Suppose $f_{n}$ in $C[0,1]$ all have Lipschitz constant $L$. Show that if $f_{n}$ converges pointwise to $f$, then the convergence is uniform and $f$ is Lipschitz with constant $L$.
Attempt
Want to show: $\forall \epsilon$ there exists a $N > 0$ s.t. $\|f_n - f\| < \epsilon$ for all $x \in [0,1]$ and for all $n \geq N$. Also that $\|f(x) - f(y)\| \leq L\|x-y\|$ for all $x,y \in [0,1]$
What we know
$f_n$ Lipschitz means $\|f(x) - f(y)\| \leq L\|x-y\|$ for all $x,y \in [0,1]$.
By Pointwise convergence: For all $\epsilon > 0$ and for all $x \in [0,1]$, there exists $N > 0$ s.t. $\lim_{n \rightarrow \infty} f(x_{0})= f(x_{0})$ for fixed $x_{0}$. Which also means $\|f_{n}(x_{0}) - f(x_{0})\| < \epsilon$ for all $n \geq N$ (maybe I should use different notation for absolute value bars here).
Proof Attempt: Let $\epsilon > 0$ and take $n \geq N$ to satisfy the sequence condition.
$$\|f_{n}(x) - f(x)\| = \|f_{n}(x) - f_{n}(x_{0}) + f_{n}(x_{0}) - f(x_{0}) + f(x_{0}) - f(x)\| \leq \|f_{n}(x) - f_{n}(x_{0})\| + \|f_{n}(x_{0}) - f(x_{0})\| + \|f(x_{0}) - f(x)\|$$.
I feel that the first and second expression I could control by the Lipschitz constant and pointwise continuity respectively, so the next line would look something like:
$$\leq L\|x-x_{0}\| +\frac{\epsilon}{3} + something $$
I also feel that if I could show all three expressions in the form of a Lipschitz constant I could take each of their limits and I would get zero which could show uniform continuity. Am I on the right track and what hints could you give me to continue solving?
