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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 $\mathbb{R}^+$?

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  • $\begingroup$ I apologize for my typo errors.. $\endgroup$ – Nitin Uniyal Oct 15 '15 at 17:43
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    $\begingroup$ $f$ has a bounded derivative, so it is a Lipschitz function, and consequently, it is uniformly continuous. $\endgroup$ – Omran Kouba Oct 15 '15 at 17:52
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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 $|f(y) - f(x)| < \epsilon$.

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  • $\begingroup$ Thanks a lot Bungo..I understand it clearly. Thanks.. $\endgroup$ – Nitin Uniyal Oct 16 '15 at 1:01
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Hint: if a function has a bounded derivative, it is uniformly continuous.

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