# Composition of two uniformly convergent sequences of functions is uniformly convergent?

I am trying to prove or provide a counter-example for the following:

Let $$f_k$$ and $$g_k$$ be sequences of continuous functions on $$[0,1]\to[0,1]$$ converging uniformly to $$f:[0,1]\to \mathbb{R}$$ and $$g:[0,1]\to \mathbb{R}$$ respectively. Does $$f_k \circ g_k$$ coverge uniformly to $$f\circ g$$?

What I've done so far:

I know I need to prove that $$\forall \epsilon>0, \exists N\in \mathbb{N}$$ such that $$||f_k(g_k(x)) - f(g(x))|| < \epsilon$$ for all $$x \in [0,1]$$ and $$k\geq N$$.

At first, I thought I can prove this easily since it follows trivially from the definition of $$f_k$$ uniformly converging to $$f$$. However, I noticed that is only true for all $$x \in [0,1]$$ and $$g$$ maps onto all of $$\mathbb{R}$$, not just $$[0,1]$$. So does that mean it's not necessarily true? Can anyone provide a counter-example?

• Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.) Commented Dec 2, 2018 at 5:52

Let $$\|h\|_\infty = \sup_{x \in [0,1]} |h(x)|$$.

Note that $$f$$ is uniformly continuous since $$[0,1]$$ is compact. Hence since $$\|g-g_k\|_\infty \to 0$$, we see that $$\|f \circ g-f \circ g_k\|_\infty \to 0$$.

Then $$\begin{eqnarray} |f \circ g(x)-f_k \circ g_k (x)| &\le& |f \circ g(x)-f \circ g_k (x)| + |f \circ g_k(x)-f_k \circ g_k (x)| \\ &\le & \|f \circ g-f \circ g_k\|_\infty + \|f-f_k\|_\infty \end{eqnarray}$$ Hence $$\|f \circ g-f_k \circ g_k\|_\infty \to 0$$.

For each $$x \in [0,1]$$, $$f_n(x) \to f(x)$$. Since $$f_n(x) \in [0,1]$$, $$f(x) \in [0,1]$$. Idem for $$g$$.

Let $$\epsilon > 0$$.

\begin{align} & \exists N_1 \in \Bbb{N}: \forall n \ge N_1, \forall y \in [0,1], |f_n(y)-f(y)| < \epsilon \tag1 \label1 \\ & \exists \delta > 0 : \forall |y - y'| \le \delta, |f(y) - f(y')| < \epsilon \tag2 \label2 \\ & \exists N_2 \in \Bbb{N}: \forall k \ge N_2, \forall x \in [0,1], |g_k(x)-g(x)| < \delta \tag3 \label3 \end{align}

\eqref{1} and \eqref{3} are the definition of uniform continuity; \eqref{2} is due to the facts that the uniform limit $$f$$ of a sequence of continuous functions $$(f_n)_n$$ is continuous, and that $$f$$ is uniformly continuous on closed and bounded interval $$[0,1]$$.

Put \eqref{1}-\eqref{3} together. Take $$N = \max\{N_1,N_2\}$$. For all $$n \ge N$$,

\begin{align} & \quad |f_n(g_n(x)) - f(g(x))| \\ &\le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \\ &\le \epsilon + \epsilon = 2\epsilon. \end{align}

In the last inequality, we applied \eqref{1} with $$y = g_n(x)$$ in the first term, and \eqref{3} with $$k = n$$ composed with \eqref{2} with $$y = g_n(x)$$ and $$y' = g(x)$$ in the second term.

Hence $$f_n \circ g_n$$ converges uniformly to $$f \circ g$$.