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A convex function $f$ defined on an interval $[a,b]$ with $a<b$ hits its maximum in one of the points a and b.

If it’s true, how to prove it? I only know the definition of convex functions and that if a functions is twice differentiable and the second derivative is nonnegative then it is convex. I don’t even know where to start. Can you please show me a complete proof so I can fully understand? Is it also available for strictly convex function? Can there exist other values $x$ such that f(x)=maximum value? Or are $a$ and $b$ the only numbers that have f(a) and f(b) maximum?

Thank you very much!

Rodrigo de Azevedo
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furfur
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2 Answers2

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So let's prove the statement. Suppose for a contradiction that $f$ is not maximal on $a$ and $b$. Assume (without loss of generality by symmetry) that $f(a)\leq f(b)$. Then $f$ is not constant. There is $c\in(a,b)$ such that $f(c)>f(b)$. $f$ is continuous (see Proof of "every convex function is continuous"). So the idea is, that $f$ goes up all the way before $t=b$, and therefore the line segment between $(a,f(a))$ and $(b,f(b))$ has to be above the graph of $f$ in some point $d$ between $a$ and $b$. Indeed, the line segment is never higher than $y=f(b)$. Hope this helps you.

The maxima/minima of $f$ need not be unique: also constant functions are convex.

If $f$ is strictly convex, as you suspected, we can say something more: $f$ is convex so $f(a)$ and $f(b)$ are maximal. Furthermore $f$ cannot attain the maximum in any other point between $a$ and $b$, because $f$ is strictly bounded from above on $(a,b)$ by the line segment from $(a,f(a))$ to $(b,f(b))$, which is never higher than the maximum.

Maurice Dekker
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Here's an elementary proof.

Consider $c\in [a,b]$ and note that $$f(c)=f\left(\frac{c-a}{b-a}b + (1-\frac{c-a}{b-a})a\right)\leq \frac{c-a}{b-a} f(b)+ (1-\frac{c-a}{b-a})f(a)\leq\max(f(a),f(b))$$

Hence $f$ reaches a maximum at $a$ or $b$.

Gabriel Romon
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