# $Lip_\alpha$ is not closed in $C[0,1]$

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

• It is better to write what is $Lip_\alpha$ here – Chinnapparaj R Apr 20 at 3:15
• @ChinnapparajR hey thanks for the heads up. just did! – Kaan Yolsever Apr 20 at 3:20
• Is your class of functions restricted to a fixed $K$ or the union over all $K$? – copper.hat Apr 20 at 4:24
• @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. – Kabo Murphy Apr 20 at 5:38
• You should tell us what $\alpha$ is. – zhw. Apr 20 at 22:09

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].

• Hi, It is not obvious to me how $f_{n}$ is bounded can you please elaborate? – Matt Apr 22 at 4:18
• 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? – Matt Apr 22 at 4:29
• @Matthieu $0 \leq f_n'(x) \leq n$. – Kabo Murphy Apr 22 at 5:07
• by that logic, can you not do the same but with $\frac{1}{\sqrt{x}}$? i.e $0$ $\leq$ $f'_{n}(x)$ $\leq$ $\frac{1}{\sqrt{x}}$ – Matt Apr 22 at 5:14

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 $$C([0,1])$$