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Let f be a convex and bounded function, meaning there is a constant $C$, such that $f(x) < C$ for every $x$.

I need to prove that $f$ is a constant function.


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marked as duplicate by 6005, Arash, azimut, Cameron Buie, Davide Giraudo Oct 7 '13 at 21:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 4 down vote accepted

Suppose f is not constant, i.e., $\exists x,y\in\mathbb{R}:f(x)>f(y)$. Since f is convex, we have: $f(x)\leq\lambda f(\frac{x-(1-\lambda)y}{\lambda})+(1-\lambda)f(y)\;\;\;\forall\lambda\in(0,1).$

(This is just the definition of convexity, $f(\lambda x'+(1-\lambda)y')\leq\lambda f(x')+(1-\lambda)f(y')\;\;\;\forall\lambda\in(0,1)$, with $x=\lambda x'+(1-\lambda)y'$ and $y=y'$.)

Hence $\frac{f(x)-(1-\lambda)f(y)}{\lambda}\leq f(\frac{x-(1-\lambda)y}{\lambda}).$

Now, since $f(x)>f(y)$, $\frac{f(x)-(1-\lambda)f(y)}{\lambda}=\frac{f(x)-f(y)}{\lambda}+f(y)\rightarrow \infty$ as $\lambda\rightarrow0^+.$

Hence f is not bounded above.

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Thank you so much. – user6163 Mar 1 '11 at 10:39

Hint: What can you say about derivatives of convex functions? Then compare the straight line joining two points (say $x$ and $x_1$) with the tangent line at $x$

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A piecewise constant function is not convex. – t.b. Mar 1 '11 at 10:13
a convex function is automatically Lipschitz. Its left and right derivatives automatically exist, and it is non-differentiable at at most countably many points. For the question asked, it is not necessary to make the assumption that $f$ is differentiable. – Willie Wong Mar 1 '11 at 10:18
@Theo, @Willie, thanks I have edited my hint – Juan S Mar 1 '11 at 10:21
Thank you qwirk. – user6163 Mar 1 '11 at 10:39