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Let $f: [0,\infty ) \to \Bbb{R}$ be defined by $$f(x)=\int_0^x \sin^2(t^2)\textrm{d}t.$$ Then how to show that $f$ is uniformly continuous on both $[0,1)$ and $(0,\infty)$. Is it uniformly continuous on $[0,\infty)$?

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    $\begingroup$ Do you know about the concept of Lipschitz continuity, and that it implies uniform continuity (even on non-compact domains)? $\endgroup$ – Ian Feb 1 '17 at 13:10
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$$\left|\int_0^x \sin^2(t^2)dt-\int_0^y \sin^2(t^2)dt\right|=\left|\int_y^x \sin^2(t^2)dt\right|\le \left(\sup_{t \in (y,x)}\sin^2(t^2)\right) |x-y| $$

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