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!
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.