Show that $\mathbb{F}=\{f\in C[0,1]:|f(x)-f(y)|\leq|x-y| \forall x,\forall y, \int_0^1f(x)\,dx=0 \}$ is closed I think
$$|\lim_{n\to\infty}f_n(x)-\lim_{n\to\infty}f_n(y)|\leq |x-y|$$ is obvious because 
$$ LHS =|\lim_{n\to\infty}(f_n(x)-f_n(y))|\leq \lim_{n\to\infty}|x-y|=|x-y|.$$
The problem is the second part:
$$\int_0^1\lim_{n\to\infty} f_n(x)\,dx=0$$
I think for that we need to interchange the limit and integral operation, but how do we go about that? Do we have to prove uniform convergence? I'm struggling to prove that. Maybe we can do with different approach than uniform convergence.
Can you give me any suggestions or comments? I would greatly appreciate it. Thank you.
 A: Two approaches:
1) As explained in the comment I wrote: it is enough to show the sequence is dominated. Let $f_n$ be a sequence of $C[0,1]$ functions converging in sup-norm to $f$. Then for large enough $N$ we have for all $n>N$: $\|f_n -f\|_\infty \leq 1$ . This, together with my comment: shows $$|f_n(x)| \leq |x-y| + |f_n(y)| \leq |x-y| + \|f_n\|_\infty \leq |x-y|+\|f\|_\infty$$
And the last expression is integrable (with respect to $x$) since $\|f\|_\infty < \infty$ (f is continuous). This shows the tail of ${f_n}$ is dominated by an integrable function.
2) Denote $V = \{ f\in C[0,1] \colon\quad $f$ \text{ is 1-lifschitz} \}  $ (This is a normed linear space when equipped with $\infty$-norm).
We then define a linear functional $\varphi\colon V \rightarrow \mathbb{C}$ by: $\varphi(f) = \int_0^1 f$. We notice $\varphi$ is continuous and hence the space $\{ f\in C[0,1] \colon\quad $f$ \text{ is 1-lifschitz and } \varphi(f)=0 \}$ is closed as it is the preimage of a closed set.
A: I assume that $\mathbb{F}$ is equipped with the supremum norm. Then the limit $f$ is continuous and by the mean value theorem for integrals there is $\xi_n\in (0,1)$ such that $\int_0^1(f-f_n)(x) \space dx = (f-f_n)(\xi_n)$ for each $n$. This yields
$$\left|\int_0^1f(x) \space dx\right| = \left|\int_0^1(f-f_n)(x) \space dx \right| = |(f-f_n)(\xi_n)|\le \|f-f_n\|_\infty\rightarrow 0,$$
thus $\int_0^1f(x) \space dx = 0$.
