Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 22I (Linear Analysis)

The Riesz representation theorem says every bounded linear functional $f$ on a Hilbert space has a unique $z$ with $f(x)=⟨x,z⟩$ and $\Vert f\Vert =\Vert z\Vert$. For $f\ne 0$, decompose relative to $\ker f$; its one-dimensional orthogonal complement supplies $z$, and Cauchy-Schwarz proves the norm statement and uniqueness.

For fixed $y$, Riesz gives a unique $Sy$ with $\theta (x,y)=⟨x,Sy⟩$. Conjugate linearity in $y$ and the inner-product convention make $S$ linear. Moreover $\Vert Sy\Vert ={\sup }_{\Vert x\Vert \le 1}|\theta (x,y)|\le \Vert \theta \Vert \Vert y\Vert$, and taking both suprema proves $\Vert S\Vert =\Vert \theta \Vert$.

The adjoint is the unique $T^{\ast }$ satisfying $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Apply Riesz to $x\mapsto ⟨Tx,y⟩$. The preceding norm calculation gives $\Vert T^{\ast }\Vert \le \Vert T\Vert$, while $\Vert Tx\Vert ={\sup }_{\Vert y\Vert =1}|⟨x,T^{\ast }y⟩|$ gives the reverse inequality.

Hermitian means $T=T^{\ast }$, unitary $T^{\ast }T=T{T}^{\ast }=I$, and normal $T^{\ast }T=T{T}^{\ast }$. If $\Vert (T-\lambda )x_n\Vert \to 0$ for unit $x_n$, then $⟨T{x}_n,x_n⟩\to \lambda$ and each term is real, so $\lambda \in \mathbb{R}$. For self-adjoint $T$, a spectral value that is not approximate would make $T-\lambda$ bounded below; its closed range has orthogonal complement $\ker (T-\bar{\lambda })=0$, hence is onto, contradicting spectrality.