Tikhonov source-condition rates balance noise at qualification one

Let $x_{\lambda }=(A^{\ast }A+\lambda I{)}^{-1}{A}^{\ast }A{x}^{†}$. By linearity, $x_{\lambda }^{\delta }-x^{†}=(x_{\lambda }-x^{†})+R_{\lambda }e$. The triangle inequality, [F1], and [F2] give $\mathrm{\parallel }{x}_{\lambda }^{\delta }-x^{†}\mathrm{\parallel }\le R{\lambda }^{\nu }+\frac{\delta }{2\sqrt{\lambda }}\mathrm{.}$

Fix $\nu >1$ and take ${\mathbb{N}}_{+}=\{1,2,\dots \}$ and $X=Y={\mathrm{\ell }}^2({\mathbb{N}}_{+},\mathbb{R})$ with standard unit vectors $e_n$. Relabelling coordinates by the bijection $k\mapsto k+1$ from $\mathbb{N}$ to ${\mathbb{N}}_{+}$ is an isometry from the complete space in [F5], so $X=Y$ is Hilbert. For $x=(x_n)\in {\mathrm{\ell }}^2$, define $(Ax{)}_n=x_n\mathrm{/}n$ and, for each $N\ge 1$, define $(A_{N}x{)}_n=x_n\mathrm{/}n$ when $n\le N$ and $0$ otherwise. These are linear and bounded because $\mathrm{\parallel }Ax{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }{A}_{N}x{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$; $A_N$ has range in $\mathrm{span}(e_1,\dots ,e_N)$. Also $\mathrm{\parallel }(A-A_N)x{\mathrm{\parallel }}_2^2={\sum }_{n>N}{n}^{-2}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le (N+1{)}^{-2}\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, with equality at $x=e_{N+1}$, so $\mathrm{\parallel }A-A_N\mathrm{\parallel }=1\mathrm{/}(N+1)\to 0$. By [F4]–[F7], $X=Y$ is Hilbert and $A$ is compact. Since every diagonal coefficient is positive, $A$ is injective; $S{e}_1=e_1$, so with $w=e_1$ the source target is $x^{†}=S^{\nu }w=e_1$ and $\mathrm{\parallel }w\mathrm{\parallel }=1$.

Substituting $\lambda =(\delta \mathrm{/}R{)}^{2\mathrm{/}(2\nu +1)}$ into the two terms in step 1.1 gives $R{\lambda }^{\nu }=R^{1\mathrm{/}(2\nu +1)}{\delta }^{2\nu \mathrm{/}(2\nu +1)}$ and $\delta \mathrm{/}(2\sqrt{\lambda })=\frac{1}{2}{R}^{1\mathrm{/}(2\nu +1)}{\delta }^{2\nu \mathrm{/}(2\nu +1)}$. Their sum is the stated rate.

For exact data $A{x}^{†}=e_1$, the first coordinate of the Tikhonov solution is $1\mathrm{/}(1+\lambda )$ and all other coordinates are zero. Hence $\mathrm{\parallel }{x}^{†}-x_{\lambda }\mathrm{\parallel }=\lambda \mathrm{/}(1+\lambda )$, and $\frac{\mathrm{\parallel }{x}^{†}-x_{\lambda }\mathrm{\parallel }}{{\lambda }^{\nu }}=\frac{{\lambda }^{1-\nu }}{1+\lambda }⟶+\mathrm{\infty }(\lambda \downarrow 0)\mathrm{.}$ Thus no constant depending only on $\nu$ and the source radius can give an $O(R{\lambda }^{\nu })$ bias bound uniformly for this compact-operator class when $\nu >1$. This proves saturation at qualification one.