- Let H be an infinite dimensional Hilbert space. Determine if the following statements are true or false. If true, provide a proof. If false, provide a counter example.
A sequence \(\left\{ f _ { n } \right\}\) in H with \(\left| \left| f _ { n } \right| \right| = 1\) for all n has a subsequence that converges in H.
A sequence \(\left\{ f _ { n } \right\}\) in H with \(| | f _ { n } | | = 1\) for all n has a subsequence that converges weakly in H.
Solution
1. Part (a) is false.
Because \(H\) is infinite dimensional, it contains an infinite orthonormal sequence \[ e_1,e_2,\dots. \] Each vector has norm \(1\), but for \(m\ne n\), \[ \|e_n-e_m\|^2 =\|e_n\|^2+\|e_m\|^2-2\operatorname{Re}\langle e_n,e_m\rangle =2. \] Thus every two distinct terms are distance \(\sqrt2\) apart. No subsequence is Cauchy, hence no subsequence converges in norm.
Therefore part (a) is false.
2. Reduce part (b) to a separable Hilbert space.
Let \((f_n)\) be any sequence with \(\|f_n\|=1\) for all \(n\), and set \[ H_0:=\overline{\operatorname{span}}\{f_n:n\ge1\}. \] Then \(H_0\) is a separable Hilbert space. Choose an orthonormal basis \[ e_1,e_2,\dots \] for \(H_0\).
For each fixed \(j\), the scalar sequence \[ \langle f_n,e_j\rangle \] is bounded by \(1\). By repeated subsequence extraction and the diagonal argument, there is a subsequence \((f_{n_k})\) such that for every \(j\) the limit \[ a_j:=\lim_{k\to\infty}\langle f_{n_k},e_j\rangle \] exists.
3. Show that the limiting coordinates define a vector of \(H_0\).
For every \(N\), Bessel’s inequality gives \[ \sum_{j=1}^N|\langle f_{n_k},e_j\rangle|^2 \le \|f_{n_k}\|^2=1. \] Letting \(k\to\infty\) yields \[ \sum_{j=1}^N|a_j|^2\le1. \] Since this holds for every \(N\), \[ \sum_{j=1}^\infty|a_j|^2\le1. \] Hence the series \[ f:=\sum_{j=1}^\infty a_j e_j \] converges in \(H_0\).
4. Prove weak convergence of the selected subsequence.
First let \[ g=\sum_{j=1}^N c_j e_j \] be a finite linear combination. Then \[ \langle f_{n_k},g\rangle \longrightarrow \sum_{j=1}^N a_j\overline{c_j} =\langle f,g\rangle. \]
Now let \(g\in H_0\) be arbitrary. Given \(\varepsilon>0\), choose such a finite linear combination \(g_N\) with \[ \|g-g_N\|<\varepsilon. \] Since \(\|f_{n_k}\|=1\) and \(\|f\|\le1\), \[ |\langle f_{n_k}-f,g-g_N\rangle| \le2\varepsilon. \] Together with convergence against \(g_N\), this gives \[ \langle f_{n_k},g\rangle\to\langle f,g\rangle \qquad(g\in H_0). \]
Finally, if \(g\in H_0^\perp\), then both \(f_{n_k}\) and \(f\) lie in \(H_0\), so all these inner products are zero. Decomposing an arbitrary \(g\in H\) into its \(H_0\) and \(H_0^\perp\) parts proves \[ f_{n_k}\rightharpoonup f \quad\text{in }H. \] Thus part (b) is true.