# For each interval $[a,b]$ contained in $I$, sequence $\{f_{n}:[a,b]\rightarrow\mathbb{R}\}$ converges uniformly to $f : [a,b] \rightarrow\mathbb{R}$

$$I$$ is an open interval

Using the following fact to show this:

$${\{f_n\}}$$ converges pointwise on $$I$$ to the function $$f$$, and $${\{f'_n}\}$$ converges uniformly on $$I$$ to the function $$g$$

Attempt:

$$|{f'_n}(x)-g(x)|<\epsilon/(2(b-a))$$, and $$|{f_n}(a)-f(a)|<\epsilon/2$$

Integrate the former equation from a to b to cancel out (b-a) so first equation $$<\epsilon/2$$

Add the two and then $$|{f_n}(x)-f(x)|<\epsilon$$