Convergence of composition of Lipschitz maps Let $f_n, g_n: \mathbb{R}^m \rightarrow \mathbb{R}^m$, $n = 1, 2, \ldots$, be two sequences of globally Lipschitz continuous maps.
Assume that $f_n \rightarrow f$ and $g_n \rightarrow g$ (convergence in the $\sup$ norm), where $f, g: \mathbb{R}^m \rightarrow \mathbb{R}^m$ are Lipschitz continuous maps as well. 
Also assume that $g_n, g$ are bounded functions.
Prove of disprove that $ f_n \circ g_n \rightarrow f \circ g $.
Comments: Clearly, $f \circ g_n \rightarrow f \circ g$ and $f_n \circ g \rightarrow f \circ g$. I think I can show that $ f_n \circ g_n \rightarrow f \circ g $ if $f_n, f, g_n, g$ are bounded, but I am not sure in the case of only $g_n, g$ being bounded.
 A: Suppose $X,Y,Z$ are metric spaces. Suppose $g_n: X\to Y$ converges uniformly to $g: X\to Y,$ and $f_n: Y\to Z$ converges uniformly to $f: Y\to Z.$ If $f$ is uniformly continuous on $Y,$ then $f_n\circ g_n \to f\circ g$ uniformly on $X.$
Proof:
$$d(f_n\circ g_n,f\circ g) \le d(f_n\circ g_n,f\circ g_n) + d(f\circ g_n,f\circ g).$$
The first summand on the right $\to 0$ uniformly by the uniform convergence of $f_n$ to $f.$ The second summand $\to 0$ uniformly by the uniform convergence of $g_n$ to $g$ and the uniform continuity of $f.$
This gives the result in the question, because we have the uniform convergence stipulated, and the function $f$ in the problem is given to be Lipschitz, hence is uniformly continuous.
A: HINT:
For all $x\in\mathbb{R}^m$ we have \begin{gather}
|(f_n\circ g_n)(x)-(f\circ g)(x)|\leq|(f_n\circ g_n)(x)-(f\circ g_n)(x)|+|(f\circ g_n)(x)-(f\circ g)(x)|. \end{gather} 
Now use Lipschitz continuity of $f$ for the second term and then the uniform convergence of both series.
A: Let $\varepsilon>0$ be given. Since $f_n \to f$ uniformly, there exists a positive integer $N_1$ so that for all $\mathbf{x} \in \mathbb{R}^m$ we have
\begin{equation} || f_n(\mathbf{x}) - f(\mathbf{x})||< \frac{\varepsilon}{2} \text{ whenever } n \geq N_1.
\end{equation}
Let $l_f$ be a Lipschitz constant for $f$, and set $\varepsilon' = \min \{ \frac{\varepsilon}{2}, \frac{\varepsilon}{2l_f} \}$. Since $g_n \to g$ uniformly, there exists a positive integer $N_2$ so that for all $\mathbf{x} \in \mathbb{R}^m$ we have
\begin{equation} || g_n(\mathbf{x}) - g(\mathbf{x})||< \varepsilon' \text{ whenever } n \geq N_2.
\end{equation}
Set $N = \max \{N_1, N_2 \}$. So if $n \geq N$, then
\begin{equation} \begin{split} || f_n(g_n(\mathbf{x})) - f(g(\mathbf{x}))|| &= ||f_n(g_n(\mathbf{x})) - f(g_n(\mathbf{x}))+f(g_n(\mathbf{x})) - f(g(\mathbf{x})) ||\\
& \leq ||f_n(g_n(\mathbf{x})) - f(g_n(\mathbf{x}))||+||f(g_n(\mathbf{x})) - f(g(\mathbf{x}))||  \\
& <\frac{\varepsilon}{2} + l_f||g_n(\mathbf{x}) - g(\mathbf{x})|| \\
& \leq \varepsilon.
\end{split}\end{equation}
