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.

  • $\begingroup$ Hint: Try taking $g(x) = f(x) - \frac{x}{2}$. Can you write the question in terms of $g$? $\endgroup$ – KSackel May 7 '14 at 13:12
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    $\begingroup$ In terms of $g$? How? $\endgroup$ – Rudstar May 7 '14 at 13:15
  • $\begingroup$ What is $g(x+y) - g(x-y)$? $\endgroup$ – KSackel May 7 '14 at 13:16


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.


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