If a sequence of unif. cont. Lipschitz functions $\{f_n\}$ converges unif. to an unif. cont. function f, then f is Lipschitz.

Question

Let $f: (X,d) \to (\Omega,\rho)$ be a uniformly continuous function such that $\{f_n\}$ is a sequence of uniformly continuous functions that converges uniformly to $f$, where each $f_n$ is Lipschitz with constant $M_n$.

Show that if $\sup(M_n) < \infty$, then the function $f$ is Lipschitz . Show that $f$ may fail to be Lipschitz when $\sup(M_n) = \infty$.

My approach:

We must show that f is Lipschitz, then

$\exists M > 0, \forall x,y \in X, \rho(f(x),f(y)) \leq M d(x,y)$

We have that each $f_n$ is Lipschitz, then $\rho(f_n(x),f_n(y)) \leq M_n d(x,y), \forall x,y \in X $

Also by the triangle inequality,

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f(y))$

Using the triangle inequality again

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f_n(y)) + \rho(f_n(y),f(y))$

Since $f_n \to f$ uniformly, we have that

$\rho(f(x),f_n(x)) \to 0$ and $\rho(f(x),f_n(x)) \to 0$ as $n \to \infty$.

then

$\displaystyle\lim_{n\to\infty}\rho(f(x),f(y)) \leq \displaystyle\lim_{n\to\infty}\rho(f(x),f_n(x)) + \displaystyle\lim_{n\to\infty}M_n d(x,y) + \displaystyle\lim_{n\to\infty}\rho(f_n(y),f(y))$

Hence,

$\rho(f(x),f(y)) \leq 0 + \displaystyle\lim_{n\to\infty}M_n d(x,y) + 0 \leq \sup M_n d(x,y)$

Taking $M = \sup M_n$, since $0 < M < \infty$, we have that f is Lipschitz.

Now if $\sup M_n = \infty$, how can I show that $f$ "may fail to be Lipschitz"?. I know that my constant $M$ will not satisfy the definition of a Lipschitz function because it must be a real number, but I do not see how to prove that it "may fail to be Lipschitz". Any help would be very appreciated. Thanks.

Answer 1

If $M := \sup_n M_n < +\infty$, then $$ \rho(f_n(x), f_n(y)) \leq M\, d(x,y) \qquad \forall x,y\in X, \quad \forall n\in\mathbb{N}. $$ Now, if the sequence $(f_n)$ converges pointwise to $f$, passing to the limit in the above relation gives $$ \rho(f(x), f(y)) \leq M\, d(x,y) \qquad \forall x,y\in X, $$ i.e. $f$ is Lipschitz continuous.

For the last part: if you consider the sequence $$ f_n(x) := \sqrt{x+\frac{1}{n}}, \qquad x\in [0,1], $$ then you see in a moment that each $f_n$ is of class $C^1([0,1])$, hence it is Lipschitz continuous in $[0,1]$ (with constant $M_n = \sqrt{n}/2$). On the other hand, the sequence $(f_n)$ converges uniformly in $[0,1]$ to the function $f(x) = \sqrt{x}$, which is not Lipschitz continuous.