Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 21F (Linear Analysis)

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

Put $A=T^{\ast }T$. Then $A^{\ast }=T^{\ast }T$ and $(x,Ax)=\Vert Tx{\Vert }^2\ge 0$, with the inner product linear in its second argument as required by the later SVD formula. For nonreal $\lambda$, self-adjointness gives $\Vert (A-\lambda I)x\Vert \ge |\mathrm{Im}\lambda |\Vert x\Vert$. For $\lambda =-a<0$, positivity gives $\Vert (A+aI)x\Vert \ge a\Vert x\Vert$. These bounds give injectivity and closed range. The range is dense because its orthogonal complement is the kernel of the adjoint, to which the same bound applies. Thus the range is all of $H$ and the inverse is bounded in either case. ConsequentlyIf $T$ is compact, composition of the compact operator $T$ with bounded $T^{\ast }$ is compact, so $T^{\ast }T$ is compact too.

Apply the spectral theorem for compact Hermitian operators to positive $A=T^{\ast }T$. Its positive eigenvalues ${\mu }_i$ have orthonormal eigenvectors $u_i$ spanning $\ker A^{\perp }$, with a finite or countable list. In the infinite case ${\mu }_i\to 0$. DefineThen $(v_i,v_j)=(u_i,T^{\ast }T{u}_j)/({\lambda }_i{\lambda }_j)={\delta }_{ij}$, so the $v_i$ are orthonormal. Also $\ker A=\ker T$, because $(x,Ax)=\Vert Tx{\Vert }^2$. Decompose $x=x_0+{\sum }_i(u_i,x)u_i$ with $x_0\in \ker T$ and apply bounded $T$ to the norm-convergent expansion. This gives the singular value decompositionThe series converges in norm; its squared tail norm is ${\sum }_{i>m}{\mu }_i|(u_i,x){|}^2\to 0$. No separability of the whole Hilbert space is necessary: the nonzero spectral subspace of a compact operator is separable. For $T=0$ use the empty list $N=0$.