Parameter rules and the noise-free veto — Examples and Counterexamples

Since $A=I_2$, the formula in [F1] gives $x_{\lambda }=(1+\lambda {)}^{-1}{e}_1$, residual norm $r_{\lambda }=\mathrm{\parallel }A{x}_{\lambda }-b{\mathrm{\parallel }}_2=\lambda \mathrm{/}(1+\lambda )$, and solution norm $s_{\lambda }=\mathrm{\parallel }{x}_{\lambda }{\mathrm{\parallel }}_2=1\mathrm{/}(1+\lambda )$. At the three grid values, $(r_{\lambda },s_{\lambda })$ equals $(1\mathrm{/}10,9\mathrm{/}10)$, $(1\mathrm{/}2,1\mathrm{/}2)$, and $(9\mathrm{/}10,1\mathrm{/}10)$, respectively. All retained points are valid because both coordinates are positive.

Their log-log points are $p_0=(-\log 10,\log (9\mathrm{/}10))$, $p_1=(-\log 2,-\log 2)$, and $p_2=(\log (9\mathrm{/}10),-\log 10)$. Thus $u=p_1-p_0=(\log 5,-\log (9\mathrm{/}5))$ and $v=p_2-p_1=(\log (9\mathrm{/}5),-\log 5)$. Their determinant is $-{\log }^{2}5+{\log }^2(9\mathrm{/}5)\mathrm{\ne }0$ because $5>9\mathrm{/}5>1$; moreover $\mathrm{\parallel }u{\mathrm{\parallel }}_2=\mathrm{\parallel }v{\mathrm{\parallel }}_2>0$ and $u+v=(\log 9,-\log 9)\mathrm{\ne }0$. The algorithm's curvature is therefore well-defined and positive.

There are exactly three retained grid indices, so the only consecutive triple is $(1,2,3)$ and its curvature is assigned to index $2$. This is the only candidate for the maximum, so the selector returns ${\lambda }_2=1$; no tie-break is needed. The computation supplies no claim about any grid outside this input.

Coordinatewise linearity gives $K$ and every $K_N$ as linear maps by [F13]. For every $x\in {\mathrm{\ell }}^2$, [F1] gives $\mathrm{\parallel }Kx{\mathrm{\parallel }}_2^2={\sum }_{j\ge 1}{j}^{-2}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }{K}_{N}x{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so [F3] makes both bounded linear operators. For $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le 1$, [F1] also gives $\mathrm{\parallel }(K-K_N)x{\mathrm{\parallel }}_2^2={\sum }_{j>N}{j}^{-2}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le (N+1{)}^{-2}$. Hence [F4] gives $\mathrm{\parallel }K-K_N\mathrm{\parallel }\le 1\mathrm{/}(N+1)$. The range of $K_N$ is exactly the span of $e_1,\dots ,e_N$: its outputs are supported in these coordinates, and $K_N(j{e}_j)=e_j$ for each $j\le N$ by [F14]. These vectors form an ordered basis by [F5], so each $K_N$ has finite rank.

By [F6], every $K_N$ is compact in the sense of [F9]. By [F2] and [F7], the codomain ${\mathrm{\ell }}^2$ is Banach; [A1] supplies the Countable Choice hypothesis of [F8]. Since $\mathrm{\parallel }K-K_N\mathrm{\parallel }\to 0$ by step 1.1, [F8] implies that $K$ is compact and hence satisfies the compact-model interface [F10].

The coordinate formula makes $K$ injective. If $y=Kx$, then $x_j=j{y}_j$ and $(j{y}_j{)}_{j\ge 1}\in {\mathrm{\ell }}^2$; conversely, whenever $(j{y}_j{)}_{j\ge 1}\in {\mathrm{\ell }}^2$, taking $x_j=j{y}_j$ gives $Kx=y$. Thus $R(K)=\{y:(j{y}_j{)}_{j\ge 1}\in {\mathrm{\ell }}^2\}$. Each $e_j\mathrm{/}j=K{e}_j$ lies in $R(K)$, and these vectors are linearly independent because they have distinct nonzero coordinates by [F14]; hence $R(K)$ is infinite-dimensional. Any vector in $R(K{)}^{\perp }$ has every coordinate zero, so $R(K{)}^{\perp }=\{0\}$ and $D(K^{†})=R(K)$. By [F11], $K^{†}y=(j{y}_j{)}_{j\ge 1}$. For each fixed $\alpha >0$, $R_{\alpha }$ is bounded since its sum has finitely many terms and $\mathrm{\parallel }{R}_{\alpha }y{\mathrm{\parallel }}_2^2\le \lfloor 1\mathrm{/}\alpha {\rfloor }^2\mathrm{\parallel }y{\mathrm{\parallel }}_2^2$. For $y\in D(K^{†})$, $R_{\alpha }y$ is the initial truncation of the square-summable sequence $K^{†}y$, so as $\alpha \downarrow 0$ these truncations converge in ${\mathrm{\ell }}^2$ to $K^{†}y$. The stated family is therefore pointwise regularizing.

The rule $\alpha$ is positive and depends only on $z$, so it is noise-level-free by [F12]. Take exact datum $y=0$ and noisy datum $z_j=e_j\mathrm{/}j=K{e}_j$. By [F1] and [F14], ${\delta }_j=\mathrm{\parallel }{z}_j-y{\mathrm{\parallel }}_2=1\mathrm{/}j\to 0$, while $\alpha (z_j)=1\mathrm{/}{j}^2$ and $\lfloor 1\mathrm{/}\alpha (z_j)\rfloor =j^2$. Therefore $R_{\alpha (z_j)}{z}_j=e_j$, whereas $K^{†}y=0$, and the reconstruction error is $1$ for every $j$. The claimed worst-case limit fails at $y=0$, since its supremum is at least $1$ along the null sequence ${\delta }_j$.

By [F6] with $\nu =1$, the equality radii ${\delta }_j=\rho s_j^2$ form a null sequence; since $\rho >0$, $s_j\to 0$. Put $y_j:=K{v}_j=s_j{u}_j$. Then $\mathrm{\parallel }{y}_j{\mathrm{\parallel }}_Y=s_j\to 0$ by [F2]. Since $v_j\in N(K{)}^{\perp }$, [F3] gives $K^{†}{y}_j=v_j$, whose norm is $1$. Thus $K^{†}$ is discontinuous at $0$ on its domain.

By [F4]–[F6] with $\nu =1$, ${\epsilon }_K(M_{\rho },\delta )\le {\delta }^{1\mathrm{/}2}{\rho }^{1\mathrm{/}2}$, and for each $j$, $x_j=\mathrm{\mid }K\mathrm{\mid }(\rho v_j)=\rho s_j{v}_j$ lies in $M_{\rho }$, has $\mathrm{\parallel }K{x}_j{\mathrm{\parallel }}_Y=\rho s_j^2={\delta }_j$, and has $\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_X=\rho s_j=\sqrt{\rho {\delta }_j}$. This attains the bound. Since ${\delta }_j\to 0$ by [F6], the upper bound tends to zero as $\delta \downarrow 0$. The estimate is restricted to $M_{\rho }$ and does not assert the universal convergence condition in [F7].