A continuous fucntion $f: \Omega\rightarrow \mathbb{C}$, with $\Omega$ subset of $\mathbb{R}^n$ or $\mathbb{C}^n$, is a $\alpha$-Hölder continuous with $0< \alpha\leq 1$, if \begin{equation} [f]_{\alpha}:=\sup_{x,y\in \Omega, x\ne y} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}<\infty. \end{equation}
Here is my problem. If $\{f_n\}$ is a sequence of $\alpha$-Hölder continuous function on a compact subset $\Omega$,and $f_n$ converges uniformly to $f$ on $\Omega$, that is, $\lim_n\|f_n-f\|_{\infty}=0$. If $f$ is also $\alpha$-Hölder continuous, is it true that $\lim_n[f_n-f]_{\alpha}=0$?
Here is my analysis. According to the definition, we have \begin{align*} [f_n-f]_{\alpha}&=\sup_{x,y\in \Omega, x\ne y}\frac{|f_n(x)-f(x)-f_n(y)+f(y)|}{|x-y|^{\alpha}}\\ &=\sup_{x,y\in \Omega, x\ne y}\frac{|f_n(x)-f(x)|+|f_n(y)-f(y)|}{|x-y|^{\alpha}}\\ &\leq \sup_{x,y\in \Omega, x\ne y}\frac{2\|f_n-f\|_{\infty}}{|x-y|^{\alpha}}\rightarrow 0. \end{align*}
It seems to me that the analysis is okay, but I need a double check in case there is some missing points.