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As the title says I am trying to show that $Lip_\alpha$ is not closed in $C[0,1]$. $Lip_\alpha$ is the class of functions on [0,1] that belong to $Lip_\alpha([0,1];K)$ where $f \in Lip_\alpha([0,1];K)$ if

$$|f(x)-f(y)| \leq K|x-y|^\alpha \text{ for all } x,y \in [0,1]$$

However, I do not understand what this statement implies. Closed means it contains all of its limit points.

So, do I need to find $f \in Lip_\alpha$ and $f \rightarrow g$ uniformly and $g \in C[0,1]$ but $g \not \in Lip_\alpha$?

I think I'm not sure about the definition of a space being closed in another space

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    $\begingroup$ It is better to write what is $Lip_\alpha$ here $\endgroup$ – Chinnapparaj R Apr 20 at 3:15
  • $\begingroup$ @ChinnapparajR hey thanks for the heads up. just did! $\endgroup$ – Kaan Yolsever Apr 20 at 3:20
  • $\begingroup$ Is your class of functions restricted to a fixed $K$ or the union over all $K$? $\endgroup$ – copper.hat Apr 20 at 4:24
  • $\begingroup$ @Matematleta The point of this question is that if $K$ can vary then the limit function need not be LIpschitz. In the definition of the space $Lip_{\alpha}$ the constant $K$ is variable. $\endgroup$ – Kabo Murphy Apr 20 at 5:38
  • $\begingroup$ You should tell us what $\alpha$ is. $\endgroup$ – zhw. Apr 20 at 22:09
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Let $f_n(x)=\int_0^{x} \min \{n, \frac 1 {\sqrt t }\} dt$. Then each $f_n$ is Lipschitz because $f_n'$ is bounded. $f_n(x) \to f(x)=\int_0^{x} \frac 1 {\sqrt t } dt=2\sqrt x$. Note that $f$ is not Lipschitz. [Of course the convergence of this sequence is uniform].

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  • $\begingroup$ Hi, It is not obvious to me how $f_{n}$ is bounded can you please elaborate? $\endgroup$ – Matt Apr 22 at 4:18
  • $\begingroup$ By FTC, we have that $f'_{n}(x)$ $=$ $min\{n,\frac{1}{\sqrt{x}}\}$. Hence, $f'_{n}(x)$ is bounded by $n$ or $\frac{1}{\sqrt{x}}$ depending on which is the larger value. Is this correct? $\endgroup$ – Matt Apr 22 at 4:29
  • $\begingroup$ @Matthieu $0 \leq f_n'(x) \leq n$. $\endgroup$ – Kabo Murphy Apr 22 at 5:07
  • $\begingroup$ by that logic, can you not do the same but with $\frac{1}{\sqrt{x}}$? i.e $0$ $\leq$ $f'_{n}(x)$ $\leq$ $\frac{1}{\sqrt{x}}$ $\endgroup$ – Matt Apr 22 at 5:14
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If $0\le \alpha\le 1$, we can use the Weierstrass theorem to prove that the claim is false:

Since $|y^n-x^n|=\sum^n_{k=0}x^ky^{n-k}\cdot|y-x|^{1-\alpha}\cdot |y-x|^{\alpha},\ $ it is easy to see that $P([0,1])\subseteq \text{Lip}_\alpha([0,1];K) $.

But then, if $\text{Lip}_\alpha([0,1];K)$ is closed in $C([0,1])$, we have $\text{Lip}_\alpha([0,1];K)\supseteq C([0,1])$ which is absurd.

On the other hand, if $\alpha>1$, then $f'=0$ on $[0,1]$ so $f$ is constant, and so in this case, $\text{Lip}_\alpha([0,1];K)$ is closed in $C([0,1])$

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