+(1-lambda)y)le lambda f(x)+(1-lambda)f(y)$$
whenever $a < x < b,; a < y < b,; 0< lambda <1$.

Prove that every convex function is continuous.




Usually it uses the fact:

If $a < s < t < u < b$ then $$frac{f(t)-f(s)}{t-s}le frac{f(u)-f(s)}{u-s}lefrac{f(u)-f(t)}{u-t}.$$



I wonder whether any other version of this proof exists or not?










share|cite|improve this question











$endgroup$





A real valued function $f$ defined in $(a,b)$ is said to be convex if
$$f(lambda x+(1-lambda)y)le lambda f(x)+(1-lambda)f(y)$$
whenever $a < x < b,; a < y < b,; 0< lambda <1$.

Prove that every convex function is continuous.




Usually it uses the fact:

If $a < s < t < u < b$ then $$frac{f(t)-f(s)}{t-s}le frac{f(u)-f(s)}{u-s}lefrac{f(u)-f(t)}{u-t}.$$



I wonder whether any other version of this proof exists or not?







real-analysis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 13 '14 at 13:55









Phil-ZXX

1,33611437




1,33611437










asked Dec 14 '12 at 7:15









cowikcowik

308145




308145








  • 2




    $begingroup$
    All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
    $endgroup$
    – copper.hat
    Dec 14 '12 at 7:39








  • 14




    $begingroup$
    Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:33








  • 3




    $begingroup$
    Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:37






  • 2




    $begingroup$
    Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:44






  • 2




    $begingroup$
    What is the "usual proof" that uses that fact?
    $endgroup$
    – MCT
    Mar 8 '15 at 22:01














  • 2




    $begingroup$
    All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
    $endgroup$
    – copper.hat
    Dec 14 '12 at 7:39








  • 14




    $begingroup$
    Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:33








  • 3




    $begingroup$
    Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:37






  • 2




    $begingroup$
    Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
    $endgroup$
    – Michael Grant
    Aug 15 '14 at 19:44






  • 2




    $begingroup$
    What is the "usual proof" that uses that fact?
    $endgroup$
    – MCT
    Mar 8 '15 at 22:01








2




2




$begingroup$
All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
$endgroup$
– copper.hat
Dec 14 '12 at 7:39






$begingroup$
All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $mathbb{R}^n$ as well.
$endgroup$
– copper.hat
Dec 14 '12 at 7:39






14




14




$begingroup$
Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
$endgroup$
– Michael Grant
Aug 15 '14 at 19:33






$begingroup$
Your title is a bit misleading. It is not the case that every convex function is continuous. What is true is that every function that is finite and convex on an open interval is continuous on that interval (including $mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.
$endgroup$
– Michael Grant
Aug 15 '14 at 19:33






3




3




$begingroup$
Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
$endgroup$
– Michael Grant
Aug 15 '14 at 19:37




$begingroup$
Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-infty,+infty]$. Continuing my example above, for instance, we could define $f(x)=+infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)inmathbb{R}^n$---even $a,b<0$!
$endgroup$
– Michael Grant
Aug 15 '14 at 19:37




2




2




$begingroup$
Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
$endgroup$
– Michael Grant
Aug 15 '14 at 19:44




$begingroup$
Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.
$endgroup$
– Michael Grant
Aug 15 '14 at 19:44




2




2




$begingroup$
What is the "usual proof" that uses that fact?
$endgroup$
– MCT
Mar 8 '15 at 22:01




$begingroup$
What is the "usual proof" that uses that fact?
$endgroup$
– MCT
Mar 8 '15 at 22:01










7 Answers
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The pictorial version. (But it is the same as your inequality version, actually.)



Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)



AA1



By convexity, the $c$ point is above the $a,b$ line, as shown:



A2



Again, the $b$ point is above the $a,c$ line, as shown:



A3



The graph lies inside the red region,



A4



so obviously we have continuity at $a$.






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    16












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    I would be careful to rephrase the query as:



    Is there an alternative proof of the fact that a real-valued convex function defined on an open interval of the reals is continuous?



    Since in general convex functions are not continuous nor are they necessarily continuous when defined on open sets in topological vector spaces.



    An alternative might be to identify the point of discontinuity as x. Then there exists a point arbitrarily close to x, denoted x', whose value f(x') is bounded away by a constant from f(x). Depending on how you want your proof structured, you may think it sufficient to note that this implies the epigraph of the function is not closed and therefore the function is not lower semicontinuous. But every convex function on the reals is lower semicontinuous on the relative interior of its effective domain, which equals the domain of definition in this case.



    A more general proof of this property is given in "Convexity and Optimization in Banach Spaces." The authors prove the proposition that every proper convex function defined on a finite-dimensional separated topological linear space is continuous on the interior of its effective domain. You can likely see the relevant proof using Amazon's or Google Book's look inside feature.






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      You can do a proof by contradiction.



      Assume $finmathbb{R}^mathbb{R}$ is convex, but not continuous at some $x_0in(a,b)$. This means that:
      $$ exists_{epsilon>0}forall_{delta>0}exists_{xin(x_0-delta,x_0+delta)} : |f(x)-f(x_0)|geepsilon$$
      This formula implies that once we fix $delta$, $f$'s graph has infinitely many points in one of the areas: I, II, III or IV, with $x_0$ as an accumulation point of their $x$ coordinates:



      enter image description here



      We split our proof into 2 cases:



      $(1)$ The area is either I or II. In this case we select some point on the function's graph from that area: $(x_1,f(x_1))$, and draw a line segment from that point to $(x_0,f(x_0))$. We then select another point on the graph from the same area: $(x_2,f(x_2))$, whose $x$ coordinate is closer to $x_0$ than the intersection of our line segment and $y=f(x_0)+epsilon$ . This contradicts the convexity of $f$, as can be seen in the following image:
      enter image description here



      $(2)$ The area is either III or IV. Assume, without loss of generality, that the area is III. In this case we select some point on the function's graph to the right of $x_0$, say: $(x_1,f(x_1))$. We then draw a ray, which starts at $(x_1,f(x_1))$, and goes through $(x_0,f(x_0))$. We use $x'$ to denote the $x$ coordinate of the intersection of our ray and $y=f(x_0)-epsilon$. If they do not intersect, we set: $x'=-infty$. Next, we select another point: $(x_2, f(x_2))$ on $f$'s graph, in area III, with $x'<x_2<x_0$. Finally, we draw a segment between $(x_2, f(x_2))$ and $(x_1, f(x_1))$. This again contradicts convexity, as can be seen in the following image:
      enter image description here






      share|cite|improve this answer











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        I presume you mean "proper convexity' as in $(1)$ above not $(2)$



        Not just mere, " "midpoint convex/jensen convexity"/ "convexity in the sense of Jensen"," as in $(2)$, below.



        Although (its hardly mere I suppose) because the $(1)$ as defined in the question and $(2)$ below are, not, but are nonetheless "almost equivalent".



        (2) $$F(frac{x+y}{2}) leq frac{F(x)}{2} +frac{F(y)}{2}$$



        That is because under relatively mild conditions measurability, regularity conditions/boundedness conditions, midpoint convex function (in the sense of Jensen) are convex in the tradition .



        Apparently a real valued midpoint convex function $(2)$ already satisfies the definition of convexity as above $(1)$, except for the restriction that $sigma$ applies only to all rational numbers in the the unit interval (not just $2$, or dy-adics). That is before continuity is applied.



        That is $(1.a)$ below according to pt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7



        $$(1.a)forall , sigma in mathbb{Q}cap[0,1];, forall (x,y), in, text{dom}(F):, F(,sigma x + [1-sigma] y,), leq, sigma F(x),+,[1-sigma]F(y).$$



        Its presumably a bit confusing the use the words 'in the sense of Jensen' for midpoint convexity.



        This is because I believe that Jensen did, or helped develop the inequalities that any 'function' must satisfy, in order to be 'properly convex',as defined above in the question $(1)$.



        Because Jensen well as the weaker notion of midpoint convexity $(2)$ which apparently is equivalent to $(1.a)$, at least if the domain is real valued, and which is named after Jensen (midpt convexity is often called jensen- convexity).






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          I like GEdgar's pictorial explanation.
          The key idea is: for $f(x)$ in $(a,b)$, we try to prove $f(x)$ is bounded above and below.
          It's easy to show $f(x)$ is bounded above by definition of convex.
          To show $f(x)$ is bounded below, we can see GEdgar's 2nd and 3rd picture.



          Once we proved $f(x)$ is bounded, $|f(x)|leq M$,



          $$a < s < t < u < b$$
          let $$s < x < u quad;quad 0<varepsilon<u-x$$
          then we can show
          $$f(x+varepsilon)-f(x)leq dfrac{f(u)-f(s)}{(u-s)}varepsilon leq dfrac{Mvarepsilon}{(u-s)}$$
          as $varepsilon rightarrow 0$ $$|f(x+varepsilon)-f(x)| rightarrow 0$$ then $f(x+)=f(x)$
          the same way we can show $f(x-)=f(x)$,
          so $f(x)$ is continuous






          share|cite|improve this answer











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            0












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            The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set. If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex. This proof used the contrapositive.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
              $endgroup$
              – Michael Grant
              Aug 15 '14 at 19:38










            • $begingroup$
              Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
              $endgroup$
              – Squirtle
              Aug 15 '14 at 19:51






            • 1




              $begingroup$
              Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
              $endgroup$
              – Michael Grant
              Aug 15 '14 at 20:07












            • $begingroup$
              As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
              $endgroup$
              – Michael Grant
              Aug 15 '14 at 20:09








            • 1




              $begingroup$
              It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
              $endgroup$
              – Michael Grant
              Aug 16 '14 at 13:55





















            0












            $begingroup$

            Here is the picture of my proof



            Let, By contrary, $c$ be a point on the domain. Choose a $d$ then there exists a sequence ${a_n}_n$ that converges to c and $f(a_n) notin (f(c)-d,f(c)+d)$.
            So from the sequence ${a_n}_n$ we can always select a one sided monotonic sequence ${c_n}_n$ which is the subsequence of the original seq. as one side of $c$ must have infinitely many points of ${a_n}_n$, WLOG let it is the left sided sequence & $f(c_1)<f(c)-d$



            Now, take $M_1=(c_1,c)$ any point $x$ in $M_1$ must be $<f(c)-d$ as $exists m$ s.t. $c_mleq xleq c_{m+1}$ hence $x$ must belong lower area of line joining of $f(c_m)$ & $f(c_{m+1})$ (By convexity of $f$) so for all $x$ in $M_1$ $f(x)<f(c)-d$.



            Now, join the point $(c_1,f(c_1))$ & $(c,f(c))$ (let the line be $l_1$). It will cut at B at the $y=f(x)+d$. Let x-cordinate of B is $c^{'}$.



            Now, note $B=(c^{'},f(c^{'})$ lie upper area of $l_1$ otherwise contradicting the point $A=(c,f(c))$ (By convexity of $f$).
            Now from B we can always draw a line such that A lie upper area of this line intersecting at B' at $y=f(c)-d$ line. So let x-cordinate of B' is $c^{''}$.
            Now lastly join $(c^{''},f(c^{''}))$ & $(c^{'},f(c^{'}))$ contradicting the point A.






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              7 Answers
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              7 Answers
              7






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              95












              $begingroup$

              The pictorial version. (But it is the same as your inequality version, actually.)



              Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)



              AA1



              By convexity, the $c$ point is above the $a,b$ line, as shown:



              A2



              Again, the $b$ point is above the $a,c$ line, as shown:



              A3



              The graph lies inside the red region,



              A4



              so obviously we have continuity at $a$.






              share|cite|improve this answer









              $endgroup$


















                95












                $begingroup$

                The pictorial version. (But it is the same as your inequality version, actually.)



                Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)



                AA1



                By convexity, the $c$ point is above the $a,b$ line, as shown:



                A2



                Again, the $b$ point is above the $a,c$ line, as shown:



                A3



                The graph lies inside the red region,



                A4



                so obviously we have continuity at $a$.






                share|cite|improve this answer









                $endgroup$
















                  95












                  95








                  95





                  $begingroup$

                  The pictorial version. (But it is the same as your inequality version, actually.)



                  Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)



                  AA1



                  By convexity, the $c$ point is above the $a,b$ line, as shown:



                  A2



                  Again, the $b$ point is above the $a,c$ line, as shown:



                  A3



                  The graph lies inside the red region,



                  A4



                  so obviously we have continuity at $a$.






                  share|cite|improve this answer









                  $endgroup$



                  The pictorial version. (But it is the same as your inequality version, actually.)



                  Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)



                  AA1



                  By convexity, the $c$ point is above the $a,b$ line, as shown:



                  A2



                  Again, the $b$ point is above the $a,c$ line, as shown:



                  A3



                  The graph lies inside the red region,



                  A4



                  so obviously we have continuity at $a$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '13 at 21:51









                  GEdgarGEdgar

                  62.5k267171




                  62.5k267171























                      16












                      $begingroup$

                      I would be careful to rephrase the query as:



                      Is there an alternative proof of the fact that a real-valued convex function defined on an open interval of the reals is continuous?



                      Since in general convex functions are not continuous nor are they necessarily continuous when defined on open sets in topological vector spaces.



                      An alternative might be to identify the point of discontinuity as x. Then there exists a point arbitrarily close to x, denoted x', whose value f(x') is bounded away by a constant from f(x). Depending on how you want your proof structured, you may think it sufficient to note that this implies the epigraph of the function is not closed and therefore the function is not lower semicontinuous. But every convex function on the reals is lower semicontinuous on the relative interior of its effective domain, which equals the domain of definition in this case.



                      A more general proof of this property is given in "Convexity and Optimization in Banach Spaces." The authors prove the proposition that every proper convex function defined on a finite-dimensional separated topological linear space is continuous on the interior of its effective domain. You can likely see the relevant proof using Amazon's or Google Book's look inside feature.






                      share|cite|improve this answer









                      $endgroup$


















                        16












                        $begingroup$

                        I would be careful to rephrase the query as:



                        Is there an alternative proof of the fact that a real-valued convex function defined on an open interval of the reals is continuous?



                        Since in general convex functions are not continuous nor are they necessarily continuous when defined on open sets in topological vector spaces.



                        An alternative might be to identify the point of discontinuity as x. Then there exists a point arbitrarily close to x, denoted x', whose value f(x') is bounded away by a constant from f(x). Depending on how you want your proof structured, you may think it sufficient to note that this implies the epigraph of the function is not closed and therefore the function is not lower semicontinuous. But every convex function on the reals is lower semicontinuous on the relative interior of its effective domain, which equals the domain of definition in this case.



                        A more general proof of this property is given in "Convexity and Optimization in Banach Spaces." The authors prove the proposition that every proper convex function defined on a finite-dimensional separated topological linear space is continuous on the interior of its effective domain. You can likely see the relevant proof using Amazon's or Google Book's look inside feature.






                        share|cite|improve this answer









                        $endgroup$
















                          16












                          16








                          16





                          $begingroup$

                          I would be careful to rephrase the query as:



                          Is there an alternative proof of the fact that a real-valued convex function defined on an open interval of the reals is continuous?



                          Since in general convex functions are not continuous nor are they necessarily continuous when defined on open sets in topological vector spaces.



                          An alternative might be to identify the point of discontinuity as x. Then there exists a point arbitrarily close to x, denoted x', whose value f(x') is bounded away by a constant from f(x). Depending on how you want your proof structured, you may think it sufficient to note that this implies the epigraph of the function is not closed and therefore the function is not lower semicontinuous. But every convex function on the reals is lower semicontinuous on the relative interior of its effective domain, which equals the domain of definition in this case.



                          A more general proof of this property is given in "Convexity and Optimization in Banach Spaces." The authors prove the proposition that every proper convex function defined on a finite-dimensional separated topological linear space is continuous on the interior of its effective domain. You can likely see the relevant proof using Amazon's or Google Book's look inside feature.






                          share|cite|improve this answer









                          $endgroup$



                          I would be careful to rephrase the query as:



                          Is there an alternative proof of the fact that a real-valued convex function defined on an open interval of the reals is continuous?



                          Since in general convex functions are not continuous nor are they necessarily continuous when defined on open sets in topological vector spaces.



                          An alternative might be to identify the point of discontinuity as x. Then there exists a point arbitrarily close to x, denoted x', whose value f(x') is bounded away by a constant from f(x). Depending on how you want your proof structured, you may think it sufficient to note that this implies the epigraph of the function is not closed and therefore the function is not lower semicontinuous. But every convex function on the reals is lower semicontinuous on the relative interior of its effective domain, which equals the domain of definition in this case.



                          A more general proof of this property is given in "Convexity and Optimization in Banach Spaces." The authors prove the proposition that every proper convex function defined on a finite-dimensional separated topological linear space is continuous on the interior of its effective domain. You can likely see the relevant proof using Amazon's or Google Book's look inside feature.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 31 '12 at 6:05









                          MikeMike

                          1612




                          1612























                              11












                              $begingroup$

                              You can do a proof by contradiction.



                              Assume $finmathbb{R}^mathbb{R}$ is convex, but not continuous at some $x_0in(a,b)$. This means that:
                              $$ exists_{epsilon>0}forall_{delta>0}exists_{xin(x_0-delta,x_0+delta)} : |f(x)-f(x_0)|geepsilon$$
                              This formula implies that once we fix $delta$, $f$'s graph has infinitely many points in one of the areas: I, II, III or IV, with $x_0$ as an accumulation point of their $x$ coordinates:



                              enter image description here



                              We split our proof into 2 cases:



                              $(1)$ The area is either I or II. In this case we select some point on the function's graph from that area: $(x_1,f(x_1))$, and draw a line segment from that point to $(x_0,f(x_0))$. We then select another point on the graph from the same area: $(x_2,f(x_2))$, whose $x$ coordinate is closer to $x_0$ than the intersection of our line segment and $y=f(x_0)+epsilon$ . This contradicts the convexity of $f$, as can be seen in the following image:
                              enter image description here



                              $(2)$ The area is either III or IV. Assume, without loss of generality, that the area is III. In this case we select some point on the function's graph to the right of $x_0$, say: $(x_1,f(x_1))$. We then draw a ray, which starts at $(x_1,f(x_1))$, and goes through $(x_0,f(x_0))$. We use $x'$ to denote the $x$ coordinate of the intersection of our ray and $y=f(x_0)-epsilon$. If they do not intersect, we set: $x'=-infty$. Next, we select another point: $(x_2, f(x_2))$ on $f$'s graph, in area III, with $x'<x_2<x_0$. Finally, we draw a segment between $(x_2, f(x_2))$ and $(x_1, f(x_1))$. This again contradicts convexity, as can be seen in the following image:
                              enter image description here






                              share|cite|improve this answer











                              $endgroup$


















                                11












                                $begingroup$

                                You can do a proof by contradiction.



                                Assume $finmathbb{R}^mathbb{R}$ is convex, but not continuous at some $x_0in(a,b)$. This means that:
                                $$ exists_{epsilon>0}forall_{delta>0}exists_{xin(x_0-delta,x_0+delta)} : |f(x)-f(x_0)|geepsilon$$
                                This formula implies that once we fix $delta$, $f$'s graph has infinitely many points in one of the areas: I, II, III or IV, with $x_0$ as an accumulation point of their $x$ coordinates:



                                enter image description here



                                We split our proof into 2 cases:



                                $(1)$ The area is either I or II. In this case we select some point on the function's graph from that area: $(x_1,f(x_1))$, and draw a line segment from that point to $(x_0,f(x_0))$. We then select another point on the graph from the same area: $(x_2,f(x_2))$, whose $x$ coordinate is closer to $x_0$ than the intersection of our line segment and $y=f(x_0)+epsilon$ . This contradicts the convexity of $f$, as can be seen in the following image:
                                enter image description here



                                $(2)$ The area is either III or IV. Assume, without loss of generality, that the area is III. In this case we select some point on the function's graph to the right of $x_0$, say: $(x_1,f(x_1))$. We then draw a ray, which starts at $(x_1,f(x_1))$, and goes through $(x_0,f(x_0))$. We use $x'$ to denote the $x$ coordinate of the intersection of our ray and $y=f(x_0)-epsilon$. If they do not intersect, we set: $x'=-infty$. Next, we select another point: $(x_2, f(x_2))$ on $f$'s graph, in area III, with $x'<x_2<x_0$. Finally, we draw a segment between $(x_2, f(x_2))$ and $(x_1, f(x_1))$. This again contradicts convexity, as can be seen in the following image:
                                enter image description here






                                share|cite|improve this answer











                                $endgroup$
















                                  11












                                  11








                                  11





                                  $begingroup$

                                  You can do a proof by contradiction.



                                  Assume $finmathbb{R}^mathbb{R}$ is convex, but not continuous at some $x_0in(a,b)$. This means that:
                                  $$ exists_{epsilon>0}forall_{delta>0}exists_{xin(x_0-delta,x_0+delta)} : |f(x)-f(x_0)|geepsilon$$
                                  This formula implies that once we fix $delta$, $f$'s graph has infinitely many points in one of the areas: I, II, III or IV, with $x_0$ as an accumulation point of their $x$ coordinates:



                                  enter image description here



                                  We split our proof into 2 cases:



                                  $(1)$ The area is either I or II. In this case we select some point on the function's graph from that area: $(x_1,f(x_1))$, and draw a line segment from that point to $(x_0,f(x_0))$. We then select another point on the graph from the same area: $(x_2,f(x_2))$, whose $x$ coordinate is closer to $x_0$ than the intersection of our line segment and $y=f(x_0)+epsilon$ . This contradicts the convexity of $f$, as can be seen in the following image:
                                  enter image description here



                                  $(2)$ The area is either III or IV. Assume, without loss of generality, that the area is III. In this case we select some point on the function's graph to the right of $x_0$, say: $(x_1,f(x_1))$. We then draw a ray, which starts at $(x_1,f(x_1))$, and goes through $(x_0,f(x_0))$. We use $x'$ to denote the $x$ coordinate of the intersection of our ray and $y=f(x_0)-epsilon$. If they do not intersect, we set: $x'=-infty$. Next, we select another point: $(x_2, f(x_2))$ on $f$'s graph, in area III, with $x'<x_2<x_0$. Finally, we draw a segment between $(x_2, f(x_2))$ and $(x_1, f(x_1))$. This again contradicts convexity, as can be seen in the following image:
                                  enter image description here






                                  share|cite|improve this answer











                                  $endgroup$



                                  You can do a proof by contradiction.



                                  Assume $finmathbb{R}^mathbb{R}$ is convex, but not continuous at some $x_0in(a,b)$. This means that:
                                  $$ exists_{epsilon>0}forall_{delta>0}exists_{xin(x_0-delta,x_0+delta)} : |f(x)-f(x_0)|geepsilon$$
                                  This formula implies that once we fix $delta$, $f$'s graph has infinitely many points in one of the areas: I, II, III or IV, with $x_0$ as an accumulation point of their $x$ coordinates:



                                  enter image description here



                                  We split our proof into 2 cases:



                                  $(1)$ The area is either I or II. In this case we select some point on the function's graph from that area: $(x_1,f(x_1))$, and draw a line segment from that point to $(x_0,f(x_0))$. We then select another point on the graph from the same area: $(x_2,f(x_2))$, whose $x$ coordinate is closer to $x_0$ than the intersection of our line segment and $y=f(x_0)+epsilon$ . This contradicts the convexity of $f$, as can be seen in the following image:
                                  enter image description here



                                  $(2)$ The area is either III or IV. Assume, without loss of generality, that the area is III. In this case we select some point on the function's graph to the right of $x_0$, say: $(x_1,f(x_1))$. We then draw a ray, which starts at $(x_1,f(x_1))$, and goes through $(x_0,f(x_0))$. We use $x'$ to denote the $x$ coordinate of the intersection of our ray and $y=f(x_0)-epsilon$. If they do not intersect, we set: $x'=-infty$. Next, we select another point: $(x_2, f(x_2))$ on $f$'s graph, in area III, with $x'<x_2<x_0$. Finally, we draw a segment between $(x_2, f(x_2))$ and $(x_1, f(x_1))$. This again contradicts convexity, as can be seen in the following image:
                                  enter image description here







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Aug 15 '14 at 19:34

























                                  answered Aug 15 '14 at 19:21









                                  SomeStrangeUserSomeStrangeUser

                                  1,6501025




                                  1,6501025























                                      2












                                      $begingroup$

                                      I presume you mean "proper convexity' as in $(1)$ above not $(2)$



                                      Not just mere, " "midpoint convex/jensen convexity"/ "convexity in the sense of Jensen"," as in $(2)$, below.



                                      Although (its hardly mere I suppose) because the $(1)$ as defined in the question and $(2)$ below are, not, but are nonetheless "almost equivalent".



                                      (2) $$F(frac{x+y}{2}) leq frac{F(x)}{2} +frac{F(y)}{2}$$



                                      That is because under relatively mild conditions measurability, regularity conditions/boundedness conditions, midpoint convex function (in the sense of Jensen) are convex in the tradition .



                                      Apparently a real valued midpoint convex function $(2)$ already satisfies the definition of convexity as above $(1)$, except for the restriction that $sigma$ applies only to all rational numbers in the the unit interval (not just $2$, or dy-adics). That is before continuity is applied.



                                      That is $(1.a)$ below according to pt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7



                                      $$(1.a)forall , sigma in mathbb{Q}cap[0,1];, forall (x,y), in, text{dom}(F):, F(,sigma x + [1-sigma] y,), leq, sigma F(x),+,[1-sigma]F(y).$$



                                      Its presumably a bit confusing the use the words 'in the sense of Jensen' for midpoint convexity.



                                      This is because I believe that Jensen did, or helped develop the inequalities that any 'function' must satisfy, in order to be 'properly convex',as defined above in the question $(1)$.



                                      Because Jensen well as the weaker notion of midpoint convexity $(2)$ which apparently is equivalent to $(1.a)$, at least if the domain is real valued, and which is named after Jensen (midpt convexity is often called jensen- convexity).






                                      share|cite|improve this answer









                                      $endgroup$


















                                        2












                                        $begingroup$

                                        I presume you mean "proper convexity' as in $(1)$ above not $(2)$



                                        Not just mere, " "midpoint convex/jensen convexity"/ "convexity in the sense of Jensen"," as in $(2)$, below.



                                        Although (its hardly mere I suppose) because the $(1)$ as defined in the question and $(2)$ below are, not, but are nonetheless "almost equivalent".



                                        (2) $$F(frac{x+y}{2}) leq frac{F(x)}{2} +frac{F(y)}{2}$$



                                        That is because under relatively mild conditions measurability, regularity conditions/boundedness conditions, midpoint convex function (in the sense of Jensen) are convex in the tradition .



                                        Apparently a real valued midpoint convex function $(2)$ already satisfies the definition of convexity as above $(1)$, except for the restriction that $sigma$ applies only to all rational numbers in the the unit interval (not just $2$, or dy-adics). That is before continuity is applied.



                                        That is $(1.a)$ below according to pt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7



                                        $$(1.a)forall , sigma in mathbb{Q}cap[0,1];, forall (x,y), in, text{dom}(F):, F(,sigma x + [1-sigma] y,), leq, sigma F(x),+,[1-sigma]F(y).$$



                                        Its presumably a bit confusing the use the words 'in the sense of Jensen' for midpoint convexity.



                                        This is because I believe that Jensen did, or helped develop the inequalities that any 'function' must satisfy, in order to be 'properly convex',as defined above in the question $(1)$.



                                        Because Jensen well as the weaker notion of midpoint convexity $(2)$ which apparently is equivalent to $(1.a)$, at least if the domain is real valued, and which is named after Jensen (midpt convexity is often called jensen- convexity).






                                        share|cite|improve this answer









                                        $endgroup$
















                                          2












                                          2








                                          2





                                          $begingroup$

                                          I presume you mean "proper convexity' as in $(1)$ above not $(2)$



                                          Not just mere, " "midpoint convex/jensen convexity"/ "convexity in the sense of Jensen"," as in $(2)$, below.



                                          Although (its hardly mere I suppose) because the $(1)$ as defined in the question and $(2)$ below are, not, but are nonetheless "almost equivalent".



                                          (2) $$F(frac{x+y}{2}) leq frac{F(x)}{2} +frac{F(y)}{2}$$



                                          That is because under relatively mild conditions measurability, regularity conditions/boundedness conditions, midpoint convex function (in the sense of Jensen) are convex in the tradition .



                                          Apparently a real valued midpoint convex function $(2)$ already satisfies the definition of convexity as above $(1)$, except for the restriction that $sigma$ applies only to all rational numbers in the the unit interval (not just $2$, or dy-adics). That is before continuity is applied.



                                          That is $(1.a)$ below according to pt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7



                                          $$(1.a)forall , sigma in mathbb{Q}cap[0,1];, forall (x,y), in, text{dom}(F):, F(,sigma x + [1-sigma] y,), leq, sigma F(x),+,[1-sigma]F(y).$$



                                          Its presumably a bit confusing the use the words 'in the sense of Jensen' for midpoint convexity.



                                          This is because I believe that Jensen did, or helped develop the inequalities that any 'function' must satisfy, in order to be 'properly convex',as defined above in the question $(1)$.



                                          Because Jensen well as the weaker notion of midpoint convexity $(2)$ which apparently is equivalent to $(1.a)$, at least if the domain is real valued, and which is named after Jensen (midpt convexity is often called jensen- convexity).






                                          share|cite|improve this answer









                                          $endgroup$



                                          I presume you mean "proper convexity' as in $(1)$ above not $(2)$



                                          Not just mere, " "midpoint convex/jensen convexity"/ "convexity in the sense of Jensen"," as in $(2)$, below.



                                          Although (its hardly mere I suppose) because the $(1)$ as defined in the question and $(2)$ below are, not, but are nonetheless "almost equivalent".



                                          (2) $$F(frac{x+y}{2}) leq frac{F(x)}{2} +frac{F(y)}{2}$$



                                          That is because under relatively mild conditions measurability, regularity conditions/boundedness conditions, midpoint convex function (in the sense of Jensen) are convex in the tradition .



                                          Apparently a real valued midpoint convex function $(2)$ already satisfies the definition of convexity as above $(1)$, except for the restriction that $sigma$ applies only to all rational numbers in the the unit interval (not just $2$, or dy-adics). That is before continuity is applied.



                                          That is $(1.a)$ below according to pt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7



                                          $$(1.a)forall , sigma in mathbb{Q}cap[0,1];, forall (x,y), in, text{dom}(F):, F(,sigma x + [1-sigma] y,), leq, sigma F(x),+,[1-sigma]F(y).$$



                                          Its presumably a bit confusing the use the words 'in the sense of Jensen' for midpoint convexity.



                                          This is because I believe that Jensen did, or helped develop the inequalities that any 'function' must satisfy, in order to be 'properly convex',as defined above in the question $(1)$.



                                          Because Jensen well as the weaker notion of midpoint convexity $(2)$ which apparently is equivalent to $(1.a)$, at least if the domain is real valued, and which is named after Jensen (midpt convexity is often called jensen- convexity).







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered May 10 '17 at 10:02









                                          William BalthesWilliam Balthes

                                          3691415




                                          3691415























                                              2












                                              $begingroup$

                                              I like GEdgar's pictorial explanation.
                                              The key idea is: for $f(x)$ in $(a,b)$, we try to prove $f(x)$ is bounded above and below.
                                              It's easy to show $f(x)$ is bounded above by definition of convex.
                                              To show $f(x)$ is bounded below, we can see GEdgar's 2nd and 3rd picture.



                                              Once we proved $f(x)$ is bounded, $|f(x)|leq M$,



                                              $$a < s < t < u < b$$
                                              let $$s < x < u quad;quad 0<varepsilon<u-x$$
                                              then we can show
                                              $$f(x+varepsilon)-f(x)leq dfrac{f(u)-f(s)}{(u-s)}varepsilon leq dfrac{Mvarepsilon}{(u-s)}$$
                                              as $varepsilon rightarrow 0$ $$|f(x+varepsilon)-f(x)| rightarrow 0$$ then $f(x+)=f(x)$
                                              the same way we can show $f(x-)=f(x)$,
                                              so $f(x)$ is continuous






                                              share|cite|improve this answer











                                              $endgroup$


















                                                2












                                                $begingroup$

                                                I like GEdgar's pictorial explanation.
                                                The key idea is: for $f(x)$ in $(a,b)$, we try to prove $f(x)$ is bounded above and below.
                                                It's easy to show $f(x)$ is bounded above by definition of convex.
                                                To show $f(x)$ is bounded below, we can see GEdgar's 2nd and 3rd picture.



                                                Once we proved $f(x)$ is bounded, $|f(x)|leq M$,



                                                $$a < s < t < u < b$$
                                                let $$s < x < u quad;quad 0<varepsilon<u-x$$
                                                then we can show
                                                $$f(x+varepsilon)-f(x)leq dfrac{f(u)-f(s)}{(u-s)}varepsilon leq dfrac{Mvarepsilon}{(u-s)}$$
                                                as $varepsilon rightarrow 0$ $$|f(x+varepsilon)-f(x)| rightarrow 0$$ then $f(x+)=f(x)$
                                                the same way we can show $f(x-)=f(x)$,
                                                so $f(x)$ is continuous






                                                share|cite|improve this answer











                                                $endgroup$
















                                                  2












                                                  2








                                                  2





                                                  $begingroup$

                                                  I like GEdgar's pictorial explanation.
                                                  The key idea is: for $f(x)$ in $(a,b)$, we try to prove $f(x)$ is bounded above and below.
                                                  It's easy to show $f(x)$ is bounded above by definition of convex.
                                                  To show $f(x)$ is bounded below, we can see GEdgar's 2nd and 3rd picture.



                                                  Once we proved $f(x)$ is bounded, $|f(x)|leq M$,



                                                  $$a < s < t < u < b$$
                                                  let $$s < x < u quad;quad 0<varepsilon<u-x$$
                                                  then we can show
                                                  $$f(x+varepsilon)-f(x)leq dfrac{f(u)-f(s)}{(u-s)}varepsilon leq dfrac{Mvarepsilon}{(u-s)}$$
                                                  as $varepsilon rightarrow 0$ $$|f(x+varepsilon)-f(x)| rightarrow 0$$ then $f(x+)=f(x)$
                                                  the same way we can show $f(x-)=f(x)$,
                                                  so $f(x)$ is continuous






                                                  share|cite|improve this answer











                                                  $endgroup$



                                                  I like GEdgar's pictorial explanation.
                                                  The key idea is: for $f(x)$ in $(a,b)$, we try to prove $f(x)$ is bounded above and below.
                                                  It's easy to show $f(x)$ is bounded above by definition of convex.
                                                  To show $f(x)$ is bounded below, we can see GEdgar's 2nd and 3rd picture.



                                                  Once we proved $f(x)$ is bounded, $|f(x)|leq M$,



                                                  $$a < s < t < u < b$$
                                                  let $$s < x < u quad;quad 0<varepsilon<u-x$$
                                                  then we can show
                                                  $$f(x+varepsilon)-f(x)leq dfrac{f(u)-f(s)}{(u-s)}varepsilon leq dfrac{Mvarepsilon}{(u-s)}$$
                                                  as $varepsilon rightarrow 0$ $$|f(x+varepsilon)-f(x)| rightarrow 0$$ then $f(x+)=f(x)$
                                                  the same way we can show $f(x-)=f(x)$,
                                                  so $f(x)$ is continuous







                                                  share|cite|improve this answer














                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  edited Dec 19 '18 at 7:51









                                                  Yadati Kiran

                                                  1,7851619




                                                  1,7851619










                                                  answered Oct 17 '17 at 14:44









                                                  Tony ATony A

                                                  212




                                                  212























                                                      0












                                                      $begingroup$

                                                      The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set. If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex. This proof used the contrapositive.






                                                      share|cite|improve this answer









                                                      $endgroup$













                                                      • $begingroup$
                                                        Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 19:38










                                                      • $begingroup$
                                                        Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                        $endgroup$
                                                        – Squirtle
                                                        Aug 15 '14 at 19:51






                                                      • 1




                                                        $begingroup$
                                                        Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:07












                                                      • $begingroup$
                                                        As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:09








                                                      • 1




                                                        $begingroup$
                                                        It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 16 '14 at 13:55


















                                                      0












                                                      $begingroup$

                                                      The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set. If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex. This proof used the contrapositive.






                                                      share|cite|improve this answer









                                                      $endgroup$













                                                      • $begingroup$
                                                        Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 19:38










                                                      • $begingroup$
                                                        Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                        $endgroup$
                                                        – Squirtle
                                                        Aug 15 '14 at 19:51






                                                      • 1




                                                        $begingroup$
                                                        Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:07












                                                      • $begingroup$
                                                        As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:09








                                                      • 1




                                                        $begingroup$
                                                        It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 16 '14 at 13:55
















                                                      0












                                                      0








                                                      0





                                                      $begingroup$

                                                      The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set. If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex. This proof used the contrapositive.






                                                      share|cite|improve this answer









                                                      $endgroup$



                                                      The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set. If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex. This proof used the contrapositive.







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered Oct 29 '13 at 3:27









                                                      SquirtleSquirtle

                                                      4,2071641




                                                      4,2071641












                                                      • $begingroup$
                                                        Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 19:38










                                                      • $begingroup$
                                                        Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                        $endgroup$
                                                        – Squirtle
                                                        Aug 15 '14 at 19:51






                                                      • 1




                                                        $begingroup$
                                                        Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:07












                                                      • $begingroup$
                                                        As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:09








                                                      • 1




                                                        $begingroup$
                                                        It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 16 '14 at 13:55




















                                                      • $begingroup$
                                                        Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 19:38










                                                      • $begingroup$
                                                        Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                        $endgroup$
                                                        – Squirtle
                                                        Aug 15 '14 at 19:51






                                                      • 1




                                                        $begingroup$
                                                        Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:07












                                                      • $begingroup$
                                                        As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 15 '14 at 20:09








                                                      • 1




                                                        $begingroup$
                                                        It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                        $endgroup$
                                                        – Michael Grant
                                                        Aug 16 '14 at 13:55


















                                                      $begingroup$
                                                      Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 19:38




                                                      $begingroup$
                                                      Consider the extended-valued function from my above comments: $f(x)=+infty$ if $x<0$, $f(x)=1$ if $x=0$, and $f(x)=-sqrt{x}$ if $x>0$. The epigraph of this function is a convex set. But $f$ is not continuous! And its epigraph is not a closed set.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 19:38












                                                      $begingroup$
                                                      Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                      $endgroup$
                                                      – Squirtle
                                                      Aug 15 '14 at 19:51




                                                      $begingroup$
                                                      Real valued function $neq$ extended real value function (the former is the setting setting of the question and my answer, the latter is your setting). In any case, your epigraph isn't even convex,... for example you can't realize it as the intersection of epigraphs of real valued functions. If you could, then I'd believe it was convex because the intersection of arbitrary number of convex sets is convex... but your set is definitely not convex
                                                      $endgroup$
                                                      – Squirtle
                                                      Aug 15 '14 at 19:51




                                                      1




                                                      1




                                                      $begingroup$
                                                      Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 20:07






                                                      $begingroup$
                                                      Oh, my epigraph is convex all right. It's not closed, but it's convex. Try and prove otherwise: give me two points in the set for whom the secant lies outside of the set. The closure of the set adds the $(x,y)in[0,0]times[0,1)$. (Asking me to prove it is the intersection of epigraphs of real valued functions is a circular argument. Besides, the unit disc in $mathbb{R}^2$ cannot be described as the intersection of epigraphs of convex functions, either.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 20:07














                                                      $begingroup$
                                                      As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 20:09






                                                      $begingroup$
                                                      As for real valued $neq$ extended real valued: extended real valued functions are simply a short-hand way of expressing real-valued functions that are not defined on the entire real line (or $mathbb{R}^n$, etc., as appropriate). For instance, consider the classic logarithmic barrier $f(x)=-log x$, defined on $(0,+infty)$. Is that not a real-valued function?
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 15 '14 at 20:09






                                                      1




                                                      1




                                                      $begingroup$
                                                      It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 16 '14 at 13:55






                                                      $begingroup$
                                                      It is always the case that a function is convex if and only if its epigraph is convex. That is true for functions defined on all of $mathbb{R}^n$, and it's true for functions whose domain $mathop{textrm{dom}}(f)$ is only a subset thereof, like the logarithmic barrier. Indeed, the epigraph definition properly implies that $mathop{textrm{dom}} f$ is a convex set. Convex functions with limited domains are simply far too important in practice to ignore. The extended reals simplify the analysis of these functions, but you can do without that if you prefer.
                                                      $endgroup$
                                                      – Michael Grant
                                                      Aug 16 '14 at 13:55













                                                      0












                                                      $begingroup$

                                                      Here is the picture of my proof



                                                      Let, By contrary, $c$ be a point on the domain. Choose a $d$ then there exists a sequence ${a_n}_n$ that converges to c and $f(a_n) notin (f(c)-d,f(c)+d)$.
                                                      So from the sequence ${a_n}_n$ we can always select a one sided monotonic sequence ${c_n}_n$ which is the subsequence of the original seq. as one side of $c$ must have infinitely many points of ${a_n}_n$, WLOG let it is the left sided sequence & $f(c_1)<f(c)-d$



                                                      Now, take $M_1=(c_1,c)$ any point $x$ in $M_1$ must be $<f(c)-d$ as $exists m$ s.t. $c_mleq xleq c_{m+1}$ hence $x$ must belong lower area of line joining of $f(c_m)$ & $f(c_{m+1})$ (By convexity of $f$) so for all $x$ in $M_1$ $f(x)<f(c)-d$.



                                                      Now, join the point $(c_1,f(c_1))$ & $(c,f(c))$ (let the line be $l_1$). It will cut at B at the $y=f(x)+d$. Let x-cordinate of B is $c^{'}$.



                                                      Now, note $B=(c^{'},f(c^{'})$ lie upper area of $l_1$ otherwise contradicting the point $A=(c,f(c))$ (By convexity of $f$).
                                                      Now from B we can always draw a line such that A lie upper area of this line intersecting at B' at $y=f(c)-d$ line. So let x-cordinate of B' is $c^{''}$.
                                                      Now lastly join $(c^{''},f(c^{''}))$ & $(c^{'},f(c^{'}))$ contradicting the point A.






                                                      share|cite|improve this answer











                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        Here is the picture of my proof



                                                        Let, By contrary, $c$ be a point on the domain. Choose a $d$ then there exists a sequence ${a_n}_n$ that converges to c and $f(a_n) notin (f(c)-d,f(c)+d)$.
                                                        So from the sequence ${a_n}_n$ we can always select a one sided monotonic sequence ${c_n}_n$ which is the subsequence of the original seq. as one side of $c$ must have infinitely many points of ${a_n}_n$, WLOG let it is the left sided sequence & $f(c_1)<f(c)-d$



                                                        Now, take $M_1=(c_1,c)$ any point $x$ in $M_1$ must be $<f(c)-d$ as $exists m$ s.t. $c_mleq xleq c_{m+1}$ hence $x$ must belong lower area of line joining of $f(c_m)$ & $f(c_{m+1})$ (By convexity of $f$) so for all $x$ in $M_1$ $f(x)<f(c)-d$.



                                                        Now, join the point $(c_1,f(c_1))$ & $(c,f(c))$ (let the line be $l_1$). It will cut at B at the $y=f(x)+d$. Let x-cordinate of B is $c^{'}$.



                                                        Now, note $B=(c^{'},f(c^{'})$ lie upper area of $l_1$ otherwise contradicting the point $A=(c,f(c))$ (By convexity of $f$).
                                                        Now from B we can always draw a line such that A lie upper area of this line intersecting at B' at $y=f(c)-d$ line. So let x-cordinate of B' is $c^{''}$.
                                                        Now lastly join $(c^{''},f(c^{''}))$ & $(c^{'},f(c^{'}))$ contradicting the point A.






                                                        share|cite|improve this answer











                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          Here is the picture of my proof



                                                          Let, By contrary, $c$ be a point on the domain. Choose a $d$ then there exists a sequence ${a_n}_n$ that converges to c and $f(a_n) notin (f(c)-d,f(c)+d)$.
                                                          So from the sequence ${a_n}_n$ we can always select a one sided monotonic sequence ${c_n}_n$ which is the subsequence of the original seq. as one side of $c$ must have infinitely many points of ${a_n}_n$, WLOG let it is the left sided sequence & $f(c_1)<f(c)-d$



                                                          Now, take $M_1=(c_1,c)$ any point $x$ in $M_1$ must be $<f(c)-d$ as $exists m$ s.t. $c_mleq xleq c_{m+1}$ hence $x$ must belong lower area of line joining of $f(c_m)$ & $f(c_{m+1})$ (By convexity of $f$) so for all $x$ in $M_1$ $f(x)<f(c)-d$.



                                                          Now, join the point $(c_1,f(c_1))$ & $(c,f(c))$ (let the line be $l_1$). It will cut at B at the $y=f(x)+d$. Let x-cordinate of B is $c^{'}$.



                                                          Now, note $B=(c^{'},f(c^{'})$ lie upper area of $l_1$ otherwise contradicting the point $A=(c,f(c))$ (By convexity of $f$).
                                                          Now from B we can always draw a line such that A lie upper area of this line intersecting at B' at $y=f(c)-d$ line. So let x-cordinate of B' is $c^{''}$.
                                                          Now lastly join $(c^{''},f(c^{''}))$ & $(c^{'},f(c^{'}))$ contradicting the point A.






                                                          share|cite|improve this answer











                                                          $endgroup$



                                                          Here is the picture of my proof



                                                          Let, By contrary, $c$ be a point on the domain. Choose a $d$ then there exists a sequence ${a_n}_n$ that converges to c and $f(a_n) notin (f(c)-d,f(c)+d)$.
                                                          So from the sequence ${a_n}_n$ we can always select a one sided monotonic sequence ${c_n}_n$ which is the subsequence of the original seq. as one side of $c$ must have infinitely many points of ${a_n}_n$, WLOG let it is the left sided sequence & $f(c_1)<f(c)-d$



                                                          Now, take $M_1=(c_1,c)$ any point $x$ in $M_1$ must be $<f(c)-d$ as $exists m$ s.t. $c_mleq xleq c_{m+1}$ hence $x$ must belong lower area of line joining of $f(c_m)$ & $f(c_{m+1})$ (By convexity of $f$) so for all $x$ in $M_1$ $f(x)<f(c)-d$.



                                                          Now, join the point $(c_1,f(c_1))$ & $(c,f(c))$ (let the line be $l_1$). It will cut at B at the $y=f(x)+d$. Let x-cordinate of B is $c^{'}$.



                                                          Now, note $B=(c^{'},f(c^{'})$ lie upper area of $l_1$ otherwise contradicting the point $A=(c,f(c))$ (By convexity of $f$).
                                                          Now from B we can always draw a line such that A lie upper area of this line intersecting at B' at $y=f(c)-d$ line. So let x-cordinate of B' is $c^{''}$.
                                                          Now lastly join $(c^{''},f(c^{''}))$ & $(c^{'},f(c^{'}))$ contradicting the point A.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited Jan 20 at 9:53

























                                                          answered Jan 20 at 3:35









                                                          Karambir.kdKarambir.kd

                                                          885




                                                          885






























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