Sequences of Lipschitz Functions

[Doom of Doom]

Let $$)<C<\infty$$ and $$a,b \in \mathbb{R}$$. Also let
$$Lip_{C}\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | \left|f(x) - f(y)\right| \leq C \left|x-y\right| \forall x,y \in \left[a,b\right]\right\}$$.

Let $$\left(f_{n}\right) _{n \in \mathbb(N)}$$ be a sequence of functions with $$f_{n} \in Lip_{C}\left(\left[a,b\right]\right)$$ for all n.

i) Show that if $$\left(f_{n}\right) _{n \in \mathbb(N)}$$ converges uniformly to a function $$f : \left[a,b\right]\rightarrow \mathbb{R}$$, then $$f \in Lip_{C}\left(\left[a,b\right]\right)$$.

ii) Is $$Lip_{C}\left(\left[a,b\right]\right)$$ a sub vector space of $$B\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | f$$ bounded }?____________________________________
Useful formula:

The sequence converges uniformly, so:
For all $$\epsilon > 0$$, there exists $$N \in \mathbb{N}$$ so that for all $$n,m > N$$, $$|f_{n}(x)-f_{m}(x)|<\epsilon$$ for all $$x \in \left[a,b\right]\$$.

For (i), I honestly have no idea.

It seems that I can just say:

$$|f(x)-f(y)|$$
$$= |\lim_{n \to \infty} f_{n}(x) - \lim_{n \to \infty} f_{n}(y)|$$
$$= |\lim_{n \to \infty} (f_{n}(x) - f_{n}(y))|$$
$$= \lim_{n \to \infty} |(f_{n}(x) - f_{n}(y))| \leq C |x-y|$$

But I know that has to be wrong. It's not even using the uniformity.For (ii) I know the answer is of course not.

Let $$f(x):= Cx$$. Then $$2 f \notin Lip_{C}\left(\left[a,b\right]\right)$$, since

$$|2f(x)-2f(y)| = 2C|x-y| > C|x-y|$$.

Help on part (i) please?

 

[Doom of Doom]

Hmm.. after a bit more thinking...
I think that the sequence only needs to converge pointwise (not uniformly) to a function in order for the limit function to also be in LipC([a,b]).

So, my proof for (i) pretty much works. (I think) Please tell me if I am wrong?!

Also, interestingly, I think that a pointwise convergent series of functions in LipC([a,b]) is also uniformly convergent.