2

$\DeclareMathOperator{\Lip}{Lip}$ I already proved that if $0 < \alpha < \beta \leq 1$, then $\Lip_{\beta}[0,1] \subset \Lip_{\alpha}[0,1]$. Recall that we say that $f \in \Lip_{\alpha}[0,1]$ if there is a $K \geq 0$ such that for every $x,y \in [0,1]$ we have that $|f(x)-f(y)| \leq K|x-y|^{\alpha}$. The infimum of all values $K$ satisfying this last inequality is denoted as $\theta_{\alpha}(f)$. This allows us to define the following norm in $\Lip_{\alpha}[0,1]$:

$$\|f\|_{\alpha}= \theta_{\alpha}(f)+ \sup \lbrace |f(x)|:x\in [0,1] \rbrace.$$

So we have a metric for $\Lip_{\alpha}[0,1]$, as $$d_{\alpha}(f,g):=\|f-g\|_{\alpha}= \theta_{\alpha}(f-g)+ \sup \lbrace |f(x)-g(x)|:x\in [0,1] \rbrace$$ for every $f,g \in \Lip_{\alpha}[0,1]$. Therefore, we have a topology of open balls with this metric. As the title says, I need to prove that $\Lip_{\beta}[0,1] \subset \Lip_{\alpha}[0,1]$ is a closed subset with the $\Lip_{\alpha}[0,1]$ topology of course. I already proved the contention $\Lip_{\beta}[0,1] \subset \Lip_{\alpha}[0,1]$ but I'm not sure how to prove this subset is closed. I have been thinking in taking a sequence of functions $\lbrace f_{n} \rbrace \subset \Lip_{\beta}[0,1]$ such that $f_{n} \to f$ pointwise, then I need to prove $f \in \Lip_{\beta}[0,1]$ or a counterexample showing that $f \notin \Lip_{\beta}[0,1]$. But I can prove neither of these.

Bernard
  • 179,256
Sok
  • 582

1 Answers1

1

I doubt this to be true. Here is an example with $\alpha=0$ (i.e., continuous functions) and $\beta=1$: $$ f_n(x):= \max(\frac1n, \sqrt x) $$ which converges uniformly to $f(x)=\sqrt x$.

Maybe one can construct a similar example with $\alpha>0$?

daw
  • 54,637
  • 2
  • 44
  • 85
  • Im run out of ideas for an example with $\alpha>0$? Maybe this work, but its for another metric https://math.stackexchange.com/questions/3091638/let-0-alpha-beta-leq-1-prove-lip-betaa-b-subset-lip-alphaa-b?rq=1 @daw – Sok May 21 '21 at 18:21