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\}<br />.

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 &gt; 0, there exists N \in \mathbb{N} so that for all n,m &gt; N, |f_{n}(x)-f_{m}(x)|&lt;\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| &gt; 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.