A constraint that implies convexity












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Let $f:mathbb{R} to mathbb{R} $ be a function such that $forall x<y, exists zin(x, y) $ with $(y-x) f(z) le (y-z) f(x) +(z-x) f(y) $.



a) Give an example of a non-convex function $f$ which has this property.



b) Prove that a continuous function $f$ which has this property is convex.




For a) it is obvious that we must search for a discontinuous $f$ with this property, but I can't find one. For b), I tried to assume that $f$ is not convex, which means that $exists u<v$ and $a in [0,1]$ such that $f(au+(1-a)v)>af(u)+(1-a)f(v)$,but here I am stuck.






real-analysis functional-analysis contest-math convexity-inequality







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    Let $f:mathbb{R} to mathbb{R} $ be a function such that $forall x<y, exists zin(x, y) $ with $(y-x) f(z) le (y-z) f(x) +(z-x) f(y) $.



    a) Give an example of a non-convex function $f$ which has this property.



    b) Prove that a continuous function $f$ which has this property is convex.




    For a) it is obvious that we must search for a discontinuous $f$ with this property, but I can't find one. For b), I tried to assume that $f$ is not convex, which means that $exists u<v$ and $a in [0,1]$ such that $f(au+(1-a)v)>af(u)+(1-a)f(v)$,but here I am stuck.






    real-analysis functional-analysis contest-math convexity-inequality







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$



      Let $f:mathbb{R} to mathbb{R} $ be a function such that $forall x<y, exists zin(x, y) $ with $(y-x) f(z) le (y-z) f(x) +(z-x) f(y) $.



      a) Give an example of a non-convex function $f$ which has this property.



      b) Prove that a continuous function $f$ which has this property is convex.




      For a) it is obvious that we must search for a discontinuous $f$ with this property, but I can't find one. For b), I tried to assume that $f$ is not convex, which means that $exists u<v$ and $a in [0,1]$ such that $f(au+(1-a)v)>af(u)+(1-a)f(v)$,but here I am stuck.






      real-analysis functional-analysis contest-math convexity-inequality







      share|cite|improve this question











      $endgroup$





      Let $f:mathbb{R} to mathbb{R} $ be a function such that $forall x<y, exists zin(x, y) $ with $(y-x) f(z) le (y-z) f(x) +(z-x) f(y) $.



      a) Give an example of a non-convex function $f$ which has this property.



      b) Prove that a continuous function $f$ which has this property is convex.




      For a) it is obvious that we must search for a discontinuous $f$ with this property, but I can't find one. For b), I tried to assume that $f$ is not convex, which means that $exists u<v$ and $a in [0,1]$ such that $f(au+(1-a)v)>af(u)+(1-a)f(v)$,but here I am stuck.






      real-analysis functional-analysis contest-math convexity-inequality




      real-analysis functional-analysis contest-math convexity-inequality






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      share|cite|improve this question













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      edited 7 hours ago









      Harry49

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      8,76331346










      asked 7 hours ago









      Math GuyMath Guy

      1117




      1117






















          2 Answers
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          $begingroup$

          Intuitively, what's happening with (b) is that the equation
          $$
          g(z) = frac{y-z}{y-x} f(x) + frac{z-x}{y-x} f(y)
          $$
          is the equation of the secant line through $(x,f(x))$ and $(y,f(y))$. Call this line the $x,y$-secant for short. So the property we're assuming is that for all $x<y$ there is a $z in (x,y)$ such that $f(z)$ lies below the $x,y$-secant. We want to prove that for all $x<y$, all $z in (x,y)$ have this property.



          Suppose for the sake of contradiction $w in [x,y]$ is such that $f(w) > g(w)$. Then let $S_1 = {z in [x,w] : f(z) le g(z)}$ and $S_2 = {z in [w,y] : f(z) le g(z)}$. We know that $sup S_1$ has to exist ($S_1$ is bounded and $x in S_1$). We know that $f(sup S_1) le g(sup S_1)$ because $S_1$ is closed (here we use continuity of $f$). We know that $sup S_1 < w$ because $f(w) > g(w)$. The corresponding things hold for $inf S_2$ as well. Also, $sup S_1 < w < inf S_2$, so $sup S_1 < inf S_2$.



          But now there is a contradiction. Strictly between $z_1 = sup S_1$ and $z_2 = inf S_2$ there is a point $z$ that lies below the $z_1,z_2$-secant. But then it lies below the $x,y$-secant as well. So $sup S_1 < z < inf S_2$ and yet $z$ should be in either $S_1$ or $S_2$, contradiction.



          So $w$ cannot exist, and therefore $f$ is convex.






          share|cite|improve this answer









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            3












            $begingroup$

            Try $f(0)=1$, $f(x) = 0$ otherwise.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you! This function solves a) indeed.
              $endgroup$
              – Math Guy
              7 hours ago












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            2 Answers
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            2 Answers
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            active

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            $begingroup$

            Intuitively, what's happening with (b) is that the equation
            $$
            g(z) = frac{y-z}{y-x} f(x) + frac{z-x}{y-x} f(y)
            $$
            is the equation of the secant line through $(x,f(x))$ and $(y,f(y))$. Call this line the $x,y$-secant for short. So the property we're assuming is that for all $x<y$ there is a $z in (x,y)$ such that $f(z)$ lies below the $x,y$-secant. We want to prove that for all $x<y$, all $z in (x,y)$ have this property.



            Suppose for the sake of contradiction $w in [x,y]$ is such that $f(w) > g(w)$. Then let $S_1 = {z in [x,w] : f(z) le g(z)}$ and $S_2 = {z in [w,y] : f(z) le g(z)}$. We know that $sup S_1$ has to exist ($S_1$ is bounded and $x in S_1$). We know that $f(sup S_1) le g(sup S_1)$ because $S_1$ is closed (here we use continuity of $f$). We know that $sup S_1 < w$ because $f(w) > g(w)$. The corresponding things hold for $inf S_2$ as well. Also, $sup S_1 < w < inf S_2$, so $sup S_1 < inf S_2$.



            But now there is a contradiction. Strictly between $z_1 = sup S_1$ and $z_2 = inf S_2$ there is a point $z$ that lies below the $z_1,z_2$-secant. But then it lies below the $x,y$-secant as well. So $sup S_1 < z < inf S_2$ and yet $z$ should be in either $S_1$ or $S_2$, contradiction.



            So $w$ cannot exist, and therefore $f$ is convex.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Intuitively, what's happening with (b) is that the equation
              $$
              g(z) = frac{y-z}{y-x} f(x) + frac{z-x}{y-x} f(y)
              $$
              is the equation of the secant line through $(x,f(x))$ and $(y,f(y))$. Call this line the $x,y$-secant for short. So the property we're assuming is that for all $x<y$ there is a $z in (x,y)$ such that $f(z)$ lies below the $x,y$-secant. We want to prove that for all $x<y$, all $z in (x,y)$ have this property.



              Suppose for the sake of contradiction $w in [x,y]$ is such that $f(w) > g(w)$. Then let $S_1 = {z in [x,w] : f(z) le g(z)}$ and $S_2 = {z in [w,y] : f(z) le g(z)}$. We know that $sup S_1$ has to exist ($S_1$ is bounded and $x in S_1$). We know that $f(sup S_1) le g(sup S_1)$ because $S_1$ is closed (here we use continuity of $f$). We know that $sup S_1 < w$ because $f(w) > g(w)$. The corresponding things hold for $inf S_2$ as well. Also, $sup S_1 < w < inf S_2$, so $sup S_1 < inf S_2$.



              But now there is a contradiction. Strictly between $z_1 = sup S_1$ and $z_2 = inf S_2$ there is a point $z$ that lies below the $z_1,z_2$-secant. But then it lies below the $x,y$-secant as well. So $sup S_1 < z < inf S_2$ and yet $z$ should be in either $S_1$ or $S_2$, contradiction.



              So $w$ cannot exist, and therefore $f$ is convex.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Intuitively, what's happening with (b) is that the equation
                $$
                g(z) = frac{y-z}{y-x} f(x) + frac{z-x}{y-x} f(y)
                $$
                is the equation of the secant line through $(x,f(x))$ and $(y,f(y))$. Call this line the $x,y$-secant for short. So the property we're assuming is that for all $x<y$ there is a $z in (x,y)$ such that $f(z)$ lies below the $x,y$-secant. We want to prove that for all $x<y$, all $z in (x,y)$ have this property.



                Suppose for the sake of contradiction $w in [x,y]$ is such that $f(w) > g(w)$. Then let $S_1 = {z in [x,w] : f(z) le g(z)}$ and $S_2 = {z in [w,y] : f(z) le g(z)}$. We know that $sup S_1$ has to exist ($S_1$ is bounded and $x in S_1$). We know that $f(sup S_1) le g(sup S_1)$ because $S_1$ is closed (here we use continuity of $f$). We know that $sup S_1 < w$ because $f(w) > g(w)$. The corresponding things hold for $inf S_2$ as well. Also, $sup S_1 < w < inf S_2$, so $sup S_1 < inf S_2$.



                But now there is a contradiction. Strictly between $z_1 = sup S_1$ and $z_2 = inf S_2$ there is a point $z$ that lies below the $z_1,z_2$-secant. But then it lies below the $x,y$-secant as well. So $sup S_1 < z < inf S_2$ and yet $z$ should be in either $S_1$ or $S_2$, contradiction.



                So $w$ cannot exist, and therefore $f$ is convex.






                share|cite|improve this answer









                $endgroup$



                Intuitively, what's happening with (b) is that the equation
                $$
                g(z) = frac{y-z}{y-x} f(x) + frac{z-x}{y-x} f(y)
                $$
                is the equation of the secant line through $(x,f(x))$ and $(y,f(y))$. Call this line the $x,y$-secant for short. So the property we're assuming is that for all $x<y$ there is a $z in (x,y)$ such that $f(z)$ lies below the $x,y$-secant. We want to prove that for all $x<y$, all $z in (x,y)$ have this property.



                Suppose for the sake of contradiction $w in [x,y]$ is such that $f(w) > g(w)$. Then let $S_1 = {z in [x,w] : f(z) le g(z)}$ and $S_2 = {z in [w,y] : f(z) le g(z)}$. We know that $sup S_1$ has to exist ($S_1$ is bounded and $x in S_1$). We know that $f(sup S_1) le g(sup S_1)$ because $S_1$ is closed (here we use continuity of $f$). We know that $sup S_1 < w$ because $f(w) > g(w)$. The corresponding things hold for $inf S_2$ as well. Also, $sup S_1 < w < inf S_2$, so $sup S_1 < inf S_2$.



                But now there is a contradiction. Strictly between $z_1 = sup S_1$ and $z_2 = inf S_2$ there is a point $z$ that lies below the $z_1,z_2$-secant. But then it lies below the $x,y$-secant as well. So $sup S_1 < z < inf S_2$ and yet $z$ should be in either $S_1$ or $S_2$, contradiction.



                So $w$ cannot exist, and therefore $f$ is convex.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Misha LavrovMisha Lavrov

                49.5k758109




                49.5k758109























                    3












                    $begingroup$

                    Try $f(0)=1$, $f(x) = 0$ otherwise.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you! This function solves a) indeed.
                      $endgroup$
                      – Math Guy
                      7 hours ago
















                    3












                    $begingroup$

                    Try $f(0)=1$, $f(x) = 0$ otherwise.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you! This function solves a) indeed.
                      $endgroup$
                      – Math Guy
                      7 hours ago














                    3












                    3








                    3





                    $begingroup$

                    Try $f(0)=1$, $f(x) = 0$ otherwise.






                    share|cite|improve this answer









                    $endgroup$



                    Try $f(0)=1$, $f(x) = 0$ otherwise.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 7 hours ago









                    Robert IsraelRobert Israel

                    331k23221478




                    331k23221478












                    • $begingroup$
                      Thank you! This function solves a) indeed.
                      $endgroup$
                      – Math Guy
                      7 hours ago


















                    • $begingroup$
                      Thank you! This function solves a) indeed.
                      $endgroup$
                      – Math Guy
                      7 hours ago
















                    $begingroup$
                    Thank you! This function solves a) indeed.
                    $endgroup$
                    – Math Guy
                    7 hours ago




                    $begingroup$
                    Thank you! This function solves a) indeed.
                    $endgroup$
                    – Math Guy
                    7 hours ago


















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