Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 21I (Linear Analysis)

$\sigma (T)=\{\lambda :T-\lambda I\text{ is not invertible}\}$ and $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$, well-defined because the spectrum is nonempty compact. If $|\lambda |>\Vert T^n{\Vert }^{1/n}$, factor $T^n-{\lambda }^{n}I$ and use a Neumann series; spectral mapping then excludes $\lambda$. Thus $r(T)\le {\inf }_n\Vert T^n{\Vert }^{1/n}$.

Complex Stone-Weierstrass requires a self-conjugate unital subalgebra separating points; apply real Stone-Weierstrass to real and imaginary parts. Finite sums $f(x)g(y)$ form such a self-conjugate unital subalgebra of $C(K\times L)$ and separate points, so they are dense.

$\Vert T_{h}f{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }\Vert f{\Vert }_{\infty }$, so $T_h$ is bounded, and $h\mapsto T_h$ is linear with operator norm at most one.

Approximate $h$ uniformly by finite sums $\sum f_j(x)g_j(y)$; their integral operators have finite-dimensional range and converge in operator norm, so $T_h$ is compact. If $h(x,y)=0$ for $x\le y$, iterated integration over an ordered simplex gives $\Vert T_h^n\Vert \le \Vert h{\Vert }_{\infty }^n/n!$. Thus $r(T_h)=0$, and compact spectral theory gives $\sigma (T_h)=\{0\}$.