Sequence in $B[0,1]\subset L^2[0,1]$ which has no convergent subseqence.

Question

In order to prove closed unit ball in $L^2[0,1]$ is not compact I want to find sequence in it which has no convergent subsequence.
I tried with some sequences but none of them turn out to be right.

Answer 1

Try the sliding bump functions $f_{n,k} = 2^{n/2}\chi_{[(k-1)/2^n, k/2^n]}$, $n \in \Bbb{N}$, $k = 1,\dots,2^n$. Then $||f_{n,k}||^2_2 = (2^{n/2})^2/2^n = 1$ for any $n,k$. Hence, the sequence $\{f_{n,k}\}_{n,k}$ arranged in increasing order of $k$ and $n$ lies in the closed unit ball $\overline{B(0,1)}$ but in has no convergent subsequence:

We claim that the squared $L^2$ distance of $f_{n,k}-f_{n',k'}$ (i.e. $||f_{n,k}-f_{n',k'}||^2_2$) is greater than $1$. If $n = n'$, since the supports of $f_{n,k}$ and $f_{n',k'}$ don't overlap, $||f_{n,k}-f_{n',k'}||^2_2 = (2^{n/2})^2 \cdot 2/2^n = 2$. Otherwise, WLOG, we assume $n' \ge n + 1$. By drawing the graph of $f_{n,k}$ and $f_{n',k'}$, we know that $||f_{n,k}-f_{n',k'}||^2_2$ is greater when the supports of $f_{n,k}$ and $f_{n',k'}$ overlap. In this case, $\require{cancel}$ \begin{align} ||f_{n,k}-f_{n',k'}||^2_2 &= \underbrace{(2^{n'/2} - 2^{n/2})^2 \left(\frac{1}{2^{n'}}\right)}_{\mbox{supports overlap}} + \underbrace{(2^{n/2})^2 \left(\frac{1}{2^{n}} - \frac{1}{2^{n'}}\right)}_{\Large \mathrm{supp}(f_{n,k}) \setminus \mathrm{supp}(f_{n',k'})} \\ &= (1 + \cancel{2^{-(n' - n)}} - 2^{1 - (n' - n)/2}) + (1 - \cancel{2^{-(n'-n)}}) \\ &= 2 - 2^{1 - (n' - n)/2} \\ &\ge 2 - \sqrt\frac12 > 1 \end{align}

because of the following inequality.

\begin{alignat}{2} & \quad & n + 1 &\le n' \\ \iff && 1 - \frac{n' - n}{2} &\le \frac12 \\ \iff && 2^{1 - (n' - n)/2} &\le \sqrt\frac12 \end{alignat}

Therefore, the distance between any two bump functions $f_{n,k}$ and $f_{n',k'}$ are (strictly) greater than $1$. Hence, it's impossible to construct a convergent subsequence from any sequence of $\{f_{n,k}\}_{n,k}$.

Answer 2

Two solutions. The first is like an already existing answer, but I want to point out that it can be simplified: Just choose pairwise disjoint intervals $I_n\subset [0,1]$ of positive length $l_n, n=1,2,\dots.$ Then the functions $f_n = l_n^{-1/2}\chi_{I_n}$ satisfy $\|f_n\|_2 = 1$ for all $n,$ and if $m\ne n,$ $\|f_n-f_m\|_2 = \sqrt 2.$ No troublesome computations needed.

The second solution is rooted in Fourier series: Define $f_n(x) = \sin (2\pi n x), n= 1,2,\dots.$ Then each $f_n\in B(0,1).$ As is well known, the functions $f_n$ are orthogonal, and $\|f_n\|_2 = 1/\sqrt 2$  for each $n.$ Thus if $m\ne n,$ then $\|f_m-f_n\|_2 = 1.$ This implies that no subsequence of $(f_n)$ is convergent in $L^2[0,1].$