Let $G$ be the following subset of $C([0,\pi])$ $$G = \{g \in C([0, \pi]); g(x) = \int_0^\pi sin(xy)f(y) dy, f \in C([0, π]), \|f\|_\infty \leq 1\}.$$ Prove that $G$ is relatively compact in $C([0,\pi])$.
I want to prove this claim by applying Arzelà-Ascoli's Theorem; by showing that G is bounded and equicontinuous, I can then conclude that G is precompact (relatively compact).
To show that $G$ is bounded I considered this:
$$\|g(x)|_\infty = \|\int_0^\pi \sin(xy)f(y)dy \|_\infty = \sup_{x,y,z\in[0,\pi]} |\int_0^\pi\sin(xy)f(y)dy|$$
Then since $\|f\|_\infty=\sup_{x\in[0,\pi]}|f(y)| \leq 1$ and $|\sin(xy)|\leq 1$, we can say that $$\sup_{x,y,z\in[0,\pi]} |\int_0^\pi\sin(xy)f(y)dy|\leq1$$
but I am not sure about this part. How could I justify better this step?
The concerning proving equicontinuity, I wanted to show that for any $\epsilon>0$ there exist a $\delta>0$ such that $|x-z|<\delta$ implies $\|g(x) - g(z)\|_\infty<\epsilon$.
I started working with this: $$\|g(x) - g(z)\|_\infty = \|\int_0^\pi \sin(xy)f(y)dy - \int_0^\pi\sin(zy)f(y)dy\|_\infty = \sup_{x,y,z\in[0,\pi]} \{|\int_0^\pi\sin(xy)f(y)dy-\int_0^\pi\sin(zy)f(y)dy|\}$$
but I am not being able to continue from this step.
First, off you want to show $G$ is uniformly bounded. This follows since $$ |g(x)| \le \int_0^\pi |\sin(xy)| |f(y)| \; dy \le \int_0^\pi 1 \; dy = \pi $$ which is independent of $x$ and of $f$. Now for equicontinuity, you have $$|g(x) - g(z)| \le \int_0^\pi |f(y)| |\sin(xy) - \sin(zy)| \; dy \le \int_0^\pi |\sin(xy)-\sin(zy)| \; dy $$
Fix $y$ and define $h(x) = \sin(xy)$. Then, by MVT
$$ h'(\xi) = \frac{h(x) - h(z)}{x-z} =\frac{\sin(xy) - \sin(zy)}{x-z} = y \cos( \xi y) $$ for some $\xi$ between $x,z$ so that $|\sin(xy) - \sin(zy)| \le y |x-z| \le \pi|x-z|$. Therefore, we have $$ |g(x)-g(z)| \le \int_0^\pi \pi |x-z| \;dy = |x-z| \pi^2 $$ So pick $|x-z|< \epsilon/\pi^2$.