Deterministic source classes and optimal recovery

By [F1], write $w=w_0+{\sum }_j{w}_j{v}_j$ in norm, with $w_0\in N(K)$ and the $v_j$ an orthonormal basis of $N(K{)}^{\perp }$. The compact spectral multiplier in [F2] gives $x=\psi (\mathrm{\mid }K\mathrm{\mid })w=\psi (0)w_0+{\sum }_j\psi (s_j)w_j{v}_j={\sum }_j\psi (s_j)w_j{v}_j$. This norm-convergent sum is a limit of finite linear combinations of the $v_j$, so [F1] places $x$ in their closed span $N(K{)}^{\perp }$; if the positive singular set is empty, the sum and closed span are both zero. Thus $x\perp N(K)$.

If $Kz=Kx$, then $n:=z-x\in N(K)$. By step 1.1, $x\perp n$, so [F3] gives $\mathrm{\parallel }z{\mathrm{\parallel }}^2=\mathrm{\parallel }x+n{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }n{\mathrm{\parallel }}^2\ge \mathrm{\parallel }x{\mathrm{\parallel }}^2$, with equality only when $n=0$, that is, $z=x$. Thus $x$ is the unique minimum-norm solution.

Let $J$ index the positive singular values in [F1]. Write $w=w_0+{\sum }_{j\in J}{w}_j{v}_j$ in norm, with $w_0\in N(K)$. By [F2]–[F3], $x={\sum }_{j\in J}{s}_j^{\nu }{w}_j{v}_j$ and $Kx={\sum }_{j\in J}{s}_j^{\nu +1}{w}_j{u}_j$, with norm-convergent expansions. For each finite $F\subseteq J$, orthonormality gives the squared norms of the partial sums as the sums of squared coefficients; taking limits by continuity of the norm gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X^2={\sum }_{j\in J}{c}_j$ and $\mathrm{\parallel }Kx{\mathrm{\parallel }}_Y^2={\sum }_{j\in J}{a}_j$. The same finite-sum identity and $w_0\perp v_j$ give $\mathrm{\parallel }w{\mathrm{\parallel }}_X^2=\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_X^2+{\sum }_{j\in J}{b}_j\ge {\sum }_{j\in J}{b}_j$, where $c_j=s_j^{2\nu }\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$, $a_j=s_j^{2\nu +2}\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$, and $b_j=\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$.

Put $\theta =\nu \mathrm{/}(\nu +1)\in (0,1)$. For a finite $F\subseteq J$, set $A_F={\sum }_{j\in F}{a}_j$, $B_F={\sum }_{j\in F}{b}_j$, and $C_F={\sum }_{j\in F}{c}_j$. If $A_F=0$ or $B_F=0$, then every $w_j$ with $j\in F$ is zero because $s_j>0$, so $C_F=0$. Otherwise, for $j\in F$ put $p_j=a_j\mathrm{/}{A}_F$ and $q_j=b_j\mathrm{/}{B}_F$. The scalar inequality $t^{\theta }\le \theta t+1-\theta$ holds for $t\ge 0$: for $t>0$ set $g(t)=\theta t+1-\theta -t^{\theta }$. By [F4], $g^{\mathrm{\prime }}(t)=\theta (1-t^{\theta -1})$. If $0<t<1$, strict increase of $u\mapsto u^{1-\theta }$ gives $t^{1-\theta }<1^{1-\theta }=1$; by [F4] and inverse order this implies $t^{\theta -1}>1$ and $g^{\mathrm{\prime }}(t)<0$. If $t>1$, the inequalities reverse and $g^{\mathrm{\prime }}(t)>0$. The mean value theorem then gives $g(t)\ge g(1)=0$ for every $t>0$, while $g(0)=1-\theta >0$. If $q_j>0$, apply the scalar inequality to $t=p_j\mathrm{/}{q}_j$ to obtain $p_j^{\theta }{q}_j^{1-\theta }\le \theta p_j+(1-\theta )q_j$; if $q_j=0$, the same inequality is immediate. Since $c_j=a_j^{\theta }{b}_j^{1-\theta }$, summing and using $\sum p_j=\sum q_j=1$ gives $C_F\le A_F^{\theta }{B}_F^{1-\theta }$.

Exhaust $J$ by its first $n$ indices when $J$ is infinite, and take $F=J$ when it is finite; if $J$ is empty, all three sums in step 1.1 are zero. Passing to the limit in step 1.2 gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X^2\le \mathrm{\parallel }Kx{\mathrm{\parallel }}_Y^{2\theta }\mathrm{\parallel }w{\mathrm{\parallel }}_X^{2(1-\theta )}$. Taking nonnegative square roots and substituting $\theta =\nu \mathrm{/}(\nu +1)$ proves the claimed inequality.

The class $X_{\nu ,\rho }(K)$ contains $0$ by taking $w=0$ in [F1]. It is symmetric: if $x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ with $\mathrm{\parallel }w{\mathrm{\parallel }}_X\le \rho$, then $-x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }(-w)$ with $\mathrm{\parallel }-w{\mathrm{\parallel }}_X\le \rho$. It is convex: for $x_i=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }{w}_i$ with $\mathrm{\parallel }{w}_i{\mathrm{\parallel }}_X\le \rho$ and $0\le t\le 1$, $t{x}_1+(1-t)x_2=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }(t{w}_1+(1-t)w_2)$, while the triangle inequality gives $\mathrm{\parallel }t{w}_1+(1-t)w_2{\mathrm{\parallel }}_X\le t\rho +(1-t)\rho =\rho$. By [F3], this class is contained in $N(K{)}^{\perp }$, so [F2] applies.

Let $x\in X_{\nu ,\rho }(K)$ satisfy $\mathrm{\parallel }Kx{\mathrm{\parallel }}_Y\le \delta$. The two admissible truths $x$ and $-x$ with noises $-Kx$ and $Kx$ both produce the same observed datum $0$. Their reconstruction errors are $\mathrm{\parallel }A(0)-x{\mathrm{\parallel }}_X$ and $\mathrm{\parallel }A(0)+x{\mathrm{\parallel }}_X$, and the triangle inequality yields $2\mathrm{\parallel }x{\mathrm{\parallel }}_X=\mathrm{\parallel }(A(0)+x)-(A(0)-x){\mathrm{\parallel }}_X\le \mathrm{\parallel }A(0)+x{\mathrm{\parallel }}_X+\mathrm{\parallel }A(0)-x{\mathrm{\parallel }}_X$. Hence the larger error is at least $\mathrm{\parallel }x{\mathrm{\parallel }}_X$, so taking the supremum over all such $x$ gives $E_{\delta }(A)\ge {\epsilon }_K(X_{\nu ,\rho }(K),\delta )$. Taking the infimum over $A$ preserves this lower bound.

If $x_1,x_2\in X_{\nu ,\rho }(K)$ and $\mathrm{\parallel }K(x_1-x_2){\mathrm{\parallel }}_Y\le 2\delta$, then $z=(x_1-x_2)\mathrm{/}2$ lies in the class by symmetry and convexity from step 1.1 and satisfies $\mathrm{\parallel }Kz{\mathrm{\parallel }}_Y\le \delta$; thus $\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_X=2\mathrm{\parallel }z{\mathrm{\parallel }}_X\le 2{\epsilon }_K(X_{\nu ,\rho }(K),\delta )$. Conversely, for every $z$ in the centered-modulus set, the pair $(z,-z)$ lies in the source ball and has $\mathrm{\parallel }K(z-(-z)){\mathrm{\parallel }}_Y=2\mathrm{\parallel }Kz{\mathrm{\parallel }}_Y\le 2\delta$, so the two-truth supremum is at least $2{\epsilon }_K(X_{\nu ,\rho }(K),\delta )$. Taking suprema proves the claimed factor of two.

If $\delta =0$ and $x$ belongs to the centered-modulus set, then $Kx=0$. By [F3], $x\in N(K{)}^{\perp }$ as well, so $x=0$ and ${\epsilon }_K(X_{\nu ,\rho }(K),0)=0$. If $K=0$, [F3] likewise forces every member of the power source ball to be zero, so the asserted bound holds with no positive singular direction.

Suppose $\delta >0$ and let $x$ be any element of the centered-modulus set. By [F1], choose $w\in X$ with $x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_X\le \rho$. Applying [F4] and using $\mathrm{\parallel }Kx{\mathrm{\parallel }}_Y\le \delta$ gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X\le {\delta }^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$; taking the supremum over the centered-modulus set proves the bound.

Fix a positive singular value $s_j$ and its unit right singular vector $v_j$ from [F5]. Set $w_j=\rho v_j$ and $x_j=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }{w}_j=\rho s_j^{\nu }{v}_j$. Then $\mathrm{\parallel }K{x}_j{\mathrm{\parallel }}_Y=\rho s_j^{\nu +1}={\delta }_j$ and $\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_X=\rho s_j^{\nu }={\delta }_j^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$. Thus $x_j$ is feasible in the centered-modulus supremum at ${\delta }_j$ and attains the upper bound from step 1.2. If there are infinitely many positive singular values, compactness gives $s_j\to 0$ and hence ${\delta }_j\to 0$; finite positive rank gives only the stated finite list, while $K=0$ was handled in step 1.1.