Q:

Uniform Convergence of Composite Functions using Uniform Continuity

Suppose that $ f_n $ converges uniformly to $f$ on $[a,b]$, and that
  $|f_n(x)|\leq M, |f(x)|\leq M$ for all $x \in [a,b] $.  Let $g$ be a
  function continuous on [-M,M].  Show that $g \circ fn$ converges to $g
 \circ f$ uniformly on $[a,b]$. Hint: Use uniform continuity. 

I know that since $g$ is continuous on [-M,M], it is uniformly continuous on [-M,M].  However, I'm not sure how to use this in this case (even though it's in the hint...).  Would I say something like:
Since $g$ is uniformly continuous on [-M,M] and  $ f_n $ converges uniformly to $f$ on $[a,b]$ and is bounded by M we can say that $g \circ fn$ converges to $g \circ f$ uniformly on $[a,b]$? 
Any help would be appreciated! 

A:

Since, as you said, $g$ is uniformly continuous, for every $\epsilon>0$ there is $\delta>0$ such that if $y,z\in [-M,M]$ with $|y-z|<\delta$ then $|g(y)- g(z)|<\epsilon.$
On the other hand, there is $N$ such that $|f_n(x) - f(x)|<\delta$ for all $n>N$ and all $x\in[a,b].$
By taking $z=f_n(x)\in[-M,M]$ and $y=f(x)\in[-M,M]$, we have shown that for every $\epsilon>0$ there is $N$ such that $|g(f_n(x)) - g(f(x))|<\epsilon$ for all $n>N$ and all $x\in[a,b].$

