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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.

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2 Answers 2

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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

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  • $\begingroup$ 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 $|.|$? $\endgroup$ Commented Mar 7, 2019 at 18:16
  • $\begingroup$ 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. $\endgroup$
    – user605486
    Commented Mar 7, 2019 at 18:25
  • $\begingroup$ 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 $\endgroup$
    – user605486
    Commented Mar 7, 2019 at 18:26
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    $\begingroup$ A quick note: you did not show $f_n\to f$ in $\|\cdot\|_{Lip_0}$. $\endgroup$
    – Jason
    Commented May 7, 2020 at 18:38
  • $\begingroup$ @Jason Yes, that has not been shown. How can you show it? $\endgroup$ Commented Apr 13, 2022 at 22:31
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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.

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