Determine all real functions $f$ such that $|f(x+h)-f(x)|\leq h^2$

Question

Determine all functions $f$ defined for all real numbers such that $$|f(x+h)-f(x)| \le h^2.$$

If we assume $f$ is not constant, then there exists some $ x$ and $y$ with $x \ne y$ such that $ f(x) \ne f(y) $.

Let $h = y - x$. Then

$$ |f(y) - f(x)| = |f(x + h) - f(x)| \le h^2 = (y - x)^2. $$

$ f(x) \ne f(y) \implies |f(y) - f(x)| > 0 $. However, as $y$ approaches $ x $, $ h = y - x $ approaches $0$, and $ h^2 $ approaches $0$, hence $ |f(y) - f(x)| $ approaches $0$, meaning $f(y) $ approaches $ f(x) $.

Hence a possible solution I've found is $f(x)=c$, since $|c-c|=0\le h^2$ is true.

Can someone verify this into a proper solution?

Cheers.

Answer 1

Your approach is in the right direction, we can verify this by considering the derivative of $f$

$\left| \frac{f(y)-f(x)}{y-x}\right|= \frac{|f(y)-f(x)|}{|y-x|}\leq \frac{|y-x|^2}{|y-x|}=|y-x|\to 0$ as $y\to x$ for all $x\in \mathbb{R}$

Hence, $\lim_{ y \to x } \frac{f(y)-f(x)}{y-x}$ exists and is zero, that is $f$ is differentiable with $f'(x)=0$ for all $x\in \mathbb{R}$

This implies that $f(x)=c$ (can prove this rigorously with mean value theorem)

Answer 2

A proof without calculus. Calculus certainly makes it simpler, but it is doable.

Note that if $x_i=x+ih$ then:

$$|f(x_{i+1})-f(x_i)|<h^2$$

So for any $n$ $$|f(x_n)-f(x_0)|\leq\sum_{i=0}^{n-1}|f(x_{i+1})-f(x_i)|\leq nh^2.$$

If $y>x$ then let $h=\frac{y-x}n$ then $x_n=y$ and we get, for any $n$: $$|f(y)-f(x)|\leq n\frac{(x-y)^2}{n^2}=\frac{(x-y)^2}{n}.\tag1$$

If $f(x)\neq f(y),$ then $|f(y)-f(x)|>0$ and we can pick $$n>\frac{(x-y)^2}{|f(y)-f(x)|}$$ then $$|f(y)-f(x)|>\frac{(x-y)^2}n,$$ which contradicts $(1).$ So our assumption that $f(x)\neq f(y)$ is false, and thus $f(x)=f(y)$ for all $x,y.$