Tikhonov regularization with a coercive penalty operator ($R_{\alpha }=(K^{\ast }K+\alpha B^{\ast }B{)}^{-1}{K}^{\ast }$)

For bounded linear operators between Hilbert spaces with $\Vert Bu\Vert \ge \beta \Vert u\Vert$ and $\beta >0$, the operator $K^{\ast }K+\alpha B^{\ast }B$ is coercive and invertible for $\alpha >0$. The reconstruction uniquely minimizes $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$. Testing its normal equation for coercive Tikhonov regularization by $u$ gives $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. If $K$ is injective, exact-data consistency and $\alpha \to 0$, $\delta /\sqrt{\alpha }\to 0$ give norm convergence to $K^{†}f$.

Let $K$ be injective, $u^{†}=K^{†}f$, $z={\beta }^{-2}{B}^{\ast }B{u}^{†}$, and ${\eta }_r={\inf }_{\Vert w\Vert \le r}\Vert z-K^{\ast }w\Vert$. Since $\overline{\mathcal{R}(K^{\ast })}=\mathcal{N}(K{)}^{\perp }=U$, ${\eta }_r\to 0$. For $e=R_{\alpha }f-u^{†}$ the normal equation for coercive Tikhonov regularization gives $\Vert Ke{\Vert }^2+\alpha \Vert Be{\Vert }^2=-\alpha {\beta }^2⟨z,e⟩$. Insert an approximate $K^{\ast }w$ and complete the square in $\Vert Ke\Vert$ to obtain $\Vert e{\Vert }^2\le \Vert z-K^{\ast }w\Vert \Vert e\Vert +\alpha {\beta }^2{r}^2/4$. This implies the displayed bound, even with coefficient $1/2$ on its last term, and proves consistency without a fixed exact source condition for quadratic regularization.

For bounded linear operators $K:U\to V$ and $B:U\to W$ between real Hilbert spaces, $\alpha >0$ and $\Vert Bu\Vert \ge \beta \Vert u\Vert$ with $\beta >0$, the unique minimizer of $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$ satisfies $(K^{\ast }K+\alpha B^{\ast }B)u_{\alpha }=K^{\ast }f$. Differentiate the objective in every direction and use the adjoint operator to obtain this equation. The coercive quadratic penalty makes its left-hand operator boundedly invertible. Setting $B=I$ gives the Tikhonov normal equation; omitting the penalty instead gives the normal equation for a linear inverse problem.