# Uniform continuity of a function given by an integral

Let $f$:$(0,\infty)\to\mathbb{R}$ be defined as

$$f(x)=\int_0^x \sin^2 t^2\,dt$$

Then how can I prove that $f$ is uniformly continuous on $[0,1)$ and on $\mathbb{R}^+$?

• I apologize for my typo errors.. – Nitin Uniyal Oct 15 '15 at 17:43
• $f$ has a bounded derivative, so it is a Lipschitz function, and consequently, it is uniformly continuous. – Omran Kouba Oct 15 '15 at 17:52

For a direct proof, suppose $x,y \geq 0$. We may assume that $y \geq x \geq 0$ (by renaming them if necessary), so \begin{aligned} |f(y) - f(x)| &= \left|\int_x^y \sin^2 t^2\ dt\right|\\ &\leq \int_x^y|\sin^2 t^2|\ dt \\ &\leq \int_x^y(1)\ dt \\ &= y - x \\ &= |y - x| \\ \end{aligned} Therefore $|y-x| < \epsilon$ implies $|f(y) - f(x)| < \epsilon$.