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Let $X$ be the vector space of Lipschitz continous functions, $[0, 1] \rightarrow \mathbb{R}$. For $x\in X$ set $$\Vert x \Vert_{Lip}=\vert x(0)\vert+sup_{s\neq t}\left\vert \frac{x(s)-x(t)}{s-t}\right\vert.$$

I need to prove:

  1. $\Vert x \Vert_{\infty}\leq\Vert x \Vert_{Lip}$ for $x\in X.$

  2. $\left ( X,\Vert.\Vert_{Lip} \right )$ is a Banach space.

I have tried to make an estimation of $\Vert x \Vert_{\infty}$ and compare it with $\Vert x \Vert_{Lip}$ but i could't reach that far. As to the Point 2. i know that you start with a Cauchy sequence and need to prove that it converges in $X.$ Unfortunately i couldnt come further either.

I will appreciate any comment or help. Thanks.

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marked as duplicate by user99914, mechanodroid, Shuhao Cao, Mark Bennet, Trevor Gunn Nov 5 '17 at 15:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know that the space of continuous function $C[0,1]$ is a Banach space with $\|x\|_\infty$ as the norm? $\endgroup$ – user99914 Nov 3 '17 at 5:07
  • $\begingroup$ Yes, i know and still it did not help me. I need to prove the space $X$ is Banach in respect to some other norm. $\endgroup$ – user249018 Nov 3 '17 at 5:42
  • $\begingroup$ It helps. The first inequality implies that if $x_n$ is a Cauchy sequence in $\|\cdot\|_{Lip}$, then it is also a Cauchy sequence with respect to the norm $\|\cdot\|_\infty$. $\endgroup$ – user99914 Nov 3 '17 at 6:27
  • $\begingroup$ So then if it is Cauchy sequence with respect to the sup norm it will imply that the limit is in $X$. Is it right ? Can you give me a hint how to prove the first inequality. As for your assertion of convergence, does it mean that the two norms are equivalent and thus convergence with respect to one norme will imply convergence with respect to the other norm ? $\endgroup$ – user249018 Nov 3 '17 at 7:02
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Take any $f \in X$, Lipschitz continuity implies that for $c= \sup_{s \neq t} \lvert \frac{f(s)-f(t)}{s-t}\rvert < +\infty$ we have $$ \lvert f(x) -f(y) \rvert \leq c \lvert x- y \rvert \leq c $$ for all $x,y \in [0,1]$. Now notice that $$ \| f \|_\infty = \sup_{x \in [0,1]} \lvert f(x)\rvert \leq \sup_{x \in [0,1]} \lvert f(x)-f(0) \rvert + \lvert f(0)\rvert. $$ Then clearly $\|f\|_\infty \leq c+\|f(0)\| = \|f \|_{\text{Lip}}$.

To prove $(X, \|\cdot \|_{\text{Lip}})$ is complete, take a Cauchy sequence $(f_n)_n$ in $(X, \|\cdot \|_{\text{Lip}})$. By the inequality we just proved, we have that $(f_n)_n$ is also a Cauchy sequence in $(C([0,1]), \|\cdot \|_\infty)$ which is a Banach space. So there's an $f \in C([0,1])$ such that $f_n$ uniformly converges to $f$. Now all you have to do is prove that $f$ is also Lipschitz continuous and that $(f_n)_n$ converges to $f$ in the Lipschitz norm.

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