Give a sequence \((f_j)\) in \(L^2(\RR)\) such that \(\norm{f_j}_2=1\) for every \(j\), but \((f_j)\) has no norm-convergent subsequence.
Solution
1. For \(j\ge1\), define\[ f_j=\mathbf1_{[j,j+1]}. \]Then \(\norm{f_j}_2=1\) for every \(j\).
Indeed, \[ \norm{f_j}_2^2=\int_j^{j+1}1\,dx=1. \]
2. Distinct terms satisfy\[ \norm{f_j-f_k}_2=\sqrt2. \]
If \(j\ne k\), the supports are disjoint, so \[ \norm{f_j-f_k}_2^2 =\norm{f_j}_2^2+\norm{f_k}_2^2 =2. \]
3. The sequence has no norm-convergent subsequence.
By step 2, distinct terms are always distance \(\sqrt2\) apart. Hence no subsequence is Cauchy in \(L^2(\RR)\), so no subsequence converges in norm.
4. Q.E.D.
Steps 1 and 3 give the required sequence and its noncompactness property.