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.
Best Answer
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.