Compact linear inverse maps and the Picard criterion — Examples and Counterexamples

The standard dot products on $\mathbb{R}$ and ${\mathbb{R}}^2$ satisfy the inner-product axioms [F3] by coordinatewise addition, multiplication and ${\sum }_i{x}_i^2=0\Rightarrow x_i=0$ for each coordinate. The standard Euclidean spaces $\mathbb{R}$ and ${\mathbb{R}}^2$ are complete: if a sequence is Cauchy in Euclidean norm, each coordinate is Cauchy because its absolute difference is bounded by the Euclidean distance; each coordinate converges by [F4], and the finite sum of squared coordinate errors tends to zero. Thus they are real Hilbert spaces by [F3] and [F5]. The map $K$ is linear and $\mathrm{\parallel }Kx\mathrm{\parallel }=\mathrm{\mid }x\mathrm{\mid }$, so it is bounded; its range is the one-dimensional span of $(1,0)$, and [F2] makes it compact. It therefore gives the stated compact linear inverse model by [F1].

For $a\in \mathbb{R}$ and $(b,c)\in {\mathbb{R}}^2$, $⟨Ka,(b,c)⟩=ab=⟨a,K^{\ast }(b,c)⟩$. Hence $N(K^{\ast })=\{(0,c):c\in \mathbb{R}\}$, while $R(K)=\{(a,0):a\in \mathbb{R}\}$. In particular $y$ is a nonzero element of $N(K^{\ast })$, and no scalar $x$ satisfies $Kx=y$.

The vector $v_1=1$ spans the input and $u_1=(1,0)$ spans $\overline{R(K)}$; they satisfy $K{v}_1=u_1$ and $K^{\ast }{u}_1=v_1$. Thus the positive singular pair has $s_1=1$. The Picard sum defined in the statement has sole term $\mathrm{\mid }⟨(0,1),(1,0)⟩{\mathrm{\mid }}^2\mathrm{/}{1}^2=0$. Step 1.2 shows that $y\in N(K^{\ast })\setminus \{0\}$, so the missing orthogonality condition allows finite Picard sum without exact solvability.

The map $K$ is linear and satisfies $\mathrm{\parallel }Kx{\mathrm{\parallel }}_2^2={\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\mathrm{/}(n+1{)}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so it is bounded. For $N\in \mathbb{N}$, let $(K_{N}x{)}_n=x_n\mathrm{/}(n+1)$ when $n\le N$ and $0$ otherwise. The estimate $\mathrm{\parallel }{K}_{N}x{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ makes $K_N$ bounded, and its range is spanned by the finite ordered family $(e_0,\dots ,e_N)$: every output has no coordinates beyond $N$, and $K_N((n+1)e_n)=e_n$ for each $n\le N$, while these coordinate vectors are linearly independent. Thus they form an ordered basis of the range, and $K_N$ is compact by [F3]. For every $x$, $\mathrm{\parallel }(K-K_N)x{\mathrm{\parallel }}_2^2={\sum }_{n>N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\mathrm{/}(n+1{)}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2\mathrm{/}(N+2{)}^2$, with equality in the operator norm at $x=e_{N+1}$. Hence $\mathrm{\parallel }K-K_N\mathrm{\parallel }=1\mathrm{/}(N+2)\to 0$. By [F2] the codomain is Banach, so [F4] gives compactness of $K$. It is a compact inverse model by [F8].

Every finitely supported $z\in H$ belongs to $R(K)$: set $x_n=(n+1)z_n$, which is finitely supported and satisfies $Kx=z$. For any $z\in H$ and $\epsilon >0$, [F1] gives a finite set $F\subseteq \mathbb{N}$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{z}_n{\mathrm{\mid }}^2<{\epsilon }^2$; choose $N$ containing $F$. Then the truncation $z^{(N)}$ supported on indices $n\le N$ obeys $\mathrm{\parallel }z-z^{(N)}{\mathrm{\parallel }}_2^2={\sum }_{n>N}\mathrm{\mid }{z}_n{\mathrm{\mid }}^2<{\epsilon }^2$. Thus finitely supported vectors are dense, and since all lie in $R(K)$, $R(K)$ is dense in $H$.

For $y_n=1\mathrm{/}(n+1)$, [F6] gives $y\in H$. Its finite truncations are in $R(K)$ by step 1.2 and converge to $y$ because the tails of its square-sum tend to zero; hence $y\in \overline{R(K)}$. If $Kx=y$, then every coordinate equation gives $x_n=1$, contradicting $x\in H$. Indeed, if the finite sums ${\sum }_{n=0}^N\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=N+1$ had an upper bound $B\ge 0$, [F7] with $\epsilon =1\mathrm{/}(B+1)$ would give $m>B+1$, and the partial sum at $N=m$ would exceed $B$. Therefore $y\mathrm{\notin }R(K)$ and the dense range is not closed.

The coordinate vectors are orthonormal because [F1] gives $⟨e_i,e_j⟩=1$ when $i=j$ and $0$ otherwise. They are complete: for any $x\in H$ and $\epsilon >0$, [F1] gives a finite set $F$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<{\epsilon }^2$; choose $N$ containing $F$, so the truncation's squared error is at most this tail. Direct calculation gives $N(K)=\{0\}$ and $\overline{R(K)}=H$ by step 1.2. For $x,z\in H$, the canonical inner product gives $⟨Kx,z⟩={\sum }_n{x}_n{z}_n\mathrm{/}(n+1)=⟨x,Kz⟩$; hence [F9] identifies $K^{\ast }=K$. Since all diagonal coefficients are positive, $N(K^{\ast })=\{0\}$. Therefore $K{e}_n=(n+1{)}^{-1}{e}_n$ and $K^{\ast }{e}_n=(n+1{)}^{-1}{e}_n$, and the family is complete on both $N(K{)}^{\perp }$ and $\overline{R(K)}$. Reindexing by $j=n+1$ gives the positive singular system $s_j=1\mathrm{/}j$, $v_j=u_j=e_{j-1}$ for $j\ge 1$. The datum has $⟨y,e_{j-1}⟩=1\mathrm{/}j$, so every Picard summand equals $1$. For any real bound $B\ge 0$, [F7] applied with $\epsilon =1\mathrm{/}(B+1)$ gives an index $m>B+1$, so the corresponding finite subsums are unbounded and the Picard sum diverges, as [F5] predicts for a datum outside the range.