5
$\begingroup$

Let $f:[0,\infty)\to \Bbb R$ be defined by

$$f(x)=\int_{0}^x \sin^2(t^2)\mathrm dt$$

Then Show that the function is uniformly continuous on $[0,1)$ and $(0,\infty)$

Attempt: Differentiating with respect to $x$ we get,

$$f'(x)=\sin^2(x^2)$$ then $$|f'(x)|=|\sin^2(x^2)|\le1,\forall x\in(0,\infty)$$

this means that derivative of $f$ is bounded, hence f is uniformly continuous on given intervals. am I right? Also provide other methods to solve this. Thank you.

$\endgroup$
6
$\begingroup$

Your solution is very good. By proving that the derivative is bounded and next using Lebesgue's mean value theorem, you can prove that $f$ is Lipschitz continuous (of Lipschitz constant $1$), hence in particular uniformly continuous not just on intervals, but even on the whole $\Bbb R$.

An alternative could be the following: assume $x \le y$; then $|f(x) - f(y)| = | \int \limits _x ^y \sin ^2 (t^2) \Bbb d x | \le \int \limits _x ^y |\sin ^2 (t^2)| \Bbb d x \le \int \limits _x ^y 1 \ \Bbb d x = y - x = |x - y|$. The advantage of this proof is that it uses integration without even touching differentiability or mean values theorems (and, in general, integrability is considered a more convenient setting than derivability, because it is weaker).

$\endgroup$
  • $\begingroup$ Thanx for quick reply :). different thoughts are also invited, so if there is another approach in your mind then edit your answer. $\endgroup$ – Chiranjeev_Kumar Nov 16 '15 at 17:08
  • $\begingroup$ @Chiranjeev: I have added a second proof based upon integrability. $\endgroup$ – Alex M. Nov 16 '15 at 17:16
  • $\begingroup$ Thanx a lot, I was waiting for such an reply. :) $\endgroup$ – Chiranjeev_Kumar Nov 16 '15 at 17:25
2
$\begingroup$

You're attempt is correct. In general if you're derivative is a priori bounded $\|f'\|_\infty\leq M$, then the function is uniformly continuous by the mean value theorem: $$f(x)-f(y)=f'(\xi)(x-y)\Longrightarrow|f(x)-f(y)|=|f'(\xi)||x-y|\leq M|x-y|.$$

$\endgroup$
2
$\begingroup$

Let $\epsilon >0$.

By MVT we have $\vert f(x)-f(y)\vert =\vert f'(c)(x-y)\vert \leq \vert x-y\vert ;\ y<c<x$.

Now set $\delta =\epsilon $ and the result follows.

$\endgroup$

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.