# composition of uniformly convergence sequence with continuous function, is uniformly convergence?

Let $(f_n)$ be a series of functions in $C[0,1]$ that uniformly converge to a continuous function $f\in C[0,1]$.

a. Let $g: [0,1]\to [0,1]$ be a continuous function. Is it true that $f_n\circ g$ uniformly converges to $f\circ g$?

b. If no, what about the special case $g(x)=x^2$? It would also be differentiable, have bounded derivative, etc.

Intuitively I find it logical that a. would be true, and intuitively I'd guess that in order to prove such claim I'd have to go directly to the $\epsilon-\delta$ definition.

• since $[0,1]$ is compact, $g$ must also be uniformly continuous, does that help? But perhaps the proof goes in the more general case, without the assumption that the domain is compact? – Mirko Nov 12 '14 at 12:15

This is actually very simple:

Let $$\epsilon>0$$, then there is $$N\in \mathbb N$$ so that

$$|f_n(y) - f(y) | < \epsilon$$

for all $$y\in [0,1]$$ and $$n\geq N$$. In particular,

$$|f_n(g(x)) - f(g(x))| < \epsilon$$

for all $$x\in [0,1]$$ and $$n\geq N$$. Thus $$f_n\circ g$$ converges to $$f\circ g$$ uniformly. Note that continuity of $$g$$ is not needed.

The harder one is the following:

Claim: If $$g: \mathbb R \to \mathbb R$$ is continuous and $$f_n: [0,1] \to \mathbb R$$ converges uniformly to $$f$$, then $$g\circ f_n$$ converges uniformly to $$g\circ f$$.

Proof: First of all, as $$f_n$$ converges to $$f$$ uniformly, there is $$M>0$$ so that $$|f_n(x)|\leq M$$ for all $$x\in [0,1]$$ and $$n\in \mathbb N$$ (Why?). Then we consider $$g:[-M, M]\to \mathbb R$$. Let $$\epsilon>0$$. Then by uniform continuity of $$g$$ (as the domain is closed and bounded), there is $$\delta>0$$ so that $$|g(y) - g(z) | < \epsilon$$

whenever $$|y-z|<\delta$$. Using this $$\delta$$ there is $$N$$ so that

$$|f_n(x) - f(x)| < \delta$$

for all $$x\in [0,1]$$ and $$n\geq N$$. Thus

$$|g(f_n(x)) - g(f(x))| <\epsilon$$

for all $$x\in [0,1]$$ and $$n\geq N$$.

• Thanks so much. Is that right that the second claim is wrong for $g$'s which are not uniformly continuous? – user188400 Nov 12 '14 at 12:36
• Uniform continuity of $g$ is not needed (please see the edit). The main point is that $f_n$ are defined on a compact interval. – user99914 Nov 12 '14 at 12:46
• can you think of an example when compactness is violated and the theorem does not hold? – fdzsfhaS Jan 9 '20 at 4:08
• @DavidWarrenKatz The idea for a counterexample is to have the $f$ be unbounded and $g$ "stretch out" the distance between points near infinity. For example: $f_n:(0,1]\to \mathbb R, f_n (x) = 1/x + 1/n$ and $g(y)=y^2$. Then $f_n$ converges uniformly on $(0,1]$ to $f(x)=1/x$, but $|g(f_n(x))-g(f(x))| = 2/(xn)+1/n^2$ is unbounded for every $n$, so $g \circ f_n$ cannot converge uniformly. – hife Jan 16 '20 at 23:11