Proof of Convexity by Contradiciton

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I have a long and complicated function $f(T)$ whose derivative is not easy to derive. So, I want to prove that $f(T)$ is a convex function over the interval $(0,infty)$ . I think it is easier by using contradiction. The convexity definition of a function states that $f(T)$ is convex over an interval $(a,b)$ if for every $T_1, T_2 in (a,b)$, $0leq lambda leq 1$ and the inequality below holds



$fleft( lambda T_1 + (1-lambda)T_2 right) leq lambda f(T_1) + (1-lambda)f(T_2)$.



Now, I assume that for every $T_1, T_2 in (0,infty)$ and $0leq lambda leq 1$, the inequality below is true:



$fleft( lambda T_1 + (1-lambda)T_2 right) > lambda f(T_1) + (1-lambda)f(T_2)$.



Then, let $T_1 = T_2 = 1$ and $lambda = 1/3$. The inequality becomes
$f(frac13 + frac23) > frac13f(1) + frac23f(1)$
$f(1) > f(1)$ which is wrong. By contradiction, $f(T)$ is convex.
But, then every function can be proven to be complex. How should I fix this proof?




convex-analysis convexity-inequality




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    I have a long and complicated function $f(T)$ whose derivative is not easy to derive. So, I want to prove that $f(T)$ is a convex function over the interval $(0,infty)$ . I think it is easier by using contradiction. The convexity definition of a function states that $f(T)$ is convex over an interval $(a,b)$ if for every $T_1, T_2 in (a,b)$, $0leq lambda leq 1$ and the inequality below holds



    $fleft( lambda T_1 + (1-lambda)T_2 right) leq lambda f(T_1) + (1-lambda)f(T_2)$.



    Now, I assume that for every $T_1, T_2 in (0,infty)$ and $0leq lambda leq 1$, the inequality below is true:



    $fleft( lambda T_1 + (1-lambda)T_2 right) > lambda f(T_1) + (1-lambda)f(T_2)$.



    Then, let $T_1 = T_2 = 1$ and $lambda = 1/3$. The inequality becomes
    $f(frac13 + frac23) > frac13f(1) + frac23f(1)$
    $f(1) > f(1)$ which is wrong. By contradiction, $f(T)$ is convex.
    But, then every function can be proven to be complex. How should I fix this proof?




    convex-analysis convexity-inequality




    share|cite|improve this question





















      up vote
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      up vote
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      down vote

      favorite











      I have a long and complicated function $f(T)$ whose derivative is not easy to derive. So, I want to prove that $f(T)$ is a convex function over the interval $(0,infty)$ . I think it is easier by using contradiction. The convexity definition of a function states that $f(T)$ is convex over an interval $(a,b)$ if for every $T_1, T_2 in (a,b)$, $0leq lambda leq 1$ and the inequality below holds



      $fleft( lambda T_1 + (1-lambda)T_2 right) leq lambda f(T_1) + (1-lambda)f(T_2)$.



      Now, I assume that for every $T_1, T_2 in (0,infty)$ and $0leq lambda leq 1$, the inequality below is true:



      $fleft( lambda T_1 + (1-lambda)T_2 right) > lambda f(T_1) + (1-lambda)f(T_2)$.



      Then, let $T_1 = T_2 = 1$ and $lambda = 1/3$. The inequality becomes
      $f(frac13 + frac23) > frac13f(1) + frac23f(1)$
      $f(1) > f(1)$ which is wrong. By contradiction, $f(T)$ is convex.
      But, then every function can be proven to be complex. How should I fix this proof?




      convex-analysis convexity-inequality




      share|cite|improve this question











      I have a long and complicated function $f(T)$ whose derivative is not easy to derive. So, I want to prove that $f(T)$ is a convex function over the interval $(0,infty)$ . I think it is easier by using contradiction. The convexity definition of a function states that $f(T)$ is convex over an interval $(a,b)$ if for every $T_1, T_2 in (a,b)$, $0leq lambda leq 1$ and the inequality below holds



      $fleft( lambda T_1 + (1-lambda)T_2 right) leq lambda f(T_1) + (1-lambda)f(T_2)$.



      Now, I assume that for every $T_1, T_2 in (0,infty)$ and $0leq lambda leq 1$, the inequality below is true:



      $fleft( lambda T_1 + (1-lambda)T_2 right) > lambda f(T_1) + (1-lambda)f(T_2)$.



      Then, let $T_1 = T_2 = 1$ and $lambda = 1/3$. The inequality becomes
      $f(frac13 + frac23) > frac13f(1) + frac23f(1)$
      $f(1) > f(1)$ which is wrong. By contradiction, $f(T)$ is convex.
      But, then every function can be proven to be complex. How should I fix this proof?





      convex-analysis convexity-inequality





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      asked Jul 25 at 10:02









      silent_man

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          If $f$ is not convex you can only say that $f(lambda T_1+(1-lambda )T_2)>lambda f(T_1)+(1-lambda )f(T_2))$ for some choices of $T_1,T_2,lambda $, not for all choices.






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          • Then, what do I need to do to prove the convexity?
            – silent_man
            Jul 25 at 10:15










          • You cannot have a universal method of proving convexity.
            – Kavi Rama Murthy
            Jul 25 at 10:19











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          up vote
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          If $f$ is not convex you can only say that $f(lambda T_1+(1-lambda )T_2)>lambda f(T_1)+(1-lambda )f(T_2))$ for some choices of $T_1,T_2,lambda $, not for all choices.






          share|cite|improve this answer





















          • Then, what do I need to do to prove the convexity?
            – silent_man
            Jul 25 at 10:15










          • You cannot have a universal method of proving convexity.
            – Kavi Rama Murthy
            Jul 25 at 10:19















          up vote
          0
          down vote













          If $f$ is not convex you can only say that $f(lambda T_1+(1-lambda )T_2)>lambda f(T_1)+(1-lambda )f(T_2))$ for some choices of $T_1,T_2,lambda $, not for all choices.






          share|cite|improve this answer





















          • Then, what do I need to do to prove the convexity?
            – silent_man
            Jul 25 at 10:15










          • You cannot have a universal method of proving convexity.
            – Kavi Rama Murthy
            Jul 25 at 10:19













          up vote
          0
          down vote










          up vote
          0
          down vote









          If $f$ is not convex you can only say that $f(lambda T_1+(1-lambda )T_2)>lambda f(T_1)+(1-lambda )f(T_2))$ for some choices of $T_1,T_2,lambda $, not for all choices.






          share|cite|improve this answer













          If $f$ is not convex you can only say that $f(lambda T_1+(1-lambda )T_2)>lambda f(T_1)+(1-lambda )f(T_2))$ for some choices of $T_1,T_2,lambda $, not for all choices.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 10:04









          Kavi Rama Murthy

          20k2829




          20k2829











          • Then, what do I need to do to prove the convexity?
            – silent_man
            Jul 25 at 10:15










          • You cannot have a universal method of proving convexity.
            – Kavi Rama Murthy
            Jul 25 at 10:19

















          • Then, what do I need to do to prove the convexity?
            – silent_man
            Jul 25 at 10:15










          • You cannot have a universal method of proving convexity.
            – Kavi Rama Murthy
            Jul 25 at 10:19
















          Then, what do I need to do to prove the convexity?
          – silent_man
          Jul 25 at 10:15




          Then, what do I need to do to prove the convexity?
          – silent_man
          Jul 25 at 10:15












          You cannot have a universal method of proving convexity.
          – Kavi Rama Murthy
          Jul 25 at 10:19





          You cannot have a universal method of proving convexity.
          – Kavi Rama Murthy
          Jul 25 at 10:19













           

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