How to show that if $f^2(x)$ is uniformly continous function then f is uniformly continous












2












$begingroup$


$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?



My attempt :



$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$



then



$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$



My problem is that how to control above difference as f may be 0 at both x and y



SO how to show above is uniformly continuous



Any help will be appreciated






real-analysis continuity uniform-continuity







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$endgroup$








  • 2




    $begingroup$
    if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
    $endgroup$
    – mathworker21
    Dec 1 '18 at 7:02










  • $begingroup$
    It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
    $endgroup$
    – Martin R
    Dec 1 '18 at 8:27
















2












$begingroup$


$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?



My attempt :



$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$



then



$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$



My problem is that how to control above difference as f may be 0 at both x and y



SO how to show above is uniformly continuous



Any help will be appreciated






real-analysis continuity uniform-continuity







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
    $endgroup$
    – mathworker21
    Dec 1 '18 at 7:02










  • $begingroup$
    It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
    $endgroup$
    – Martin R
    Dec 1 '18 at 8:27














2












2








2


1



$begingroup$


$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?



My attempt :



$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$



then



$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$



My problem is that how to control above difference as f may be 0 at both x and y



SO how to show above is uniformly continuous



Any help will be appreciated






real-analysis continuity uniform-continuity







share|cite|improve this question











$endgroup$




$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?



My attempt :



$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$



then



$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$



My problem is that how to control above difference as f may be 0 at both x and y



SO how to show above is uniformly continuous



Any help will be appreciated






real-analysis continuity uniform-continuity




real-analysis continuity uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 7:36









twnly

542110




542110










asked Dec 1 '18 at 6:47









MathLoverMathLover

49310




49310








  • 2




    $begingroup$
    if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
    $endgroup$
    – mathworker21
    Dec 1 '18 at 7:02










  • $begingroup$
    It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
    $endgroup$
    – Martin R
    Dec 1 '18 at 8:27














  • 2




    $begingroup$
    if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
    $endgroup$
    – mathworker21
    Dec 1 '18 at 7:02










  • $begingroup$
    It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
    $endgroup$
    – Martin R
    Dec 1 '18 at 8:27








2




2




$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02




$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02












$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27




$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27










3 Answers
3






active

oldest

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2












$begingroup$

Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.



$sqrt x$ is uniformly continuous



composition of two uniformly continuous functions.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    To follow up on Martin R's comment:



    Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.

    Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.



    Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.



    Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
        $endgroup$
        – Martin R
        Dec 1 '18 at 10:08










      • $begingroup$
        @MartinR I mean you can remove the non-negative condition provided f is continuous.
        $endgroup$
        – MiGang
        Dec 1 '18 at 12:05













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.



      $sqrt x$ is uniformly continuous



      composition of two uniformly continuous functions.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.



        $sqrt x$ is uniformly continuous



        composition of two uniformly continuous functions.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.



          $sqrt x$ is uniformly continuous



          composition of two uniformly continuous functions.






          share|cite|improve this answer











          $endgroup$



          Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.



          $sqrt x$ is uniformly continuous



          composition of two uniformly continuous functions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 8:59

























          answered Dec 1 '18 at 8:19









          simplemathsimplemath

          363




          363























              0












              $begingroup$

              To follow up on Martin R's comment:



              Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.

              Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.



              Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.



              Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                To follow up on Martin R's comment:



                Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.

                Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.



                Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.



                Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To follow up on Martin R's comment:



                  Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.

                  Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.



                  Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.



                  Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.






                  share|cite|improve this answer









                  $endgroup$



                  To follow up on Martin R's comment:



                  Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.

                  Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.



                  Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.



                  Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 8:55









                  jmerryjmerry

                  3,447413




                  3,447413























                      0












                      $begingroup$

                      I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                        $endgroup$
                        – Martin R
                        Dec 1 '18 at 10:08










                      • $begingroup$
                        @MartinR I mean you can remove the non-negative condition provided f is continuous.
                        $endgroup$
                        – MiGang
                        Dec 1 '18 at 12:05


















                      0












                      $begingroup$

                      I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                        $endgroup$
                        – Martin R
                        Dec 1 '18 at 10:08










                      • $begingroup$
                        @MartinR I mean you can remove the non-negative condition provided f is continuous.
                        $endgroup$
                        – MiGang
                        Dec 1 '18 at 12:05
















                      0












                      0








                      0





                      $begingroup$

                      I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.






                      share|cite|improve this answer











                      $endgroup$



                      I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 1 '18 at 9:43

























                      answered Dec 1 '18 at 9:31









                      MiGangMiGang

                      636




                      636












                      • $begingroup$
                        Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                        $endgroup$
                        – Martin R
                        Dec 1 '18 at 10:08










                      • $begingroup$
                        @MartinR I mean you can remove the non-negative condition provided f is continuous.
                        $endgroup$
                        – MiGang
                        Dec 1 '18 at 12:05




















                      • $begingroup$
                        Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                        $endgroup$
                        – Martin R
                        Dec 1 '18 at 10:08










                      • $begingroup$
                        @MartinR I mean you can remove the non-negative condition provided f is continuous.
                        $endgroup$
                        – MiGang
                        Dec 1 '18 at 12:05


















                      $begingroup$
                      Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                      $endgroup$
                      – Martin R
                      Dec 1 '18 at 10:08




                      $begingroup$
                      Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
                      $endgroup$
                      – Martin R
                      Dec 1 '18 at 10:08












                      $begingroup$
                      @MartinR I mean you can remove the non-negative condition provided f is continuous.
                      $endgroup$
                      – MiGang
                      Dec 1 '18 at 12:05






                      $begingroup$
                      @MartinR I mean you can remove the non-negative condition provided f is continuous.
                      $endgroup$
                      – MiGang
                      Dec 1 '18 at 12:05




















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                      Ellipse (mathématiques)

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                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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                      Quarter-circle Tiles

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                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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