Question. (True/False) Let $V=C_c(\mathbb{R})$, the space of continuous functions on $\mathbb{R}$ with continuous support endowed with the metric $$d(f,g)=(\int_{\mathbb{R}}|f(t)-g(t)|^2dt)^{\frac{1}{2}}$$ Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function which vanishes outside the interval $[0,1]$. Define $f_n(x)=f(x-n)$ for $n\in \mathbb{N}$. Then $\{f_n\}$ has a convergent subsequence in $V$.
My attempt: Now here, $\|f_n\|_2=\|f\|_2$ for all $n\in \mathbb{N}$.Also $f_n$ is basically same as $f$, just shifted in the interval $[n,n+1]$ and outside this interval $f_n$ is zero.Hence, $\lim_{n\to \infty}f_n(x)=0$ for all $x \in \mathbb{R}$ (i.e., pointwise limit is zero function on $\mathbb{R}$). Also $f_n$ is not uniformly convergent on $\mathbb{R}$.
But how can I conclude the convergency of its subsequence in $\| ~\|_2$ norm? Please help. Thank you.
It would seem to me that you do not have convergent sub-sequence. If you were to prove that the distance between any two functions in the sequence is bounded from below, then there can't be a Cauchy sub-sequence, implying also there can't be a convergent sub-sequence.
I would argue that since any two functions in the sequence are supported on 'disjoint' sets, then:
$\Vert f_n-f_m\Vert_2= \sqrt{ \Vert f_n\Vert_2^2+\Vert f_m \Vert_2^2 }\geq \Vert f\Vert_2 $
Should $f$ not be $0$, you have no Cauchy subsequence, and the statement is false.