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Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous and midpoint convex function, which means that

$$\forall x,y\in \mathbb{R}\;,\;\;\displaystyle{ f\left(\frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2} .}$$

Assume there exists $a,b\in \mathbb{R}$ such that $f(a)=f(b)=0$.

Show that $f\leq 0$ on $[a,b]$. And in a second time, deduce that $f$ is convex.

Any hint would appreciated.

DonAntonio
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anonymus
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  • What you wrote in the second line is the definition of convex function over an interval (or the whole real line) that I know of. What is yours? – DonAntonio Nov 14 '16 at 12:19
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    Possibly that $f((1-t)x + t y) \le (1-t) f(x) + t f(y)$ for $0 \le t \le 1$ (i.e. the line segment from $(x, f(x))$ to $(y, f(y))$ lies entirely below the graph of $f$. – John Hughes Nov 14 '16 at 12:22
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    Hint: look at the menu sidebar Related to the right. – A.Γ. Nov 14 '16 at 12:26

2 Answers2

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Hint: To make it simple, look at the case $a = 0, b = 1$; any other case reduces to this via $$ g(x) = f(\frac{x-a}{b-a}). $$ Can you show that if $f(x), f(y) \le 0$, then $f(p) \le 0$ where $p$ is the average of $x$ and $y$?

If so, then you know that $f$ is negative at every point of the form $$ \frac{n}{2^k} $$ where $n$ is an integer and $0 \le \frac{n}{2^k} \le 1$.

Now: in what way have we used the continuity of $f$ so far?

John Hughes
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  • thank you very much ! First I prove that $f(2^{-k})\leq 0$ for all $k$. Then, I show that $f(\frac{n}{2^k}) \leq \sum_{i=1}^n 2^{-i}f(\frac{1}{2^{k-i}})$ by iterating on the hypothesis. The last quantity is nonpositive. Finally, by continuity and density, $f(x)\leq 0$ holds for all $x\in [0,1]$. – anonymus Nov 14 '16 at 18:18
  • That looks right to me. – John Hughes Nov 14 '16 at 18:19
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John Hughes gave a very nice path to the first part. For the second part, use the transformation $$ g(x)=f(x)-mx-k $$ where $y=mx+k$ is the line through $(a,f(a))$ and $(b,f(b))$. Then $g(a)=g(b)=0$.

String
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  • Nice. Thanks alot ! The idea is to show that $f$ is below any ray linking two points of the graph of $f$. In order to prove that, we show that $g$ fulfilles the hypothesis of the first part and hence is always nonpositive. – anonymus Nov 14 '16 at 18:46