2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.