# Showing there is a subsequence that uniformly converges?

Let $g$ be continuous on $[0,1]$. Let $(f_n)$ be a uniformly bounded sequence of Riemann-integrable functions on $[0,1]$ and let $$F_n(x) = g(x)\int_0^x f_n(t)\,dt.$$ Prove that there is a subsequence $(F_{n_k})$ that is uniformly convergent on $[0,1]$.

By the Arzelà–Ascoli theorem, it suffices to show that $\{F_n\}$ is pointwise bounded and equicontinuous. Note that there exists $M > 0$ such that $|g(x)| \leq M$ for all $x\in [0,1]$. Since $\{f_n\}$ is uniformly bounded, there exists some $N > 0$ such that $|f_n(t)| < N$ for all $t\in [0,1]$ and for all $n\in\mathbb{N}$. We have that $$|F_n(x)| = \left|g(x)\int_0^x f_n(t)\,dt\right| \leq |g(x)|\int_0^x |f_n(t)|\,dt \leq M \int_0^x N\,dt = MNx$$ for all $n\in\mathbb{N}$, so $\{F_n\}$ is pointwise bounded.

I'm stuck on proving the equicontinuity condition. Fix $\epsilon > 0$. For any $x,y\in[0,1]$ with $|x-y|<\delta$, $$|F_n(x) - F_n(y)| = \left|g(x)\int_0^x f_n(t)\,dt - g(y)\int_0^y f_n(t)\,dt \right|,$$ but how do I proceed?

Hint, prove this: Suppose $E \subset C[0,1]$ is equicontinuous and uniformly bounded. Let $g\in C[0,1].$ Then $\{gf: f \in E\}$ is equicontinuous.
• It helps because the integrals are easily shown to be equicontinuous. (To prove the hint, recall that $g$ is uniformly continuous and how to prove the product of continuous functions is continuous.) – zhw. Dec 14 '15 at 2:13