Vtyjkyuk

Question. (True/False) Let $V=C_c(mathbbR)$, the space of continuous functions on $mathbbR$ with continuous support endowed with the metric $$d(f,g)=(int_mathbbR|f(t)-g(t)|^2dt)^frac12$$
Let $f:mathbbRto mathbbR$ be a continuous function which vanishes outside the interval $[0,1]$. Define $f_n(x)=f(x-n)$ for $nin mathbbN$. Then $f_n$ has a convergent subsequence in $V$.

My attempt: Now here, $|f_n|_2=|f|_2$ for all $nin mathbbN$.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, $lim_nto inftyf_n(x)=0$ for all $x in mathbbR$ (i.e., pointwise limit is zero function on $mathbbR$). Also $f_n$ is not uniformly convergent on $mathbbR$.

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_mVert_2= sqrt Vert f_nVert_2^2+Vert f_m Vert_2^2 geq Vert fVert_2 $

Should $f$ not be $0$, you have no Cauchy subsequence, and the statement is false.