- Let H be a Hilbert space equipped with an inner product \(( \cdot , \cdot )\) and a norm \(| | \cdot | | = ( \cdot , \cdot ) ^ { \frac { 1 } { 2 } }\) Recall the following: A sequence \(\{ f _ { k } \} \subset { \mathcal { H } }\) is said converge to \(f \in \mathcal H\) if \(\vert \vert f _ { k } - f \vert \vert \to 0\) . A sequence \(\{ f _ { k } \} \subset { \mathcal { H } }\) is said converge weakly to \(f \in { \mathcal { H } }\) if \(( f _ { k } , g ) \to ( f , g )\) for any \(g \in { \mathcal { H } }\) . Prove the following statements:
\(\{ f _ { k } \}\) converges to f if and only if \(\vert \vert f _ { k } \vert \vert \to \vert \vert f \vert \vert\) and \(\{ f _ { k } \}\) converges weakly to \(f .\)
If H is a finite dimensional Hilbert space, then the weak convergence implies convergence. Give a counter example to show that weak convergence does not necessarily imply convergence in an infinite dimensional Hilbert space.
If a sequence \(\{ f _ { k } \}\) converges weakly to \(f ,\) then there exists a subsequence \(\{ f _ { k _ { n } } \}\) such that
\[ \frac { f _ { k _ { 1 } } + \cdots + f _ { k _ { n } } } { n } \]
converges to \(f .\) (You may use the fact that a weakly convergent sequence is a bounded sequence.)
Solution
1. Prove part (a).
If \(f_k\to f\) in norm, then continuity of the norm gives \[ \|f_k\|\to\|f\|, \] and for every \(g\in H\), \[ |\langle f_k-f,g\rangle| \le \|f_k-f\|\,\|g\|\to0, \] so \(f_k\rightharpoonup f\).
Conversely, assume \(f_k\rightharpoonup f\) and \[ \|f_k\|\to\|f\|. \] Then \[ \begin{aligned} \|f_k-f\|^2 &=\|f_k\|^2+\|f\|^2 -2\operatorname{Re}\langle f_k,f\rangle. \end{aligned} \] Weak convergence gives \[ \langle f_k,f\rangle\to\|f\|^2, \] and the norm hypothesis gives \(\|f_k\|^2\to\|f\|^2\). Hence \[ \|f_k-f\|\to0. \] This proves the equivalence.
2. Prove part (b).
Suppose first that \(H\) is finite dimensional, and let \[ e_1,\dots,e_N \] be an orthonormal basis. If \(f_k\rightharpoonup f\), then for each \(j\), \[ \langle f_k-f,e_j\rangle\to0. \] Therefore \[ \|f_k-f\|^2 =\sum_{j=1}^N|\langle f_k-f,e_j\rangle|^2 \longrightarrow0. \] Thus weak convergence implies norm convergence in finite dimension.
In infinite dimension, let \((e_n)\) be an orthonormal sequence. Then \[ \|e_n\|=1 \] for all \(n\), so \((e_n)\) cannot converge to \(0\) in norm. But for every \(g\in H\), Bessel’s inequality implies \[ \sum_{n=1}^\infty|\langle g,e_n\rangle|^2\le\|g\|^2, \] hence \[ \langle e_n,g\rangle\to0. \] Thus \(e_n\rightharpoonup0\) but not strongly.
3. Reduce part (c) to a weakly null bounded sequence.
Set \[ u_k:=f_k-f. \] Then \(u_k\rightharpoonup0\). A weakly convergent sequence is bounded, so there is \(M<\infty\) such that \[ \|u_k\|\le M \qquad\text{for all }k. \]
We inductively select a subsequence \((u_{k_n})\). Choose \(k_1\) arbitrarily. Having chosen \[ k_1<\cdots<k_{n-1}, \] weak convergence to \(0\) implies \[ \langle u_k,u_{k_j}\rangle\to0 \qquad(k\to\infty) \] for each \(j<n\). Hence we can choose \(k_n>k_{n-1}\) so large that \[ |\langle u_{k_n},u_{k_j}\rangle|\le\frac1n \qquad(1\le j<n). \]
4. Show that the Cesàro means of the selected subsequence converge in norm.
Let \[ A_n:=\frac1n\sum_{j=1}^n u_{k_j}. \] Then \[ \begin{aligned} \|A_n\|^2 &=\frac1{n^2}\left( \sum_{j=1}^n\|u_{k_j}\|^2 +2\operatorname{Re}\sum_{1\le j<\ell\le n} \langle u_{k_j},u_{k_\ell}\rangle \right)\\ &\le \frac{M^2}{n} +\frac2{n^2}\sum_{\ell=2}^n\frac{\ell-1}{\ell}\\ &\le \frac{M^2}{n}+\frac2n. \end{aligned} \] Hence \(\|A_n\|\to0\). Since \(u_{k_j}=f_{k_j}-f\), \[ \frac{f_{k_1}+\cdots+f_{k_n}}n-f=A_n\to0 \] in norm. Therefore \[ \boxed{ \frac{f_{k_1}+\cdots+f_{k_n}}n\longrightarrow f.} \]