Past exam of the functional analysis II course of the University of Oxford / 2023 / 1 / a

In a real Hilbert space, weak convergence in a Hilbert space meansBy the Riesz representation theorem, every continuous linear functional on $X$ has the form $u\mapsto ⟨u,z⟩$. Thus this is exactly weak convergence, defined by convergence under every continuous linear functional.

Define the continuous linear functionals $f_n(z)=⟨x_n,z⟩$. For each fixed $z\in X$, weak convergence makes $f_n(z)$ a convergent, hence bounded, scalar sequence. The Uniform boundedness principle states that a pointwise bounded family of bounded linear operators from a Banach space to a normed space has uniformly bounded operator norms. Since a Hilbert space is a Banach space, it applies to $(f_n)$.

The Cauchy-Schwarz inequality gives $\Vert f_n\Vert \le \Vert x_n\Vert$, and evaluation at $z=x_n/\Vert x_n\Vert$ gives equality when $x_n\ne 0$; equality also holds at zero. Consequently

By weakly convergent sequence is bounded, $M={\sup }_n\Vert x_n{\Vert }^2$ is finite. Choose any $n_1$. Once $n_1<\cdots <n_k$ have been chosen, put $s_k={\sum }_{j=1}^k{x}_{n_j}$. Since $s_k$ is fixed, weak convergence to zero gives $⟨s_k,x_n⟩\to 0$. Choose $n_{k+1}>n_k$ with $|⟨s_k,x_{n_{k+1}}⟩|\le 1$. This gives the required one-sided inequality as well.

The useful estimate is for the squared norm. Indeed, $\Vert s_1{\Vert }^2\le M$, and the inner product expansion givesBy mathematical induction, $\Vert s_k{\Vert }^2\le Mk+2(k-1)\le (M+2)k$. HenceThis proves norm convergence of the averages, the weak Banach–Saks theorem in a Hilbert space. The printed inequality omits the square on its left side. Our stronger estimate also proves that printed bound, since $\sqrt{(M+2)k}\le (M+2)k$ for $k\ge 1$, but the printed bound alone only bounds the averages and cannot imply their norm convergence to zero.