Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 22H (Linear Analysis)

Assume $x_n\rightharpoonup x_{\infty }$. For each $n$, define the bounded functionalWeak convergence makes $(T_n(h){)}_n$ convergent, hence bounded, for every $h\in H$. The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n\Vert$, so $(x_n)$ is bounded.

Taking $h=e_i$ in the definition of weak convergence in a Hilbert space givesfor every $i$. Thus (i) implies (ii).

Conversely, supposeand defineFor every $N$, Bessel inequality and passage to the limit giveHence ${\sum }_i|a_i{|}^2\le C^2$. The partial sums ${\sum }_{i=1}^N{a}_i{e}_i$ are therefore Cauchy; completeness of the Hilbert space gives an element

Let $P_{N}h={\sum }_{i=1}^N⟨h,e_i⟩e_i$. For fixed $N$, coordinate convergence givesThe Hilbertian basis property gives $P_{N}h\to h$, andFirst choose $N$ large and then $n$ large. This provesfor every $h\in H$, so $x_n\rightharpoonup x_{\infty }$. This establishes the coordinate criterion for weak convergence in a separable Hilbert space.

Let $\Vert x_n\Vert \le C$. For each $i$, Bessel inequality givesso the $i$th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\varphi (n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The coordinate criterion for weak convergence in a separable Hilbert space supplies an $x_{\infty }\in H$ such thatThis proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Suppose first that $\Vert x_n-x_{\infty }\Vert \to 0$. The reverse triangle inequality givesso (i) implies (ii).

It also implies (iii). Given $ϵ>0$, choose $N$ so thatBy the Parseval identity for a Hilbertian basis, choose $I_0$ so that the basis tail of $x_{\infty }$ beyond $I_0$ is less than $ϵ/8$. For $n\ge N$ and $I\ge I_0$,There are only finitely many $n<N$, so increasing $I$ makes each of their tails less than $ϵ$. Thus one $I$ works for every $n$, proving (iii).

Assume (ii). Weak convergence givesUsing the inner product identity,Hence (ii) implies (i); this is the Radon-Riesz theorem in a Hilbert space.

Assume (iii), and let $\delta >0$. Choose $I$ such thatfor every $n$. Coordinate convergence implies, for every finite $J\ge I$,Letting $J\to \infty$ shows that the corresponding tail of $x_{\infty }$ is also at most $\delta$.

By the Parseval identity for a Hilbertian basis,The finite first sum tends to zero by weak coordinate convergence, while the second is at mostSince $\delta$ is arbitrary, $\Vert x_n-x_{\infty }\Vert \to 0$. Thus (iii) implies (i), completing the equivalence and proving the uniform basis-tail criterion for strong convergence.