Let $K: [0,1] \times [0,1] \to \mathbb{R}$ be continuous. Let $f_n : [0,1] \to \mathbb{R}^n$ be continuous and uniformly bounded. Define
$$g_n(x) = \int_{0}^{1} K(x,y) f_n(y) \ dy, \ \ x \in [0,1], n \geq 1.$$ Show that the sequence $\{g_n(x) \}$ is uniformly bounded and equicontinuous.
Attempt:
To show that the sequence is uniformly bounded, we must show there exists an $M$ such that $|g_n(x)| \leq M,\ \forall n \in \mathbb{N}$ and for all x.
Since $f_n(x)$ are uniformly bounded, we have that $|f_n(x)| \leq M'$. Since $K(x,y)$ is continuous on a compact interval, we have that $K(x,y)$ reaches a maximum := $N$. Thus, $$|g_n(x)| = \left|\int_{0}^{1} K(x,y) f_n(y) \ dy\right| \leq \left|M' \int_{0}^{1} K(x,y) \ dy\right| \leq |M'N(1-0)| := M.$$ Thus, the sequence $\{g_n(x)\}$ is uniformly bounded.
To show that the sequence is equicontinuous, we must show that for every $\epsilon > 0$ there exists a $\delta$ (which can depend on $x_0$) such that $|g_n(x) - g_n(x_0)| < \epsilon$ when $|x-x_0| < \delta.$ Well, $$|g_n(x) - g_n(x_0)| = \left|\int_{0}^{1} K(x,y) f_n(y) \ dy - \int_{0}^{1} K(x_0,y) f_n(y) \ dy\right| \leq M' \left|\int_{0}^{1} K(x,y) - K(x_0, y) \ dy \right|.$$
Since $K$ is uniformly continuous (because $K$ is continuous on a compact set), we have that $|K(x,y) - K(x_0, y)| < \epsilon/M'$ whenever $|x-x_0| < \delta$.
So, we get $$|g_n(x) - g_n(x_0)| \leq M' \left|\int_{0}^{1} K(x,y) - K(x_0, y) \ dy \right| \leq M' \int_{0}^{1} |K(x,y) - K(x_0, y)| \ dy \leq \ M'\epsilon/M' = \epsilon$$ whenever $|x-x_0| < \delta.$ Hence, the sequence is equicontinuous.
