Landweber discrepancy stopping converges under a source condition

Put $p=\nu +\frac{1}{2}$. By [F2], $x_k-x^{†}=-r_k(T)T^{\nu }w$. The compact spectral expansion [F6], the power convention [F3], and $\mathrm{\parallel }Az{\mathrm{\parallel }}^2=⟨A^{\ast }Az,z⟩$ from [F4] give $\mathrm{\parallel }A{x}_k-y\mathrm{\parallel }=\mathrm{\parallel }{T}^p{r}_k(T)w\mathrm{\parallel }$. The scalar filter bound [F1] therefore gives $\mathrm{\parallel }A{x}_k-y\mathrm{\parallel }\le C_{p,A,\omega }R{k}^{-p}$. Choose $K_{\delta }=\max \text{\hspace{0.17em}⁣}\{1,\lceil {\left(\frac{C_{p,A,\omega }R}{(\tau -1)\delta }\right)}^{1\mathrm{/}p}\rceil \}\mathrm{.}$ Then $\mathrm{\parallel }A{x}_{K_{\delta }}-y\mathrm{\parallel }\le (\tau -1)\delta$. By [F2], the noisy residual differs from the exact-data residual by $r_{K_{\delta }}(A{A}^{\ast })(y_{\delta }-y)$. [F7] makes $A{A}^{\ast }$ compact positive self-adjoint, and [F4] gives $\mathrm{\parallel }A{A}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }A{\mathrm{\parallel }}^2$. Every eigenvalue $t$ of $A{A}^{\ast }$ therefore satisfies $0\le t\le \mathrm{\parallel }A{\mathrm{\parallel }}^2$, so $\mathrm{\mid }1-\omega t\mathrm{\mid }\le 1$ by the step-size hypothesis. [F5] bounds the noise residual by $\delta$. Hence $\mathrm{\parallel }A{x}_{K_{\delta }}^{\delta }-y_{\delta }\mathrm{\parallel }\le \tau \delta$, so the first crossing exists and $k_{\delta }\le K_{\delta }$ (provided the finite budget reaches $K_{\delta }$).

At the crossing, [F2] and the same multiplier estimate [F5] give $\mathrm{\parallel }A{x}_{k_{\delta }}-y\mathrm{\parallel }\le \mathrm{\parallel }A{x}_{k_{\delta }}^{\delta }-y_{\delta }\mathrm{\parallel }+\delta \le (\tau +1)\delta$.

Let $v=r_{k_{\delta }}(T)w$ and $\theta =2\nu \mathrm{/}(2\nu +1)$. For a finite set $E\subseteq \mathrm{\Sigma }$, put $a_t=t^{2\nu +1}\mathrm{\parallel }{P}_{t}v{\mathrm{\parallel }}^2$ and $b_t=\mathrm{\parallel }{P}_{t}v{\mathrm{\parallel }}^2$. Then $t^{2\nu }\mathrm{\parallel }{P}_{t}v{\mathrm{\parallel }}^2=a_t^{\theta }{b}_t^{1-\theta }$. If either ${\sum }_E{a}_t$ or ${\sum }_E{b}_t$ is zero, every summand on the left is zero. Otherwise normalize ${\tilde{a}}_t=a_t\mathrm{/}({\sum }_E{a}_t)$ and ${\tilde{b}}_t=b_t\mathrm{/}({\sum }_E{b}_t)$. The scalar inequality $u^{\theta }{v}^{1-\theta }\le \theta u+(1-\theta )v$ follows by minimizing $f(s)=\theta s+(1-\theta )-s^{\theta }$: $f^{\mathrm{\prime }}(s)=\theta (1-s^{\theta -1})$ is negative for $0<s<1$ and positive for $s>1$, so $f(s)\ge f(1)=0$. Thus the sum of the normalized products is at most $1$, and $\mathrm{\parallel }{T}^{\nu }v\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\nu +1\mathrm{/}2}v{\mathrm{\parallel }}^{2\nu \mathrm{/}(2\nu +1)}\mathrm{\parallel }v{\mathrm{\parallel }}^{1\mathrm{/}(2\nu +1)}\mathrm{.}$ Since [F6] gives the norm-convergent spectral expansion, increasing finite sets $E$ to $\mathrm{\Sigma }$ passes this inequality to the full sums. Also $\mathrm{\parallel }v\mathrm{\parallel }\le R$ because $\mathrm{\mid }{r}_{k_{\delta }}(t)\mathrm{\mid }\le 1$. Finally [F2], [F3], [F4], [F6], and [step 1.2] give $x^{†}-x_{k_{\delta }}=T^{\nu }v$. For every $z\in X$, $\mathrm{\parallel }Az{\mathrm{\parallel }}^2=⟨Tz,z⟩={\sum }_{t\in \mathrm{\Sigma }}t\mathrm{\parallel }{P}_{t}z{\mathrm{\parallel }}^2=\mathrm{\parallel }{T}^{1\mathrm{/}2}z{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }{T}^{\nu +1\mathrm{/}2}v\mathrm{\parallel }=\mathrm{\parallel }A{T}^{\nu }v\mathrm{\parallel }=\mathrm{\parallel }A{x}_{k_{\delta }}-y\mathrm{\parallel }\le (\tau +1)\delta$. Hence $\mathrm{\parallel }{x}_{k_{\delta }}-x^{†}\mathrm{\parallel }=\mathrm{\parallel }{T}^{\nu }v\mathrm{\parallel }\le ((\tau +1)\delta {)}^{2\nu \mathrm{/}(2\nu +1)}{R}^{1\mathrm{/}(2\nu +1)}\mathrm{.}$

From [F1], $\mathrm{\parallel }{x}_{k_{\delta }}^{\delta }-x_{k_{\delta }}\mathrm{\parallel }\le \sqrt{2\omega K_{\delta }}\delta$. Put $B=C_{p,A,\omega }R\mathrm{/}(\tau -1)$, choosing $C_{p,A,\omega }>0$. For sufficiently small $\delta$, $B\mathrm{/}\delta \ge 1$, so $K_{\delta }=\lceil (B\mathrm{/}\delta {)}^{1\mathrm{/}p}\rceil \le 2(B\mathrm{/}\delta {)}^{1\mathrm{/}p}$. Therefore $\sqrt{2\omega K_{\delta }}\delta \le 2\sqrt{\omega }{B}^{1\mathrm{/}(2p)}{\delta }^{1-1\mathrm{/}(2p)}=C^{\mathrm{\prime }}{R}^{1\mathrm{/}(2\nu +1)}{\delta }^{2\nu \mathrm{/}(2\nu +1)},$ because $2p=2\nu +1$. Combining this with [step 2.1] by the triangle inequality proves the claim.