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I have the question that is $sin^2x$ uniformly continuous on $x \in [0,\infty]$ ?
My approach:

Let $\left|x-y\right|<\delta$ we have:- $$\left|sin^2x-sin^2y\right|=\left|(\sin x+\sin y)(sin x-sin y)\right|$$ $$=4\left|\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right|\lt4\left|\sin \left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\tag{1}$$ $$\lt \left|(x-y)(x+y)\right|<\left|(x+y)\right|\delta,$$which is dependent on $x$ so $\sin^2\!x$ is not uniformly continuous.
Is this solution correct or not? I have some doubt about validity of inequality $(1)$ also, if it is correct then why?

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  • $\begingroup$ In case it's useful, a continuous, $\ell$-periodic function $f$ on the real line is uniformly continuous. (Proving this is a good exercise. Hint: Look at the restriction of $f$ to the interval $[0, 2\ell]$ rather than $[0, \ell]$.) $\endgroup$ – Andrew D. Hwang Jan 13 '16 at 2:47
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I'm not sure if you are "allowed" to use this method, but for a place to start:

Note that the derivative $2\sin x\cos x$ is bounded between $[-2,2]$

Now we want to show that for any $\epsilon$, we can pick $\delta$ such that $|x-y|<\delta$ implies $|\sin^2x-\sin^2y|<\epsilon$.

And since the derivative is bounded, we have that for all $x,y$ $$\left|\frac{\sin^2x-\sin^2y}{x-y}\right|\le 2$$

And you can probably take it from here.

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Carrying your idea further, we have for $|x-y| < \delta(\epsilon) = \epsilon/2$

$$|\sin^2 x - \sin^2 y|= |\sin x + \sin y||\sin x - \sin y| \\ \leqslant 2 |\sin x - \sin y|\\ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right| \\ \leqslant 4\left|\sin\left(\frac{x-y}{2}\right)\right|\\ \leqslant 2|x-y| \\< \epsilon.$$

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  • $\begingroup$ how is $|\sin x + \sin y||\sin x - \sin y| \le 2 |\sin x - \sin y|$? $\endgroup$ – Mayank Deora Jan 13 '16 at 2:48
  • $\begingroup$ @Mayank Deora: By the triangle inequality -- $|\sin x + \sin y| \leqslant |\sin x| + |\sin y| \leqslant 2$ $\endgroup$ – RRL Jan 13 '16 at 2:53
  • $\begingroup$ So you were on the right track. $\endgroup$ – RRL Jan 13 '16 at 2:54
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There is another elegant aprouch that is more general:

Lemma: If $f:[a,b]\to\mathbb{R}$ is continuous, then is uniformly continous Proof: Check any Calculus book.

Corolary: If $f:\mathbb{R}\to\mathbb{R}$ is continous and periodic, then is uniformly continous.

Proof: Suppose $f(x)=f(x+p)$ for all $x$ and $p>0$. In $[0,p]$, by Lemma, $f$ is uniformly continous. Let $\epsilon>0$. Then, there is $\delta>0$ such that if $|x-y|<\delta$ and $x,y\in[0,p]$, then $|f(x)-f(y)|<\epsilon$. Now, if $a,b\in\mathbb{R}$ satisfied $|a-b|<\delta$, there are $x,y\in[0,p]$ such that $f(a)=f(x)$, $f(b)=f(y)$ and $|a-b|=|x-y|$ and we're done.

Therefore, $\sin^2$ is uniformly continuous.

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