# Show that the sequence $\phi_{n}(t)=\int_{0}^{1}{\sin^{2}(t-ns)g(s)ds}$ has convergent subsequence.

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!

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.