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Let $X$ be a compact metric space, and $F$ the space of all lipschitz functions $X \to \mathbf{C}$. Let $|f|_L$ be the least Lipschitz constant. We endow $F$ with the norm $||f|| = |f|_L + |f|_\infty$. I want to show that $F$ is a Banach space. Ascoli's theorem implies that any Cauchy sequence for this norm has a limit for the $| \cdot |_\infty$ norm. What about the $|\cdot|_L$ norm? Obviously a function that is small for this norm may have a big Lipschitz constant so I fail to conclude. Any clue?

Thanks.

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Fix $\varepsilon>0$. We can find $N(\varepsilon)$ such that for all $n,m\geq N(\varepsilon)$ and all $x,y\in X$ with $x\neq y$: $\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{d(x,y)}\leq \varepsilon$. Since $f_n$ converges to $f$ pointwise, we have, taking the limit $m\to\infty$ for $x,y$ fixed that \begin{equation} (*) \quad \forall n\geq N(\varepsilon),\forall x,y\in X,x\neq y\quad\frac{|(f_n-f)(x)-(f_n-f)(y)|}{d(x,y)}\leq \varepsilon. \end{equation} In particular for $n=N(\varepsilon)$ we have $\frac{|f(x)-f(y)|}{d(x,y)}\leq \varepsilon+\frac{|f_{N(\varepsilon)}(x)-f_{N(\varepsilon)}(y)|}{d(x,y)}$, which shows that $f$ is Lipschitz. Now we have to show that $|f-f_n|_L$ converges to $0$. But this is a consequence of $(*)$, since we only have to take the supremum over these $x$ and $y$.

Note that we don't need to use Ascoli's theorem to show that $\{f_n\}$ has a limit (just use the same proof when you show that the set of the continuous real functions on a compact space for the uniform norm is a Banach space).

Here compactness is needed to be sure that each Lipschitz function is bounded, since such a function is continuous.

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