Show that a sequence of Lipschitz functions converges to a Lipschitz function.

Question

Let $(X,d),(Y,d')$ be metric spaces and $(f_n)_n$ a sequence of k-Lipschitz functions $X \to Y$ such that $f_n \to^u f$ (converges uniformly). Show that $f$ is k-Lipschitz as well.

My attempt:

Let $\epsilon > 0$. Choose $n_0$ such that $\forall n \geq n_0: \forall x \in X: d'(f_n(x),f(x)) < \epsilon.$

Let $x,y \in X$ be given.

Then, we have:

$$d'(f(x),f(y)) \leq d'(f(x), f_{n_0}(x)) + d'(f_{n_0}(x), f(y))$$ $$\leq d'(f(x),f_{n_0}(x)) + d'(f_{n_0}(x),f_{n_0}(y)) + d'(f_{n_0}(y),f(y)) $$ $$< \epsilon + kd(x,y) + \epsilon = 2\epsilon + kd(x,y) $$

and because $\epsilon$ can be made as small as possible, the result follows. Is this correct?

Answer 1

Put $\epsilon>0$ instead of $1$. Then \begin{align*} d'(f(x),f(y))\leq d'(f_{n_{0}}(x),f(x))+d'(f_{n_{0}}(y),f(y))+d'(f_{n_{0}}(x),f_{n_{0}}(y))\leq 2\epsilon +kd(x,y), \end{align*} since this is true for all $\epsilon>0$, so $d'(f(x),f(y))\leq kd(x,y)$.

Answer 2

We don't need uniform convergence; pointwise convergence is enough. We just need this little lemma: In any metric space $(Z,\rho),$ if $x_n \to x, y_n\to y,$ then $\rho(x_n,y_n) \to \rho(x,y).$ You prove this by pushing the triangle inequality around.

So in the problem at hand, let $x,y \in X.$ We have, from pointwise convergence and the result above,

$$d'(f(x),f(y)) = \lim d'(f_n(x),f_n(y)).$$

Each term on the right is bounded above by $kd(x,y)$ by hypothesis, hence so is the limit, and we're done.