Is $(Lip((a,b),\lVert\cdot\rVert_{\infty})$ closed?

Question

Let $Lip((a,b))$ be the space of Lipschitz functions on $(a,b)$: it is obvious the inclusion $Lip((a,b))\subset \mathcal{C}^0([a,b])$, but I was wondering if maybe this subspace is closed with respct to the supremum norm.

I've studied that $B_{Lip((a,b))}$, that is the unit ball of $Lip((a,b))$ is compact in $\mathcal{C}^0([a,b])$, but my feeling is that the whole subspace is not closed.

My idea: we know that, on $(0,1)$, $\sqrt{x}$ is not Lipschitz, thus we should be able to construct a sequence $(f_h)_h$ of Lipschitz functions such that $$ \lim_{h\rightarrow \infty} \lVert f_h-f \rVert_{\infty}=0,$$ which in particolar yelds the non-closedness of $(Lip((a,b))$.

My question: How to create such a sequence, if it exists?

Any hint, help or answer would be much apppreciate, thanks in advance.

Answer 1

No. Take $f(x) = \sqrt{x}$ for $x \ge 0$ and let $f_n(x) = \min(nx,\sqrt{x})$. Then $f_n$ is Lipschitz but the limit is not. Note that the Lipschitz constant of $f_n$ is $n$.

If the $f_n$ have a uniform constant, say $L$, then since $\|f_n(x)-f_n(y)\| \le L \|x-y\|$ and $f_n \to f$ in the $\sup$ norm then we have $\|f(x)-f(y)\| \le L \|x-y\|$ by continuity.

Answer 2

The space of all Lipschitz functions on $[a,b]$ is actually dense in the continuous functions in $C^0$ with the supremum norm. This follows from Weierstrass's approximation theorem for instance.

The space of functions on $[a,b]$ with a given nonstrict uniform bound on the Lipschitz constant is closed in $C^0$ with the supremum norm. If you additionally impose a uniform bound on $f(a)$ then the space is compact in $C^0$ by the Arzela Ascoli theorem.