I am aware that the following proposition is similar to Theorem.3.7.1. in Analysis II by Terence Tao and Theorem.7.17 in Baby Rudin.
Given a sequence of functions $(f_n)_{n \geq 0}, f \in C^1 [-1, 1]$ and $f_n \longrightarrow f$ uniformly.
Define $F_n(t) := \displaystyle \int_{-t}^{t} f_n(s) ds$ and $F(t) := \displaystyle \int_{-t}^{t} f(s) ds$.
How do we prove that $F_n \longrightarrow F$ uniformly and $F'_n \longrightarrow F'$ uniformly please?
The professor gave the following sketch of the proof:
By FTC I, $(F_n - F)(x) := \displaystyle \int_{x_0}^{x} (f_n - f)(t) dt$
And by fundamental estimate, $|(F_n - F)(x)| \leq \text{Length}(I) \cdot \displaystyle \sup_{I} |f_n - f|$, where $I$ is an interval.
However I am unsure how to bound $|F'(x)|$ or $|(F'_n - F')(x)|$ (in order to show the uniform convergence of $F'_n$)
Many thanks in advance!
We have $$F(t) := \displaystyle \int_{-t}^{t} f(s) ds =\int_{-1}^{t} f(s) ds - \int_{-1}^{-t} f(s) ds $$
Hence $$F'(t) = f(t) - f(-t)$$ and so we get $$|(F'_n - F')(x)| = |f_n(x) - f_n(-x) - f(x) + f(-x)| \le 2\sup_{[-1,1]}|f_n - f|$$