# Show that collection of Lipschitz functions with Lipschitz norm is a Banach space

Question: Let $$\mathcal{L}$$ be the normed space of all Lipschitz functions on a Banach space $$X$$ that are equal to $$0$$ at the origin, under the norm $$\|f\| = \sup\left\{ \frac{|f(x)-f(y)|}{\|x-y\|}:x,y\in X \right\}.$$ Show that $$\mathcal{L}$$ is a Banach space.

My attempt:

Let $$(f_n)_{n=1}^\infty$$ be a Cauchy sequence in $$(\mathcal{L},\|\cdot\|).$$ Define $$f:X\to \mathbb{R}$$ by $$f(x) = \lim_{n\to\infty} f_n(x).$$ We claim that $$f$$ is well-defined, that is, the limit exists. Fix $$x\in X\setminus \{0\}$$ and $$\epsilon>0.$$ Since $$(f_n)_{n=1}^\infty$$ is Cauchy, there exists $$N\in \mathbb{N}$$ such that for any $$m,n\geq N,$$ $$\|f_m-f_n\| <\frac{\epsilon}{\|x\|}.$$ Note that $$\|f_m-f_n\| = \sup_{x,y\in X} \frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{\|x-y\|} \geq \frac{|f_m(x) - f_n(x)|}{\|x\|}.$$ So we have $$|f_m(x) - f_n(x)| < \epsilon.$$ Therefore, $$(f_n(x))_{n=1}^\infty$$ is Cauchy in $$\mathbb{R}.$$ Since $$\mathbb{R}$$ is complete, so $$\lim_{n\to\infty} f_n(x)$$ exists, that is, $$f$$ is well-defined.

We claim that $$f\in \mathcal{L}.$$ By construction, $$f(0) = 0.$$ Since $$(f_n)_{n=1}^\infty$$ is Cauchy in $$(\mathcal{L},\|\cdot\|),$$ therefore it is bounded, that is, there exists $$B>0$$ such that $$\|f_n\|\leq B$$ for all $$n\geq 1.$$ Since $$f = f_N + (f-f_N),$$ so for any $$x,y\in X,$$ \begin{align*} |f(x)-f(y)| & \leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \\ & \leq \|f_N\|\| x-y\| + \|f-f_N\| \|x-y\| \\ & \leq \|f_N\|\| x-y\| + \|x-y\| \\ & = (\|f_N\|+1)\|x-y\|. \end{align*} Therefore, $$\|f\|$$ is finite and hence $$f\in \mathcal{L}.$$

Lastly, we wish to prove that $$(f_n)$$ converges to $$f$$ in $$(\mathcal{L},\|\cdot\|).$$ Indeed, for every $$\epsilon>0,$$ by Cauchyness of $$(f_n),$$ there exists $$N\in\mathbb{N}$$ such that for any $$m,n\geq N,$$ $$\|f_n-f_m\|<\frac{\epsilon}{2}.$$ Then $$\|f-f_n\| = \|\lim_{m\to\infty} f_m - f_n\| = \lim_{m\to\infty} \|f_m-f_n\| < \epsilon.$$ We conclude that $$\mathcal{L}$$ is complete.

Is my proof above correct?

• Why define $f(0) = 0$ instead of $f(0) = \lim_{n \to \infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz. Feb 3, 2019 at 5:50
• @TheoBendit Yes, you are right. Edited my proof. Feb 3, 2019 at 5:52
• Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place! Feb 3, 2019 at 5:54
• @TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion. Feb 3, 2019 at 5:59

Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $$f \in \mathcal{L}$$ before you've proved it and confusing the pointwise limit with the limit in $$\mathcal{L}$$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $$f \in \mathcal{L}$$, on the second line you write down $$\|f-f_N\|$$ and then bound this by $$1$$. Doing this already requires you to assume $$f-f_N \in \mathcal{L}$$ and hence $$f \in \mathcal{L}$$ and also something like $$f_n \to f$$ in $$\mathcal{L}$$.
Instead, write \begin{align*} |f(x) - f(y)| &= |\lim_{n \to \infty} (f_n(x) - f_n(y))| \\&= \lim_{n \to \infty} |f_n(x) - f_n(y)| \\& \leq \lim_{n \to \infty} \|f_n\| |x-y| \\& \leq C|x-y| \end{align*} where in the last line we used that $$\|f_n\|$$ is a bounded sequence. Note that in the second line, I only use that $$f_n \to f$$ pointwise and not in $$\mathcal{L}$$.
This is important because when trying to prove that $$f_n \to f$$, you take a pointwise limit and take it outside of the $$\mathcal{L}$$-norm. This doesn't work since you can only do that once you know that the limit converges in $$\mathcal{L}$$ and not just pointwise. Instead, work again with a pointwise limit by writing \begin{align*} |f(x) - f(y) - f_n(x) + f_n(y)| &= \lim_{m \to \infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\\ & \leq \lim_{m \to \infty} \|f_m - f_n\| |x-y| \end{align*} Now since $$(f_n)$$ is Cauchy in $$\mathcal{L}$$, there is $$N$$ independent of $$x,y$$ such that for $$n,m \geq N$$ $$\|f_n - f_m\| \leq \varepsilon$$. So we get $$|f(x) - f(y) - f_n(x) + f_n(y)| \leq \varepsilon |x-y|$$ which implies for $$n \geq N$$, $$\|f-f_n\| \leq \varepsilon$$ so $$f_n \to f$$ in $$\mathcal{L}$$.
• Maybe some limits have to be replaced by limsup? For example $\lim\|f_n(x)-f_n(y)\|\leq\limsup\|f_n\||x-y|$ since we don't know whether the limit of the norms exists... Jan 27, 2021 at 19:35
• @LorenzoCecchi $f_n$ is a Cauchy sequence for $\|\cdot\|$ and so by the reverse triangle inequality $\|f_n\|$ is Cauchy in $\mathbb{R}$ and hence converges. Jan 27, 2021 at 20:19