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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 $fcirc g$?

What I've done so far:

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

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?

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 = 2epsilon.
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 $f circ g$.