$f''(x)leq 0$ implies $f$ concave












0












$begingroup$


I'm using the following definition:



$f:[0,infty)rightarrow[0,infty)$ is concave if $forall x,yin[0,infty)$ and $sin[0,1]$, we have



$$f(sx+(1-s)y)geq sf(x)+(1-s)f(y) $$



I need to prove that every function $f:[0,infty)rightarrow[0,infty)$ twice differentiable satisfying $f''(x)leq 0$ for all $xin[0,infty)$ is concave.



I found the reciprocal, but not this statement.



Well, I know that $f''(x)leq 0$ implies that for every $x<y$, $f'(y)leq f'(x)$. Can someone give me just some hints?






functions derivatives







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  • 1




    $begingroup$
    See this. The answers use techniques you should use as well.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:49
















0












$begingroup$


I'm using the following definition:



$f:[0,infty)rightarrow[0,infty)$ is concave if $forall x,yin[0,infty)$ and $sin[0,1]$, we have



$$f(sx+(1-s)y)geq sf(x)+(1-s)f(y) $$



I need to prove that every function $f:[0,infty)rightarrow[0,infty)$ twice differentiable satisfying $f''(x)leq 0$ for all $xin[0,infty)$ is concave.



I found the reciprocal, but not this statement.



Well, I know that $f''(x)leq 0$ implies that for every $x<y$, $f'(y)leq f'(x)$. Can someone give me just some hints?






functions derivatives







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See this. The answers use techniques you should use as well.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:49














0












0








0





$begingroup$


I'm using the following definition:



$f:[0,infty)rightarrow[0,infty)$ is concave if $forall x,yin[0,infty)$ and $sin[0,1]$, we have



$$f(sx+(1-s)y)geq sf(x)+(1-s)f(y) $$



I need to prove that every function $f:[0,infty)rightarrow[0,infty)$ twice differentiable satisfying $f''(x)leq 0$ for all $xin[0,infty)$ is concave.



I found the reciprocal, but not this statement.



Well, I know that $f''(x)leq 0$ implies that for every $x<y$, $f'(y)leq f'(x)$. Can someone give me just some hints?






functions derivatives







share|cite|improve this question









$endgroup$




I'm using the following definition:



$f:[0,infty)rightarrow[0,infty)$ is concave if $forall x,yin[0,infty)$ and $sin[0,1]$, we have



$$f(sx+(1-s)y)geq sf(x)+(1-s)f(y) $$



I need to prove that every function $f:[0,infty)rightarrow[0,infty)$ twice differentiable satisfying $f''(x)leq 0$ for all $xin[0,infty)$ is concave.



I found the reciprocal, but not this statement.



Well, I know that $f''(x)leq 0$ implies that for every $x<y$, $f'(y)leq f'(x)$. Can someone give me just some hints?






functions derivatives




functions derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 17:31









Mateus RochaMateus Rocha

812117




812117








  • 1




    $begingroup$
    See this. The answers use techniques you should use as well.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:49














  • 1




    $begingroup$
    See this. The answers use techniques you should use as well.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:49








1




1




$begingroup$
See this. The answers use techniques you should use as well.
$endgroup$
– Don Thousand
Dec 4 '18 at 17:49




$begingroup$
See this. The answers use techniques you should use as well.
$endgroup$
– Don Thousand
Dec 4 '18 at 17:49










1 Answer
1






active

oldest

votes


















2












$begingroup$

Use Rolle's Theorem.



Define $z = sx + (1-s)y$.



$exists z_1 in (x, z), quadtext{s.t.}quad f(z) - f(x) = (z-x)f'(z_1)$



$exists z_2 in (z, y), quadtext{s.t.}quad f(y) - f(z) = (y-z)f'(z_2)$



so
$$
begin{aligned}
sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \
= & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \
= & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\
= & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]
end{aligned}
$$



Apparently
$s>0, 1-s>0, y-x>0$ and also $f'(z_2) leq f'(z_1)$ because $f''(x)$ always negative.



So
$[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] leq 0$






share|cite|improve this answer









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    1 Answer
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    2












    $begingroup$

    Use Rolle's Theorem.



    Define $z = sx + (1-s)y$.



    $exists z_1 in (x, z), quadtext{s.t.}quad f(z) - f(x) = (z-x)f'(z_1)$



    $exists z_2 in (z, y), quadtext{s.t.}quad f(y) - f(z) = (y-z)f'(z_2)$



    so
    $$
    begin{aligned}
    sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \
    = & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \
    = & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\
    = & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]
    end{aligned}
    $$



    Apparently
    $s>0, 1-s>0, y-x>0$ and also $f'(z_2) leq f'(z_1)$ because $f''(x)$ always negative.



    So
    $[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] leq 0$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Use Rolle's Theorem.



      Define $z = sx + (1-s)y$.



      $exists z_1 in (x, z), quadtext{s.t.}quad f(z) - f(x) = (z-x)f'(z_1)$



      $exists z_2 in (z, y), quadtext{s.t.}quad f(y) - f(z) = (y-z)f'(z_2)$



      so
      $$
      begin{aligned}
      sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \
      = & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \
      = & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\
      = & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]
      end{aligned}
      $$



      Apparently
      $s>0, 1-s>0, y-x>0$ and also $f'(z_2) leq f'(z_1)$ because $f''(x)$ always negative.



      So
      $[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] leq 0$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Use Rolle's Theorem.



        Define $z = sx + (1-s)y$.



        $exists z_1 in (x, z), quadtext{s.t.}quad f(z) - f(x) = (z-x)f'(z_1)$



        $exists z_2 in (z, y), quadtext{s.t.}quad f(y) - f(z) = (y-z)f'(z_2)$



        so
        $$
        begin{aligned}
        sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \
        = & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \
        = & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\
        = & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]
        end{aligned}
        $$



        Apparently
        $s>0, 1-s>0, y-x>0$ and also $f'(z_2) leq f'(z_1)$ because $f''(x)$ always negative.



        So
        $[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] leq 0$






        share|cite|improve this answer









        $endgroup$



        Use Rolle's Theorem.



        Define $z = sx + (1-s)y$.



        $exists z_1 in (x, z), quadtext{s.t.}quad f(z) - f(x) = (z-x)f'(z_1)$



        $exists z_2 in (z, y), quadtext{s.t.}quad f(y) - f(z) = (y-z)f'(z_2)$



        so
        $$
        begin{aligned}
        sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \
        = & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \
        = & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\
        = & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]
        end{aligned}
        $$



        Apparently
        $s>0, 1-s>0, y-x>0$ and also $f'(z_2) leq f'(z_1)$ because $f''(x)$ always negative.



        So
        $[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] leq 0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 17:49









        MoonKnightMoonKnight

        1,359611




        1,359611






























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