Let \(\{e_n\}\) be an orthonormal basis for a Hilbert space \(H\).
Show that \(e_n \to 0\) weakly. (Explain what weak convergence means.)
Show that \(e_n\) does not tend to zero strongly. (Explain what strong convergence means.)
Let \(v_n = \frac{1}{n} \sum_{j=1}^{n} e_j\). Show that \(v_n \to 0\) strongly.
Solution
A sequence \(x_n\) in \(H\) converges weakly to \(x\) if \(\phi(x_n)\to\phi(x)\) for every bounded linear functional \(\phi\) on \(H\); by the Riesz representation theorem, if and only if \(\langle x_n,y\rangle\to\langle x,y\rangle\) for every \(y\in H\). It converges strongly to \(x\) if \(\norm{x_n-x}\to0\).
1. (a) \(e_n\to0\) weakly.
For \(y\in H\), Bessel’s inequality gives \(\sum_n\abs{\langle y,e_n\rangle}^2\le\norm y^2<\infty\), so the terms \(\abs{\langle e_n,y\rangle}\) tend to \(0\).
2. (b) \(e_n\) does not tend to \(0\) strongly.
\(\norm{e_n-0}=\norm{e_n}=1\) for every \(n\).
3. (c) \(\norm{v_n}=1/\sqrt n\), so \(v_n\to0\) strongly.
By orthonormality, \(\norm{v_n}^2=\frac1{n^2}\sum_{j,k=1}^n\langle e_j,e_k\rangle=\frac{n}{n^2}=\frac1n\).
4. Q.E.D.
Steps 1, 2 and 3 answer parts (a), (b) and (c).