Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 78 / 4

Use the inner product convention linear in its first argument. A singular value system consists of positive singular values ${\sigma }_n$ and orthonormal bases $u_n$ of $(\ker A{)}^{\perp }\subset X$ and $v_n$ of $\overline{\mathrm{ran}A}\subset Y$, withFor an infinite-rank compact operator, ${\sigma }_n\to 0$; a finite-rank operator has a finite system. Applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$ constructs $u_n$, with eigenvalues ${\sigma }_n^2$, and then $v_n=A{u}_n/{\sigma }_n$. This solution uses the input/output naming of this problem, which is the reverse of another common singular-vector convention.

The Moore–Penrose inverse of an operator isIts domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$, equivalently the data satisfying the Picard criterionIt ignores the component in $\ker A^{\ast }$ and lies in $(\ker A{)}^{\perp }$. The series then converges in $X$ by orthonormality and is the minimum-norm least-squares solution. If $y\in \mathrm{ran}A$, it is an exact solution; all other exact solutions are $x^{†}+z$ with $z\in \ker A$. Compactness does not make the inverse bounded when there are infinitely many positive singular values.

A linear regularization is a family of bounded operators $R_{\alpha }:Y\to X$ such that $R_{\alpha }y\to A^{†}y$ for every $y$ in the domain of the Moore–Penrose inverse of an operator as $\alpha \downarrow 0$. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, a convergent regularization of an inverse problem also specifies a parameter choice $\alpha (\delta ,y^{\delta })$ giving convergence to $A^{†}y$, uniformly over the allowed noise at each fixed admissible exact datum.

Expand $x_{\alpha }$ in the input singular vectors. The Tikhonov normal equation impliesThe component in $\ker A$ is zero because $\alpha >0$. Comparing with the given filter convention givesThe singular-system Tikhonov filter is often described by its gain $g_{\alpha }(\sigma )=\sigma /(\alpha +{\sigma }^2)$; the requested $f_{\alpha }$ includes one extra factor of $\sigma$. For admissible data, $f_{\alpha }\to 1$ and $0\le f_{\alpha }\le 1$, so dominated convergence theorem applied to the squared Picard criterion coefficients proves exact-data convergence. AlsoBy the noise-bias decomposition for linear regularization, $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$ suffice for noisy-data convergence; for example $\alpha =\delta$ as $\delta \downarrow 0$. This establishes both consistency and stability of Tikhonov regularization.

For $f,g\in L^2(0,1)$, interchange the integrals in the inner product of the Volterra integration operator:Thus the adjoint operator is $A^{\ast }g(t)={\int }_t^{1}g(x)dx$. The triangular integration kernel is square-integrable, so $A$ is a Hilbert-Schmidt operator and hence a compact operator.

Set $a_n=(n-\frac{1}{2})\pi$ and ${\sigma }_n=1/a_n$. Direct integration givessince $\cos a_n=0$. Product-to-sum identities give $⟨u_n,u_j⟩=⟨v_n,v_j⟩={\delta }_{nj}$. For completeness, $g=A^{\ast }Af$ satisfies $g^{\prime \prime }=-f$, $g^{\prime }(0)=0$, $g(1)=0$. The associated Sturm-Liouville problem has precisely the complete mixed-boundary cosine basis $u_n$; $A{A}^{\ast }$ gives the mixed-boundary sine basis $v_n$, with $v_n(0)=0$, $v_n^{\prime }(1)=0$. Both $A$ and $A^{\ast }$ are injective, since differentiation of their zero integrals recovers their inputs. Their singular vectors therefore cover the entire two Hilbert spaces, rather than just proper orthogonal complements.

The given functions form the complete singular value system, with ${\sigma }_n=2/((2n-1)\pi )$. This is the mixed-boundary singular system of the Volterra operator.

Substitution into the Tikhonov regularization series gives an explicit reconstruction for every $y\in L^2(0,1)$:Equivalently the summand is ${\sigma }_n⟨y,v_n⟩u_n/(\alpha +{\sigma }_n^2)$. The Tikhonov stability bound makes the series converge in $L^2$ for every positive $\alpha$. It suppresses high-frequency components of unstable differentiation, rather than differentiating arbitrary noisy data pointwise.

The range of the Volterra integration operator is $\{y\in H^1(0,1):y(0)=0\}$. On exactly these data, $A^{†}y=y^{\prime }$ and the series converges to $y^{\prime }$ as $\alpha \downarrow 0$. For arbitrary $L^2$ data outside that range, a finite regularized solution still exists, but a finite-norm unregularized solution need not exist. This is Spectral Tikhonov differentiation for the Volterra operator.