Hi everyone I'd like to know if the following is correct, I'd appreciate your opinion and also any suggestion to improve my argument. Thanks in advance for your time.
If $(K,d)$ is a compact metric space and $u\in K$ show that for any finite $M$ and $\alpha\in (0,1]$, $\mathcal{F}=\{f\in \text{Lip}(\alpha,M): \lvert f(u)\rvert\ \le M \}$ is compact with respect to the $d_{sup}$, where $\text{Lip}(\alpha,M)$ is the set of all real valued function on $S$, s.t. $\lvert f(x)-f(y)\rvert\le Md(x,y)^{\alpha}$
Proof: $\mathcal{F}$ is uniformly equicontinuous. Given $\varepsilon>0$, let $\delta =[\varepsilon/{(M+1)]}^{1/\alpha}$. Then for $x,y\in K$, $d(x,y)<\delta$ and $f\in \mathcal{F}$ we thus have $\rvert f(x)-f(y)\lvert\le Md(x,y)^{\alpha}<\varepsilon$. Also $\mathcal{F}$ is uniformly bounded, since $\lvert f(u)\rvert\ \le M $ for any $u\in K$, hence $\sup_{u\in K}\lvert f(u)\rvert\le M$. By Arzela- Ascoli $\mathcal{F}$ is totally bounded by $d_{sup}$.
To conclude, we will show that $\mathcal{F}$ is complete. It sufficient to show that $\mathcal{F}$ is closed, since $\mathcal{F}\subset C_b(K)$ which is complete [$C_b(K)$ are the continuous bounded functions from $K$ to $\mathbb{R}$].
Suppose $f_n\to f$ for $f_n\in \mathcal{F}$. Given $\varepsilon>0$ for a sufficient large $n\in\mathbb{N}$, $d_{sup}(f,f_n)<\varepsilon$. Let $x\in K$, then $\lvert f(x)-f_n(x) \rvert \le d_{sup}(f,f_n)$, it follows that $\vert f(x)\rvert <\lvert f_i(x) \rvert+\varepsilon\le M+\varepsilon $, since $\varepsilon$ was arbitrary then $\lvert f(x) \rvert\le M $.
Also we have the following: \begin{align}\lvert f(x)-f(y) \rvert \le \lvert f(x)-f_n(x) \rvert+\lvert f_n(x)-f_n(y) \rvert +\lvert f_n(y)-f(y) \rvert \\< Md(x,y)^{\alpha}+2\varepsilon\end{align}
Thus $\lvert f(x)-f(y) \rvert \le Md(x,y)^{\alpha}$ and hence $ \mathcal{F}$ is closed as we claimed.
Is the above OK? this is the version of Arzela Ascoli that I use
Nobody could help me, please?