# composition of bounded uniformly convergence sequences

I'm hoping to make a generalization of the answer to this question. Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $$f \circ g$$.

Here's why I think it does.

Consider $$\left( f _ { k } \right) _ { k = 1 } ^ { \infty }$$ and $$( g_k ) \stackrel { \infty } { k } = 1$$ to be sequences of continuous functions $$[ 0,1 ] \rightarrow [ 0,1 ]$$ converging uniformly to $$f$$ and $$g : [ 0,1 ] \rightarrow \mathbb { R }$$ respectively.

By applying the Weierstrass M-Test, $$\exists \quad M > 0$$ so that $$|f_k(x)| \leq M \quad \forall x \in [0,1]$$ and $$k \in \mathbb{N}$$

Now restrict the domain of $$(g_k)$$ to $$[ - M , M ]$$. Since the codomain of $$(f_k)$$ is $$[0,1]$$, this $$[ - M , M ]$$ will be within that domain. Therefore, there exists $$N \in \mathbb{N}$$ such that $$(g_k)$$ converges uniformly within this restricted domain.

Therefore

$$$$\left\| g _ { k } ( f_k(x )) - g(f ( x )) \right\| < \varepsilon \text { for all } x \in S \text { and } k \geq N$$$$ so $$f _ { k } \circ g _ { k }$$ converge uniformly to $$f \circ g$$.

Is it that simple, or am I missing something?

• It is not that simple: math.stackexchange.com/questions/1680370/… – Matt A Pelto Dec 4 '18 at 0:35
• How do I show Lipshitz then? Your link doesn't make sense... – Andrew Hardy Dec 4 '18 at 0:40
• See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read. – Matt A Pelto Dec 4 '18 at 0:51

Suppose $$g_n: [0,1] \longrightarrow [0,1]$$, $$\,f_n:[0,1] \longrightarrow \mathbb R$$ $$(n=1,2,\ldots)$$ are two sequences of continuous functions that converge uniformly to $$g$$ and $$f$$, respectively.
Let $$\varepsilon>0$$ be given. By the uniform limit theorem and by the Heine-Cantor theorem, $$f$$ is uniformly continuous on $$[0,1]$$. Thus there is $$\delta>0$$ so that $$|\,f(u)-f(v)|<\frac12 \varepsilon$$ whenever $$|u-v|<\delta$$ and $$u,v \in [0,1]$$. Since $$g_n \to g$$ uniformly, there is a positive integer $$N_1$$ so that for all $$x \in [0,1]$$ we have $$|g_n(x)-g(x)|<\delta \text{ whenever } n \geq N_1.$$ Since $$f_n \to f$$ uniformly, there is a positive integer $$N_2$$ so that for all $$x \in [0,1]$$ we have $$|\,f_n(x)-f(x)|<\frac{\varepsilon}{2} \text{ whenever} n \geq N_2.$$ Set $$N=\max\{N_1,N_2\}$$. So if $$n \geq N$$ and $$x \in [0,1]$$, then we have $$$$\begin{split} | \,f_n(g_n(x)) - f(g(x))| &= |\,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\\ & \leq |\,f_n(g_n(x)) - f(g_n(x))|+|\,f(g_n(x)) - f(g(x))| \\ & <\frac{\varepsilon}{2} +\frac{\varepsilon}{2}\\ & = \varepsilon. \end{split}$$$$
Therefore, $$f_n \circ g_n \to f \circ g$$ uniformly on $$[0,1]$$ (definition of uniform convergence).
• Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/… – Matt A Pelto Dec 4 '18 at 2:06
• For the record, having every function in the sequence of functions $\{g_n\}_{n=1}^\infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous. – Matt A Pelto Dec 4 '18 at 3:06