= Solution
Write $u^\dagger=K^\dagger f$, $e=R_\alpha f-u^\dagger$, and $z=\beta^{-2}B^*Bu^\dagger$, with $\beta>0$ as required by coercivity and by the printed inverse power. The <Tikhonov regularization with a coercive penalty operator> normal equation gives
$$
(K^*K+\alpha B^*B)e=-\alpha B^*Bu^\dagger.
$$
This also holds for inconsistent data in $\mathcal D(K^\dagger)$, because $f-Ku^\dagger\perp\mathcal R(K)$ and therefore $K^*f=K^*Ku^\dagger$. Taking the inner product with $e$ gives
$$
\|Ke\|^2+\alpha\|Be\|^2=-\alpha\beta^2\langle z,e\rangle.
$$
For any $w$ with $\|w\|\leq r$, put $d=\|z-K^*w\|$ and use <Cauchy-Schwarz inequality> to bound the right-hand side by $\alpha\beta^2(d\|e\|+r\|Ke\|)$. Since $\|Be\|\geq\beta\|e\|$, completing the square in $\|Ke\|$ yields
$$
\alpha\beta^2\|e\|^2\leq\alpha\beta^2d\|e\|+\frac{\alpha^2\beta^4r^2}{4}.
$$
The positive root of this quadratic inequality is at most $d+\sqrt\alpha\beta r/2$. Infimizing over $w$ proves the <approximate source bound for coercive Tikhonov regularization>, in fact slightly more strongly:
$$
\boxed{\|R_\alpha f-K^\dagger f\|\leq\eta_r+\tfrac12\sqrt\alpha\beta r\leq\eta_r+\sqrt\alpha\beta r.}
$$
Injectivity gives $\overline{\mathcal R(K^*)}=\mathcal N(K)^\perp=U$. Given any positive tolerance, approximate $z$ by $K^*w$ for some finite-norm $w$; for all sufficiently large $r$ it is admissible in the infimum. Thus
$$
\boxed{\eta_r\downarrow0\quad(r\to\infty).}
$$
No attainment of that infimum is needed.
Back to article page
