Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 72 / 3

For a bounded linear operator $A:X\to Y$ between Hilbert spaces, let $N=\ker A$ and $R=\mathrm{ran}A$. The Moore–Penrose inverse of an operator is defined onFor $y=Ax+z$, with $z\in R^{\perp }$, defineThis is independent of the chosen preimage $x$, since two preimages differ by a vector in $N$. Equivalently, it is the inverse of the restriction of $A$ to $N^{\perp }$, applied to the component of the data in $R$, and is zero on $R^{\perp }$.

The generalized solution $x^{†}=A^{†}y$ is the unique minimum-norm least-squares solution. Its residual is orthogonal to the range, giving the operator normal equationThe operator normal equation alone leaves an arbitrary null-space component; the second condition fixes the minimum-norm representative. The identities $A^{†}A=P_{N^{\perp }}$ and $A{A}^{†}=P_{\overline{R}}$ hold on their appropriate domains.

For an infinite-rank compact operator, the range need not be closed, and $A^{†}$ is generally unbounded. Data outside $R\oplus R^{\perp }$ need not have any least-squares solution at all, even though they can be approximated by range elements. This domain qualification is crucial in part (d); the Moore–Penrose notation does not turn an ill-posed inverse into an everywhere-defined bounded operator.

A singular value system of a compact operator consists of positive numbers ${\sigma }_j$ and orthonormal families $v_j\in X$, $u_j\in Y$ satisfyingThus $v_j$ are positive-eigenvalue eigenvectors of $A^{\ast }A$ and $u_j$ of $A{A}^{\ast }$, with eigenvalue ${\sigma }_j^2$. The families are complete in $(\ker A{)}^{\perp }$ and $\overline{\mathrm{ran}A}$ respectively. The singular values can be listed nonincreasingly with multiplicities, and tend to zero in the infinite-rank case. For finite rank there are only finitely many positive singular values; zero-kernel directions are handled separately.

Use the convention that $⟨f,g⟩$ is conjugate-linear in its first entry. Then the singular value system givesThe latter series converges precisely on the admissible range component specified by the Picard criterion:with $R^{\perp }$ components annihilated by the inverse. Merely writing a formal singular expansion does not imply it converges in $X$.

Use the orthonormal Fourier series basis $e_n(x)=(2\pi {)}^{-1/2}{e}^{inx}$, indexed by $n\in \mathbb{Z}$, and the Hilbert space inner product $⟨f,g⟩={\int }_0^{2\pi }\overline{f(x)}g(x)dx$. The integral operator is a periodic convolution operator. Changing variables $s=x-x^{\prime }$ and using periodicity givesThere is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x^{\prime }-x)}$, so $A^{\ast }{e}_n=\overline{c_n}{e}_n$.

A singular system of a periodic convolution operator is thereforeIndeed $A{v}_n={\sigma }_n{u}_n$ and $A^{\ast }{u}_n=(c_n/|c_n|)\overline{c_n}{e}_n={\sigma }_n{v}_n$. The unit factor determined by the complex argument of $c_n$ in $u_n$ is necessary when the complex Fourier coefficients are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne 0$. One may enumerate $\mathbb{Z}$ by $\mathbb{N}$, or order the positive singular values by decreasing magnitude.

The continuous kernel on a finite square makes $A$ a Hilbert-Schmidt operator, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne 0$,by the Riemann-Lebesgue lemma. In particular the singular values tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^{\ast }=0$: the range is dense, but the inverse is unbounded and the range is not closed.

Let $y_n=⟨e_n,y⟩$. Using the phase of $u_n$ in the singular value system, the Moore–Penrose inverse of an operator becomesThis gives the admissible data for a periodic convolution inverse. Because the range is dense and the kernel is zero here, the domain of the inverse is exactly the range, not all of $L^2$.

For $y(x)=e^{\alpha x}$ and $\alpha \notin i\mathbb{Z}$,The formal expression would consequently beHowever, the high-frequency obstruction for nonperiodic exponential data prevents this from being a Hilbert-space solution. The numerator is nonzero, $|y_n|$ is asymptotic to a nonzero constant divided by $|n|$, and part (c) proved $|c_n|=o(1/|n|)$. Hence $|y_n/c_n|\to \infty$, so the Picard criterion fails. For $\alpha \notin i\mathbb{Z}$, $A^{†}y$ is not defined in the specified Hilbert space. The formal series is not a convergent generalized solution.

If $\alpha =im$ with $m\in \mathbb{Z}$, the data are a single periodic Fourier mode: $y_n=\sqrt{2\pi }{\delta }_{nm}$. Then there is an exact unique solution,In particular, if the intended parameter is real, only $\alpha =0$ is admissible, giving $f^{†}=1/c_0$.

For the inadmissible cases the inverse problem still has approximate solutions $f_N={\sum }_{|n|\le N}(y_n/c_n)e_n$: their images are Fourier projections converging to $y$ in $L^2$, while their norms diverge. Thus the least-squares residual has infimum zero but no minimizer. This distinguishes an undefined exact inverse from a regularized truncated reconstruction; it is the necessary qualification to the question's unrestricted constant $\alpha$.