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

Question

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

Answer 1

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].

Answer 2

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 $C([0,1])$