Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 22H (Linear Analysis)

For fixed $y$, the map $x\mapsto ⟨Tx,y⟩$ is a bounded functional. The Riesz representation theorem gives a unique vector $T^{\ast }y$ representing it. Linearity, uniqueness, and $\Vert T^{\ast }y\Vert \le \Vert T\Vert \Vert y\Vert$ make $T^{\ast }$ the unique bounded adjoint operator.

A countable orthonormal basis is a frame with $A=B=1$. For any frame, a vector orthogonal to every $e_i$ makes the lower frame bound force its norm to vanish. Therefore the span is dense.

The upper frame bound gives $\Vert Ux{\Vert }_{{\ell }^2}\le \sqrt{B}\Vert x\Vert$. For $c\in {\ell }^2$, the adjoint is the norm-convergent synthesis seriesbecause its inner product with $x$ is ${\sum }_i{c}_i⟨e_i,x⟩$.

Given $\epsilon >0$, choose $N$ with ${\sum }_{i>N}|⟨a,e_i⟩{|}^2<{\epsilon }^2$. Every cube point has uniformly bounded tail by this sum, while its first $N$ coordinates lie in a compact finite-dimensional box. A finite net for that box is therefore an $\epsilon$-net for the cube. The coordinate inequalities define a closed set, so the cube is complete and totally bounded, hence compact.