Let $f_n \in C[0,1], \forall n \in \Bbb N$, and let $f:[0,1] \rightarrow \Bbb R$. Suppose, $\exists C$ such that $|f_n(x) - f_n(y)| \le C|x-y|$, $\forall n \in \Bbb N$. Then, if $f_n \rightarrow f$ pointwise, then $f_n \rightarrow f$ uniformly.
This has been my approach so far:
Since $f_n$ is continuous on $[0,1]$, and $[0,1]$ is compact, $f_n$ is uniformly continuous on $[0,1]$.
Fix $\epsilon \gt 0$. Then, $\exists \delta \gt 0$ such that $|f_n(x) - f_n(y)| \lt \epsilon, \forall x,y\in [0,1]$ for which $|x-y| \lt \delta$.
So, choose $x,y \in [0,1]$ such that $|x-y| \lt \delta$. We have: $$ |f_n(x) - f(x)| \le |f_n(x) - f_n(y)| + |f_n(y) - f(x)| \lt \epsilon + |f_n(y) - f(x)|$$
Now, this is where I'm stuck as if I apply triangle inequality again to get:
$$ |f_n(x) - f(x)| \lt \epsilon + |f_n(y) - f_n(x)| + |f_n(x) - f(x)| \le \epsilon + C|x-y| + |f_n(x) - f(x)|$$ I end up erasing the LHS and getting nowhere.
I have seen an approach where we "temporarily" fix $x$ and set $N$ satisfying the pointwise of convergence of $f_n$ to $f$ at the point $x$. But this doesn't really make sense to me, as then we cannot show uniform convergence since our inequality chain above will depend on the $N$, which itself depends on the chosen $x$.
Any hints are greatly appreciated!!