5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.