# Space of lipschitz functions form a Banach space

Let $$X$$ be a Banach space. Show that $$L=\{f:X\to\mathbb{R}: f \mbox{ is Lipschitz}, f(0) = 0\}$$ with the norm

$$||f||_{Lip_0} = \sup\left\{\frac{|f(x)-f(y)|}{||x-y||}, x\neq y\in X\right\}$$

is a Banach space.

I've found Banach space of p-Lipschitz functions but I did not understand the proof given.

I have a few questions first. Which norm is $$||x-y||$$?

So I need to prove that every Cauchy sequence in $$L$$ converges to an element of $$L$$, right?

In other words, $$\forall \epsilon>0$$ there exists $$n_0$$ such that $$m,n>n_0\implies ||f_m-f_n||_{Lip_0}<\epsilon$$

$$||f_m-f_n||_{Lip_0} = \sup\left\{\frac{|(f_m-f_n)(x)-(f_m-f_n)(y)|}{||x-y||}, x\neq y\in X\right\} = \sup\left\{\frac{|f_m(x)-f_m(y)|}{||x-y||}+\frac{f_n(y)-f_n(x)}{||x-y||}, x\neq y\in X\right\}$$

both $$f_m$$ and $$f_n$$ are Lipschitz so they're continous, which means something I don't know what.

It would be a good exercise to show that $$||f||_{Lip_0}$$ is indeed a norm. Otherwise, here are the general steps:

Assume $$f_n$$ is Cauchy.

1) First step is to show pointwise convergence so we can define a limit. $$|f_m(x) - f_n(x)| = |(f_m-f_n)(x) - (f_m-f_n)(0)| \leq ||f_m-f_n||_{Lip_0} ||x||.$$ This shows that $$f_m(x)$$ is Cauchy and therefore has a limit, i.e., f(x).

2) It remains to show that $$f$$ is in $$L$$. It is obvious that $$f(0) = 0$$. It therefore only remains to show that $$f$$ as defined above is Lipschitz. We have:

$$|f_n(x) - f_n(y)| \leq ||f_n||_{Lip_0} ||x-y||.$$

But $$f_n$$ is Cauchy, hence bounded by say $$M$$. Hence, $$|f_n(x) - f_n(y)| \leq M ||x-y||.$$ At the limit, we have: $$|f(x) - f(y)| \leq M ||x-y||.$$ Hence $$f$$ is Lipschitz, and the result is proven

• I have many questions in 1). Did you use that sum of lipschitz is lipschitz? Why $||x||$? Which norm is that and which norm is $|.|$? Commented Mar 7, 2019 at 18:16
• In $X$, the norm is $\|.\|$. We do not need to know which norm it is, or how it is defined. Now, the function $f$ maps in $\mathbb{R}$, so here we talking about absolut value. Commented Mar 7, 2019 at 18:25
• Here I just showed completeness. Now, you would have to show that the space is question is a vector space, and this is where you would use that the sum of tow lipschitz functions is lipschitz Commented Mar 7, 2019 at 18:26
• A quick note: you did not show $f_n\to f$ in $\|\cdot\|_{Lip_0}$. Commented May 7, 2020 at 18:38
• @Jason Yes, that has not been shown. How can you show it? Commented Apr 13, 2022 at 22:31

To complete Pebeto's answer, one needs to show that $$\|f_n-f\|_{Lip_0}\to0$$ as $$n\to\infty$$. Let $$\varepsilon>0$$. Then there exists $$N$$ such that $$\|f_n-f_m\|_{Lip_0}<\frac\varepsilon2$$ for all $$m,n\ge N$$. Fix $$n\ge N$$. We will be done if we can show $$\|f_n-f\|_{Lip_0}\le\varepsilon$$.

Fix $$x,y\in X$$ with $$x\neq y$$. There exists $$M_{x,y}$$ such that $$\frac{|(f_m-f)(x)-(f_m-f)(y)|}{\|x-y\|}<\frac\varepsilon2$$ for all $$m\ge M_{x,y}$$. Let $$m=\max\{N,M_{x,y}\}$$. Then

\begin{align*}\frac{|(f_n-f)(x)-(f_n-f)(y)|}{\|x-y\|} &\le \frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{\|x-y\|} + \frac{|(f_m-f)(x)-(f_m-f)(y)|}{\|x-y\|} \\ &\le \|f_n-f_m\|_{Lip_0}+\frac\varepsilon2 \\ &<\varepsilon. \end{align*} Taking suprema of both sides over $$x,y$$, we find $$\|f_n-f\|_{Lip_0}\le\varepsilon$$ for all $$n\ge N$$, completing the proof.