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.


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

Is it that simple, or am I missing something?

  • $\begingroup$ It is not that simple: math.stackexchange.com/questions/1680370/… $\endgroup$ – Matt A Pelto Dec 4 '18 at 0:35
  • $\begingroup$ How do I show Lipshitz then? Your link doesn't make sense... $\endgroup$ – Andrew Hardy Dec 4 '18 at 0:40
  • $\begingroup$ 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. $\endgroup$ – 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{equation} \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}\end{equation}

Therefore, $f_n \circ g_n \to f \circ g$ uniformly on $[0,1]$ (definition of uniform convergence).

  • 1
    $\begingroup$ OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge? $\endgroup$ – Andrew Hardy Dec 4 '18 at 1:57
  • $\begingroup$ 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/… $\endgroup$ – Matt A Pelto Dec 4 '18 at 2:06
  • $\begingroup$ 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. $\endgroup$ – Matt A Pelto Dec 4 '18 at 3:06

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