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Suppose that $a\lt b$ and $f:[a,b]\longrightarrow\mathbb{R}$.

We say that f is strictly convex on $[a,b]$ if, for each $a,y\in[a,b]$ with $x\neq y$, $$f(0.5x+0.5y) \lt 0.5f(x)+0.5f(y)$$.

If $f$ is continuous and strictly convex (by the above definition) on $[a,b]$ then I want to prove that $f$ attains a unique minimum in $[a,b]$ and a maximum value at either $a,b $ or both.

This is what I have so far:

The existence of both follows from the min-max theorem as $f$ is continuous on a closed interval.

The minimum was straight forward, I did a proof by contradiction, assuming two minima and then showing, by the definition of strictly convex, that we got another point $x$, where $f(x) \lt f(c)$ (when c was one of our minima).

With the maxima I couldn't get anywhere. I tried assuming a maxima on $(a,b)$, again trying for contradiction but I had no luck. Any help would be great!

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Suppose $c \in (a,b)$ is the global maximum over $[a,b]$.

Let $d= \min(c-a, b-c)$, then $c-d,c+d \in [a,b]$.

Let $x=c-d$ and $y=c+d$, then we can conclude that

$$f(c)< 0.5f(x)+0.5f(y)$$

Try to argue that $f(x)>f(c)$ or $f(y)>f(c)$ to get a contradiction.

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  • $\begingroup$ Thanks a lot, this makes sense! $\endgroup$ Commented Nov 12, 2017 at 8:13

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