Show that $f(x)={x\over2}+c$, where $c$ is a constant

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying $|f(x+y)-f(x-y)-y|$$\le y^2$

for all $x,y\in\mathbb{R}$. Show that $f(x)={x\over2}+c$, where $c$ is a constant.

Answer 1

Hint:

Set $x-y = z$ and $2y = h$.

Let $\epsilon > 0$ be given.

We have $\bigg|\dfrac{f(z+ h) - f(z)}{h} - \dfrac{1}{2} \bigg| \le \dfrac{h^2}{4}$.

So if we take $|h| < 2 \sqrt{\epsilon}$, then $\bigg|\dfrac{f(z+ h) - f(z)}{h} - \dfrac{1}{2}\bigg | < \epsilon$, this means $f'(z)$ exists and equals $\dfrac{1}{2}$. Rest is trivial.