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Let $g\in\mathcal{C}([0,1])$ and $$\phi_{n}(t)=\int_{0}^{1}{\sin^{2}(t-ns)g(s)ds}$$

Show that the sequence $\{\phi_{n}\}_n$ has a uniformly convergent subsequence on $[0,1]$.

I tried to use the arzela-Ascoli theorem; First of all, we have that $\{\phi_{n}\}_n$ is bounded, since

$$|\phi_{n}(t)|\leq\int_{0}^{1}{|\sin^{2}(t-ns)||g(s)|ds}\leq M\int_{0}^{1}{|\sin^{2}(t-ns)|ds}=M$$

Where $M:=\sup\{g(s)\mid s\in[0,1]\}$ ($g$ is a continuous function over a compact set). And the second condition, namely the equicontinuity, we have

$$|\phi_{n}(x)-\phi_{n}(y)|\leq M\int_{0}^{1}{|\sin^{2}(x-ns)-\sin^{2}(y-ns)|ds}\qquad\qquad(*)$$

Now, note that $$|\sin^2 (x) - \sin^2 (y)|= |\sin (x) + \sin (y)||\sin (x) - \sin (y)|\leq 2 |\sin (x) - \sin (y)|$$ $$ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right|\leq 4\left|\sin\left(\frac{x-y}{2}\right)\right|\leq 2|x-y|$$

This implies that, given $\epsilon>0$, if $|x-y|<\delta$ with $\delta=\frac{\epsilon}{2M}$, then $$|\sin^2(x-ns) - \sin^2(y-ns)|\leq 2|x-y|<\frac{\epsilon}{M}$$

Therefore, (*) is equal to

$$|\phi_{n}(x)-\phi_{n}(y)|\leq M\int_{0}^{1}{|\sin^{2}(x-ns)-\sin^{2}(y-ns)|ds}<\epsilon\int_{0}^{1}{ds}=\epsilon$$

So we have the hypothesis of Arzela-Ascoli theorem, then there exist a uniformly convergent subsequence on $[0,1]$. You think this approach is correct? any hint will be appreciated. Thanks!

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Looks fine.

You might like to use the term uniformly bounded.

I think the following should be an inequality $$M\int_{0}^{1}{|\sin^{2}(t-ns)|ds}\le M.$$

Also, we can also obtain $|\sin(x)-\sin(y)| \le |x-y|$ by using mean value theorem.

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