proving inequality through convexity and continuity











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Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})]
$$



if we move all the terms to the left hand side, we'll have:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge 0
$$
I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
$$f(frac{x+y}{2})le frac{1}{2}f(x)+frac{1}{2}f(y)$$
$$f(frac{y+z}{2})le frac{1}{2}f(y)+frac{1}{2}f(z)$$
$$f(frac{z+x}{2})le frac{1}{2}f(z)+frac{1}{2}f(x)$$
$$Longrightarrow f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2}) le f(x)+f(y)+f(z)
$$
Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge f(frac{x+y+z}{3})-frac{f(x)+f(y)+f(z)}{3}
$$
According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:



first part:$$frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge 2f(frac{x+y+z}{3})$$
second part:$$2[frac{1}{3}f(frac{x+y}{2}) +frac{1}{3}f(frac{y+z}{2}) +frac{1}{3}f(frac{z+x}{2})] ge 2f(frac{x+y+z}{3})$$



I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.






proof-verification convex-analysis convexity-inequality







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    Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
    $$
    frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})]
    $$



    if we move all the terms to the left hand side, we'll have:
    $$
    frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge 0
    $$
    I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
    $$f(frac{x+y}{2})le frac{1}{2}f(x)+frac{1}{2}f(y)$$
    $$f(frac{y+z}{2})le frac{1}{2}f(y)+frac{1}{2}f(z)$$
    $$f(frac{z+x}{2})le frac{1}{2}f(z)+frac{1}{2}f(x)$$
    $$Longrightarrow f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2}) le f(x)+f(y)+f(z)
    $$
    Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
    $$
    frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge f(frac{x+y+z}{3})-frac{f(x)+f(y)+f(z)}{3}
    $$
    According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:



    first part:$$frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge 2f(frac{x+y+z}{3})$$
    second part:$$2[frac{1}{3}f(frac{x+y}{2}) +frac{1}{3}f(frac{y+z}{2}) +frac{1}{3}f(frac{z+x}{2})] ge 2f(frac{x+y+z}{3})$$



    I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.






    proof-verification convex-analysis convexity-inequality







    share|cite|improve this question


























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      Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})]
      $$



      if we move all the terms to the left hand side, we'll have:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge 0
      $$
      I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
      $$f(frac{x+y}{2})le frac{1}{2}f(x)+frac{1}{2}f(y)$$
      $$f(frac{y+z}{2})le frac{1}{2}f(y)+frac{1}{2}f(z)$$
      $$f(frac{z+x}{2})le frac{1}{2}f(z)+frac{1}{2}f(x)$$
      $$Longrightarrow f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2}) le f(x)+f(y)+f(z)
      $$
      Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge f(frac{x+y+z}{3})-frac{f(x)+f(y)+f(z)}{3}
      $$
      According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:



      first part:$$frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge 2f(frac{x+y+z}{3})$$
      second part:$$2[frac{1}{3}f(frac{x+y}{2}) +frac{1}{3}f(frac{y+z}{2}) +frac{1}{3}f(frac{z+x}{2})] ge 2f(frac{x+y+z}{3})$$



      I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.






      proof-verification convex-analysis convexity-inequality







      share|cite|improve this question















      Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})]
      $$



      if we move all the terms to the left hand side, we'll have:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge 0
      $$
      I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
      $$f(frac{x+y}{2})le frac{1}{2}f(x)+frac{1}{2}f(y)$$
      $$f(frac{y+z}{2})le frac{1}{2}f(y)+frac{1}{2}f(z)$$
      $$f(frac{z+x}{2})le frac{1}{2}f(z)+frac{1}{2}f(x)$$
      $$Longrightarrow f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2}) le f(x)+f(y)+f(z)
      $$
      Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
      $$
      frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge f(frac{x+y+z}{3})-frac{f(x)+f(y)+f(z)}{3}
      $$
      According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:



      first part:$$frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge 2f(frac{x+y+z}{3})$$
      second part:$$2[frac{1}{3}f(frac{x+y}{2}) +frac{1}{3}f(frac{y+z}{2}) +frac{1}{3}f(frac{z+x}{2})] ge 2f(frac{x+y+z}{3})$$



      I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.






      proof-verification convex-analysis convexity-inequality




      proof-verification convex-analysis convexity-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 22 at 10:12

























      asked Nov 20 at 15:45









      F.Chi

      143




      143






















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          This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:



          Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.






          share|cite|improve this answer





















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            This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:



            Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:



              Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:



                Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.






                share|cite|improve this answer












                This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:



                Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 13:41









                daw

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