Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 2 / b

Let $T_{\alpha }=A^{\ast }A+\alpha I$. For every $f\in X$,Thus $T_{\alpha }$ is a coercive operator. If it is invertible and $T_{\alpha }f=g$, the lower bound and the Cauchy-Schwarz inequality giveand therefore $\Vert T_{\alpha }^{-1}g\Vert \le {\alpha }^{-1}\Vert g\Vert$. Hence

Subtracting $T_{\alpha }{u}_n=z_n$ and $T_{\alpha }{u}_m=z_m$ and applying the coercive estimate from part i yieldsThus $(z_n)$ Cauchy implies $(u_n)$ Cauchy. Completeness of the Hilbert space $X$ gives $u_n\to u\in X$, and boundedness of $T_{\alpha }$ gives $z_n=T_{\alpha }{u}_n\to T_{\alpha }u$.

The lower bound in part i shows that $T_{\alpha }u=0$ implies $u=0$, so $T_{\alpha }$ is injective. Since $T_{\alpha }$ is self-adjoint,so its range is dense. Part ii shows that its range is also closed: a convergent sequence $z_n=T_{\alpha }{u}_n$ has Cauchy preimages, whose limit $u$ satisfies $T_{\alpha }u=\lim z_n$. Hence $\mathcal{R}(T_{\alpha })=X$. The operator is bijective, and the estimate in part i proves that its inverse is bounded. Therefore $A^{\ast }A+\alpha I$ is invertible for every $\alpha >0$.

The equation $(A^{\ast }A+{\alpha }_{\ast }I)u_{\ast }=A^{\ast }f$ is the Tikhonov normal equation. For $u=u_{\ast }+h$, expand the Tikhonov regularization functional:The normal equation makes the linear term zero. The final two terms are nonnegative and are strictly positive for $h\ne 0$ because ${\alpha }_{\ast }>0$. Thus $u_{\ast }$ is the unique global minimizer.