Value of a function that is uniformly convergent.

Question

"Let $f$ be a function of $\mathbb{R}$ into $\mathbb{R}$ such that $\vert f(x)-f(y) \vert\leq\frac{\pi}{2}\vert x-y\vert^2$ for all $x,y\in\mathbb{R}$, and such that $f(0)=0$. What is $f(\pi)$?".

Since we know nothing about the differentiability of $f$, I don't think we can apply the mean value theorem here. With a choice of $\delta =\sqrt{\frac{2\epsilon}{\pi}}$, it can be shown this function is uniformly continuous in $\Bbb{R}$ $\implies$ $f$ is continuous in each point $x \in \Bbb{R}$. Therefore $$f(\pi) = \lim_{x\to \pi} f(x).$$

Geometrically speaking, the secant line between $x,y$ is bounded by $\frac{\pi}{2} \vert x-y \vert$. This is all the meaningful information I could find. Can anyone explain what this problem is about?

If you divide on both sides by $|x-y|$, you do know something about the differentiability of $f$. — Stefan, May 25 '20 at 16:56
Yes, we get that $f'(x)$ is bounded. Does this mean that $f$ is differentiable? — In the blind, May 25 '20 at 16:58
It becomes clear that you misunderstand parts of the problem statement. It's also written down somewhat sloppily. I'll suggest a better version by editing your question. — Stefan, May 25 '20 at 18:17

score 3 · Accepted Answer · answered May 25 '20 at 18:23

Fix $a \in \mathbb R$.

Then $$\left| \frac{f(x)-f(a)}{x-a} \right|\leq \frac{\pi}{2} |x-a|$$

Taking the limit when $x \to a$ you get that $f$ is differentiable at $a$ and $$f'(a)=0$$

Since $a$ is arbitrary, you get that $f$ is differentiable everywhere and $f'=0$. This implies that $f$ is constant, and since $f(0)=0$ you get $f\equiv 0$.

score 2 · Answer 2 · answered May 25 '20 at 16:57

2

The answer is 0. Indeed $f$ is 2-Holder continuous and any function $\alpha$-Holder continuous for $\alpha$>1 is a constant.

If $f$ is holder continuous for $\alpha >1$ then $f$ is constant.

answered May 25 '20 at 16:57

ECL

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Value of a function that is uniformly convergent.

2 Answers2