TITLE: Sequence of elements having a convergent subsequence -NBHM $2014$
QUESTION [5 upvotes]: Question is to find which of the following are true?

Let $V$ be the space of continuous functions on $\mathbb{R}$ with compact support endowed with metric $$d(f,g)=\bigg(\int_{-\infty}^{\infty}|f(t)-g(t)|^2\bigg)^{\frac{1}{2}}$$Let $f:\mathbb{R}\rightarrow \mathbb{R}$ which vanish outside $[0,1]$. define $f_n(x)=f(x-n)$ for $x\in \mathbb{N}$. Then $(f_n)$ has a convergent sub sequence in $V$.
Let $\varphi, \psi$ be continuous function on $[0,1]$. Let $(f_n)$ be a sequence in $\mathcal{C}[0,1]$ with its sup norm topology suchthat, for all $n\in \mathbb{N}$ the functions $f_n$ are continuously diffrerentiable and for all $x\in[0,1]$ and for all $n\in \mathbb{N}$ we have $|f_n(x)|\leq \varphi(x)$ and $|f_n'(x)|\leq \psi(x)$ Then there exists a sub sequence of $(f_n)$ which converges in $\mathcal{C}[0,1]$.
Let $\{A_n\}$ be a sequence of orthogonal matrices in $M_n(\mathbb{R})$. then it has a convergent subsequence.

I am sure that $3$ is true as Orthogonal matrices are compact and every sequence in compact space has a convergent subsequence.
I see that $2$ is also correct by Arzela Ascoli Theorem.
I am not so sure about $1$ and actually i did not understand the idea behind that. 
Please help me to see this question more clearly and help me to answer this.
Thank you.

REPLY [3 votes]: For any supposed limit $g$ (continuous function with compact support), take $n_0$ large enough s.t. for $n\ge n_0$: ${\rm supp}\,f_n\cap{\rm supp}\,g=\emptyset$. Now, if $n_k\ge n_0$:
$$
d(f_{n_k},g)^2=\int_{-\infty}^{\infty}|f_{n_k}(t)-g(t)|^2dt=
\int_{-\infty}^{\infty}|f_{n_k}(t)|^2dt+\int_{-\infty}^{\infty}|g(t)|^2dt\ge
$$
$$
\ge\int_{-\infty}^{\infty}|f_{n_k}(t)|^2dt
=\int_{-\infty}^{\infty}|f(t)|^2dt,
$$
and this implies $f=0$. The property is only true in this trivial case.