Deterministic source classes and optimal recovery — Examples and Counterexamples

For $0<t\le 1$, $1\mathrm{/}t\ge 1$, so [F8] gives $\log (1\mathrm{/}t)\ge \log 1=0$. Thus $1+\log (1\mathrm{/}t)\ge 1$; since $q>0$, the positive power of this base is at least $1$, and its reciprocal ${\psi }_q(t)$ lies in $(0,1]$. Together with ${\psi }_q(0)=0$, this verifies that the logarithmic index is bounded and is admissible under [F6].

The bijection $Ux=(x_{n+1}{)}_{n\in \mathbb{N}}$ from ${\mathrm{\ell }}^2(I,\mathbb{R})$ to ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{R})$ preserves the coordinate inner product, so [F1] transfers completeness to ${\mathrm{\ell }}^2(I,\mathbb{R})$. For $x=(x_j)\in {\mathrm{\ell }}^2(I)$, $\mathrm{\parallel }{K}_{a}x{\mathrm{\parallel }}_2^2={\sum }_{j\in I}{j}^{-2a}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so $K_a$ is a well-defined bounded linear map with $\mathrm{\parallel }{K}_a\mathrm{\parallel }\le 1$; $K_a{e}_1=e_1$ gives $\mathrm{\parallel }{K}_a\mathrm{\parallel }\ge 1$. For $x,y\in {\mathrm{\ell }}^2(I)$, [F2] makes ${\sum }_{j\in I}{j}^{-a}{x}_j{y}_j$ absolutely convergent, and the coordinate inner product gives $⟨K_{a}x,y⟩={\sum }_{j\in I}{j}^{-a}{x}_j{y}_j=⟨x,K_{a}y⟩$. Uniqueness of the adjoint in [F3] yields $K_a^{\ast }=K_a$, and $⟨K_{a}x,x⟩={\sum }_{j\in I}{j}^{-a}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\ge 0$, so $K_a$ is positive.

For $N\ge 1$ define $(K_{a,N}x{)}_j=j^{-a}{x}_j$ for $j\le N$ and $0$ for $j>N$. The coordinate estimate from step 1.2 gives $\mathrm{\parallel }{K}_{a,N}x{\mathrm{\parallel }}_2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2$, so $K_{a,N}$ is bounded; since each $j^{-a}>0$ for $j\le N$, its range is exactly $\mathrm{span}\{e_1,\dots ,e_N\}$, so [F4] makes it compact. For every $x$, $\mathrm{\parallel }(K_a-K_{a,N})x{\mathrm{\parallel }}_2^2={\sum }_{j>N}{j}^{-2a}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le (N+1{)}^{-2a}\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, with equality at $x=e_{N+1}$; therefore $\mathrm{\parallel }{K}_a-K_{a,N}\mathrm{\parallel }=(N+1{)}^{-a}$. Given $\epsilon >0$, [F8] supplies a threshold $N$ with $1\mathrm{/}j<{\epsilon }^{1\mathrm{/}a}$ for all $j\ge N+1$; monotonicity of $t\mapsto t^a$ gives $(N+1{)}^{-a}<\epsilon$. Thus $K_{a,N}\to K_a$ in norm, and [F4] makes $K_a$ compact. Each diagonal coefficient is positive, so $K_{a}x=0$ forces every $x_j=0$ and $K_a$ is injective.

Since $K_a$ is compact, self-adjoint and positive by steps 1.2–2.1, $K_a$ is the unique positive square root of $K_a^{\ast }{K}_a=K_a^2$; [F5] therefore gives $\mathrm{\mid }{K}_a\mathrm{\mid }=K_a$. Each $e_j$ is an eigenvector of $\mathrm{\mid }{K}_a\mathrm{\mid }$ with eigenvalue $j^{-a}$. Applying the compact spectral multiplier [F5] coordinatewise yields $(\mathrm{\mid }{K}_a{\mathrm{\mid }}^{\nu }w{)}_j=j^{-a\nu }{w}_j$ and $({\psi }_q(\mathrm{\mid }{K}_a\mathrm{\mid })w{)}_j={\psi }_q(j^{-a})w_j=(1+a\log j{)}^{-q}{w}_j$, proving the stated power and logarithmic coordinates.

For each $j\ge 1$ take $w^{(j)}=\rho e_j$ and $x^{(j)}=\mathrm{\mid }{K}_a{\mathrm{\mid }}^{\nu }{w}^{(j)}=\rho j^{-a\nu }{e}_j$. Then $\mathrm{\parallel }{K}_a{x}^{(j)}{\mathrm{\parallel }}_2=\rho j^{-a(\nu +1)}={\delta }_j$ and $\mathrm{\parallel }{x}^{(j)}{\mathrm{\parallel }}_2=\rho j^{-a\nu }={\delta }_j^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$. This feasible source-ball element attains the universal upper bound from [F7], so equality holds at ${\delta }_j$.

For $0<t\le 1$, [F8] gives $\log (1\mathrm{/}t)\ge 0$, so $1+\log (1\mathrm{/}t)\ge 1$; since $q>0$, $(1+\log (1\mathrm{/}t){)}^q\ge 1$ and ${\psi }_q(t)\le 1$. Together with ${\psi }_q(0)=0$, this verifies that the logarithmic index is bounded and admissible under [F6].

The bijection $Ux=(x_{n+1}{)}_{n\in \mathbb{N}}$ from ${\mathrm{\ell }}^2(I,\mathbb{R})$ to ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{R})$ preserves the coordinate inner product, so [F1] transfers completeness to ${\mathrm{\ell }}^2(I,\mathbb{R})$. Since ${\gamma }_j\ge 0$, strict increase of the exponential [F10] gives $0<\exp (-{\gamma }_j)\le 1$. Thus for $x=(x_j)\in {\mathrm{\ell }}^2(I)$, $\mathrm{\parallel }{K}_{\gamma }x{\mathrm{\parallel }}_2^2={\sum }_{j\in I}\exp (-2{\gamma }_j)\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so $K_{\gamma }$ is bounded with $\mathrm{\parallel }{K}_{\gamma }\mathrm{\parallel }\le 1$; $K_{\gamma }{e}_1=e_1$ gives $\mathrm{\parallel }{K}_{\gamma }\mathrm{\parallel }=1$. For $x,y\in {\mathrm{\ell }}^2(I)$, [F2] makes ${\sum }_{j\in I}\exp (-{\gamma }_j)x_j{y}_j$ absolutely convergent, and the coordinate inner product gives $⟨K_{\gamma }x,y⟩={\sum }_{j\in I}\exp (-{\gamma }_j)x_j{y}_j=⟨x,K_{\gamma }y⟩$. By uniqueness in [F3], $K_{\gamma }^{\ast }=K_{\gamma }$; moreover $⟨K_{\gamma }x,x⟩={\sum }_{j\in I}\exp (-{\gamma }_j)\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\ge 0$, so $K_{\gamma }$ is positive.

Since ${\gamma }_j\to \mathrm{\infty }$, [F9] gives $s_j=\exp (-{\gamma }_j)\to 0$. For $N\ge 1$, truncate $K_{\gamma }$ by $(K_{\gamma ,N}x{)}_j=s_j{x}_j$ for $j\le N$ and $0$ for $j>N$. By step 1.2, $K_{\gamma ,N}$ is bounded; since each $s_j>0$ for $j\le N$, its range is exactly $\mathrm{span}\{e_1,\dots ,e_N\}$, so [F4] makes it compact. Since $s_j$ decreases to $0$, $\mathrm{\parallel }{K}_{\gamma }-K_{\gamma ,N}\mathrm{\parallel }={\sup }_{j>N}{s}_j=s_{N+1}\to 0$; [F4] then makes $K_{\gamma }$ compact, and positivity of each $s_j$ makes it injective. The choices ${\gamma }_j=j-1$ and ${\gamma }_j=j(j-1)\mathrm{/}2$ are strictly increasing from $0$ to infinity; for the second, ${\gamma }_j\ge j-1$ for $j\ge 2$ and [F8] makes the natural numbers cofinal. Its gaps are ${\gamma }_{j+1}-{\gamma }_j=j$, so [F10] gives $s_{j+1}\mathrm{/}{s}_j=\exp (-{\gamma }_{j+1})\mathrm{/}\exp (-{\gamma }_j)=\exp (-j)\to 0$ by [F8]–[F9].

By steps 1.2 and 2.1, $K_{\gamma }$ is compact, self-adjoint, and positive, and $K_{\gamma }^{\ast }{K}_{\gamma }=K_{\gamma }^2$. Thus $K_{\gamma }$ is itself a compact positive square root of $K_{\gamma }^{\ast }{K}_{\gamma }$, so uniqueness in [F5] gives $\mathrm{\mid }{K}_{\gamma }\mathrm{\mid }=K_{\gamma }$. Hence each coordinate vector has singular value $s_j=\exp (-{\gamma }_j)$; [F5] gives $(\mathrm{\mid }{K}_{\gamma }{\mathrm{\mid }}^{\nu }w{)}_j=\exp (-\nu {\gamma }_j)w_j$ and $({\psi }_q(\mathrm{\mid }{K}_{\gamma }\mathrm{\mid })w{)}_j=(1+\log (1\mathrm{/}{s}_j){)}^{-q}{w}_j=(1+{\gamma }_j{)}^{-q}{w}_j$.

For each $j$, take $w^{(j)}=\rho e_j$ and $x^{(j)}=\mathrm{\mid }{K}_{\gamma }{\mathrm{\mid }}^{\nu }{w}^{(j)}=\rho s_j^{\nu }{e}_j$ by step 3.1. Then $\mathrm{\parallel }{K}_{\gamma }{x}^{(j)}{\mathrm{\parallel }}_2=\rho s_j^{\nu +1}$ and $\mathrm{\parallel }{x}^{(j)}{\mathrm{\parallel }}_2=\rho s_j^{\nu }=(\rho s_j^{\nu +1}{)}^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$, so [F7] gives equality at that noise radius.

Take ${\gamma }_j=j(j-1)\mathrm{/}2$, write $r_j=s_{j+1}\mathrm{/}{s}_j=\exp (-j)$, and set ${\delta }_j^{\star }=\rho (s_j{s}_{j+1}{)}^{(\nu +1)\mathrm{/}2}$. For any $x=\mathrm{\mid }{K}_{\gamma }{\mathrm{\mid }}^{\nu }w$ with $\mathrm{\parallel }w{\mathrm{\parallel }}_2\le \rho$ and $\mathrm{\parallel }{K}_{\gamma }x{\mathrm{\parallel }}_2\le {\delta }_j^{\star }$, step 3.1 gives $x_k=s_k^{\nu }{w}_k$ and $(K_{\gamma }x{)}_k=s_k^{\nu +1}{w}_k$. Since $s_k\ge s_j$ for $k\le j$, ${\sum }_{k\le j}\mathrm{\mid }{x}_k{\mathrm{\mid }}^2\le s_j^{-2}{\sum }_{k\le j}\mathrm{\mid }(K_{\gamma }x{)}_k{\mathrm{\mid }}^2\le ({\delta }_j^{\star }{)}^2\mathrm{/}{s}_j^2$; since $s_k\le s_{j+1}$ for $k>j$, ${\sum }_{k>j}\mathrm{\mid }{x}_k{\mathrm{\mid }}^2\le s_{j+1}^{2\nu }{\sum }_{k>j}\mathrm{\mid }{w}_k{\mathrm{\mid }}^2\le {\rho }^2{s}_{j+1}^{2\nu }$. Thus $\mathrm{\parallel }x{\mathrm{\parallel }}_2^2\le ({\delta }_j^{\star }{)}^2\mathrm{/}{s}_j^2+{\rho }^2{s}_{j+1}^{2\nu }$. With $M_j=({\delta }_j^{\star }{)}^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}=\rho (s_j{s}_{j+1}{)}^{\nu \mathrm{/}2}$, this gives $\mathrm{\parallel }x{\mathrm{\parallel }}_2^2\mathrm{/}{M}_j^2\le r_j+r_j^{\nu }$. By [F8], $r_j^{\nu }=\exp (-\nu j)\to 0$ by [F9], since $\nu >0$ and the natural numbers are cofinal; likewise $r_j=\exp (-j)\to 0$. Taking the supremum over such feasible $x$ using [F7] gives ${\epsilon }_{K_{\gamma }}(X_{\nu ,\rho }(K_{\gamma }),{\delta }_j^{\star })\mathrm{/}{M}_j\le \sqrt{r_j+r_j^{\nu }}\to 0$. By step 2.1, ${\gamma }_j\to \mathrm{\infty }$; [F8] and the real-power definition give ${\delta }_j^{\star }=\rho \exp (-\frac{\nu +1}{2}({\gamma }_j+{\gamma }_{j+1}))\to 0$ by [F9]. Therefore, for every $c>0$ and ${\delta }_0>0$, some $j$ has ${\delta }_j^{\star }<{\delta }_0$ and ${\epsilon }_{K_{\gamma }}(X_{\nu ,\rho }(K_{\gamma }),{\delta }_j^{\star })<c{M}_j$, ruling out a positive constant lower bound by the power scale at every sufficiently small noise radius.