Why the space of Lipschitz functions from $[0,1]$ to $\mathbb{R}$ with the uniform norm

Question

is not a Banach space ?

We have to find a Cauchy sequence of Lipschitz functions $(f_n)_{n>0}$ such that this sequence does not converge to a Lipschitz function $f$ (for the norm $\vert \vert g \vert \vert_{\infty} = \sup_{t\in[0,1]} \vert g(t) \vert$ with $g$ a Lipschitz function).

How to build such a sequence ?

Thanks in advance !

Answer 1

Basic idea: pick a continuous function that's not Lipschitz, uniformly approximate it by Lipschitz functions. For instance if the function is only non-Lipschitz because of bad behavior near one point, just flatten it out near that point and leave it be everywhere else.

With that in mind, you can look at $f_n(x)=\left ( \max \{ 1/n,x \} \right )^{1/2}$, which converges uniformly to $f(x)=x^{1/2}$ which is not Lipschitz.

Answer 2

Consider $f_n(x)=\begin{cases} \frac{1}{x}\sin(x^2) \text{ if } \frac{1}{\sqrt{n\pi}}\leq x \leq 1, \\ 0 \text{ if } 0\leq x<\frac{1}{\sqrt{n\pi}}. \end{cases}$

These are Lipschitz and you can check that they converge uniformly to $f(x)=\begin{cases} \frac{1}{x}\sin(x^2) \text{ if } 0< x \leq 1, \\ 0 \text{ if } x=0. \end{cases}$, but $f$ is not Lipschitz since $f'(x) \rightarrow \infty$ as $x\rightarrow 0$.