I have been trying to prove that $C^1[0,1]$, i.e the space of continuously differentiable functions is closed under the $C^{0,1}[0,1]$ norm. $$||f||_{C^{0,1}}=||f||_{C^0}+\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|}$$ So what I have been able to see is that if $f_n \to f$ in the $\text{Lip}$ norm for $f_n \in C^1$ then $f$ is differentiable with bounded derivative.
I cannot see why this $f$ must have a continuous derivative. More precisely $$\left|\frac{f(x+h)-f(x)}{h}-\frac{f(y+h)-f(y)}{h}\right|\lesssim ||f_n-f||_{C^{0,1}}+|f_n'(x)-f'_n(y)|$$ But of course we just have uniform convergence for $f_n$ in $C^0$. So I can't see why we can make the second term small independent of $n$. Perhaps I'm missing something.
Any help is appreciated
If $f_n \to f$ uniformly and $f_n' \to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| \leq \|f_n-f_m||_{C^{0,1}} \to 0$ so $\{f_n'\}$ converges uniformly].