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Which of the following functions are uniformly continuous on the whole real line: $\sin x, x\sin x,x\sin(x^2),\sqrt{|x|}\sin x$?

To prove something like this, we must prove, for example, that given any $\varepsilon$ there exists $\delta$ such that $|x-y|<\delta\rightarrow|\sin x-\sin y|<\varepsilon$ for all $x,y\in\mathbb{R}$. But we can't express $\sin x-\sin y$ easily in terms of $x-y$. So how can we prove (or disprove) such a thing?

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You can use mean value theorem, for instance. – njguliyev Aug 28 '13 at 21:56
As @njguliyev mentions, for continuously differentiable functions on the line, uniform continuity amounts to bounded derivative. This implies that we only really need to work with the last example (or do we?) – Jonathan Y. Aug 28 '13 at 22:02
@copper.hat you cannot use unboundness of derivative to show non uniform continuous – Neeraj Bhauryal Aug 29 '13 at 17:10
@NeerajBhauryal: You are correct, I was thinking of a Lipschitz constant. I will delete my comment above. – copper.hat Aug 29 '13 at 18:27
up vote 1 down vote accepted

The mean value theorem, along with the fact that $\sin'=\cos$ is bounded allows you to conclude that $x \mapsto \sin x$ is Lipschitz and hence uniformly continuous.

Now consider a function of the form $f(x) = \phi(x) \sin x$, where $\phi$ satisfies $\lim_{x \to \infty} \phi(x) = \infty$. Note that there exists a $\delta>0$ such that if $x \in [0,\delta)$, then $\sin x \ge \frac{1}{2} x$. Since $\sin$ is $2 \pi$-periodic, we see that if $x \in [0,\delta)$ then $\sin (x+2 \pi n) \ge \frac{1}{2} x$. The condition on $\phi$ shows that if $n$ is sufficiently large, then $\phi(x+2 \pi n) > 0$ for all $x \in [0,\delta)$. Then $f(x+2 \pi n)-f(2 \pi n) \ge \frac{1}{2} x \, \phi(x+2 \pi n)$. Now let $x_k = \frac{1}{k}$ ans suppose $k$ is large enough so that $x_k < \delta$. Choose $n_k$ such that $\phi(x_k+2 \pi n_k) > k$. Then $f(x_k+2 \pi n_k)-f(2 \pi n_k) \ge \frac{1}{2}$ for all $k$ sufficiently large. Since $x_k \to 0$, it follows that $f$ cannot be uniformly continuous. Hence $x \mapsto x \sin x$ and $x \mapsto \sqrt{|x|} \sin x$ are not uniformly continuous.

Finally consider $f(x) = x \sin x^2$. Then if $n$ is a positive integer, we have$f(\sqrt{2 \pi n + \frac{\pi}{2}}) - f(\sqrt{2 \pi n }) =\sqrt{2 \pi n + \frac{\pi}{2}} > \sqrt{\frac{\pi}{2}}$, whereas $\lim_ {n \to \infty} ( \sqrt{2 \pi n + \frac{\pi}{2}} - \sqrt{2 \pi n }) = 0$. Hence $f$ is not uniformly continuous.

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