Compact linear inverse maps and the Picard criterion

If $K=0$, then $N(K)=X$ is closed. Otherwise choose a bound $C>0$ from [F1]. For any $x\in \overline{N(K)}$, [F5] lets us choose, for every $\epsilon >0$, an $n_{\epsilon }\in N(K)$ with $\mathrm{\parallel }x-n_{\epsilon }\mathrm{\parallel }<\epsilon$. Hence $\mathrm{\parallel }Kx\mathrm{\parallel }=\mathrm{\parallel }K(x-n_{\epsilon })\mathrm{\parallel }\le C\epsilon$ for every $\epsilon >0$, so $\mathrm{\parallel }Kx\mathrm{\parallel }=0$ and $Kx=0$ by [F6]. Thus $\overline{N(K)}\subseteq N(K)$, and [F5] implies that $N(K)$ is closed.

Fix $r_0\in R(K)$ and choose $x_0\in X$ with $K{x}_0=r_0$. By [F3] and step 1.1, write $x_0=p+n$ with $p\in N(K{)}^{\perp }$ and $n\in N(K)$; then $Kp=K{x}_0-Kn=r_0$. The restriction $\tilde{K}$ is therefore surjective. If $p,q\in N(K{)}^{\perp }$ and $Kp=Kq$, then $p-q\in N(K)\cap N(K{)}^{\perp }$, so $p=q$ by [F2]. Thus $\tilde{K}$ is bijective.

If $x$ is any solution of $Kx=r_0$, then $x-p\in N(K)$. Since $p\perp (x-p)$, [F4] gives $\mathrm{\parallel }x{\mathrm{\parallel }}^2=\mathrm{\parallel }p{\mathrm{\parallel }}^2+\mathrm{\parallel }x-p{\mathrm{\parallel }}^2\ge \mathrm{\parallel }p{\mathrm{\parallel }}^2$, with equality only when $x=p$. So $p={\tilde{K}}^{-1}{r}_0$ is the unique minimum-norm solution.

For $y=r+z\in R(K)\oplus R(K{)}^{\perp }$, the decomposition is unique because $R(K)\cap R(K{)}^{\perp }=\{0\}$ by [F2], so the definition is single-valued. If $z\in R(K{)}^{\perp }$, then $r=0$ and the unique minimum-norm preimage of $0$ is $0$ by step 3.1; hence $K^{†}z=0$. For $y_i=r_i+z_i$ and scalars $a,b$, the vector $a{K}^{†}{y}_1+b{K}^{†}{y}_2$ lies in $N(K{)}^{\perp }$ and maps under $K$ to $a{r}_1+b{r}_2$, so uniqueness in step 2.1 gives $K^{†}(a{y}_1+b{y}_2)=a{K}^{†}{y}_1+b{K}^{†}{y}_2$. The definition is therefore a well-defined linear operator on the stated domain.

Suppose the positive singular system is finite, with index set $J=\{1,\dots ,r\}$. If $r=0$, the SVD expansion gives $K=0$, so $R(K)=\{0\}$ is closed by [F7] and $K^{†}=0$ is bounded on its domain. If $r>0$, each $u_j=s_j^{-1}K{v}_j$ lies in $R(K)$, while the SVD expansion places every $Kx$ in $\mathrm{span}\{u_1,\dots ,u_r\}$; hence the range equals this finite-dimensional span and is closed by [F7]. For $y=q+z\in D(K^{†})$, $q\in R(K)$ and $z\in R(K{)}^{\perp }$, so $K^{†}y={\sum }_{j=1}^r{s}_j^{-1}⟨q,u_j⟩v_j$ by [F2] and [F3]. Pythagoras and Bessel give $\mathrm{\parallel }{K}^{†}y{\mathrm{\parallel }}^2={\sum }_{j=1}^r{s}_j^{-2}\mathrm{\mid }⟨q,u_j⟩{\mathrm{\mid }}^2\le s_r^{-2}\mathrm{\parallel }q{\mathrm{\parallel }}^2\le s_r^{-2}\mathrm{\parallel }y{\mathrm{\parallel }}^2$; here $s_r$ is the smallest positive singular value, and the last inequality follows from $q\perp z$. Thus finite rank implies closed range and bounded $K^{†}$.

Suppose instead that the positive singular index set is $J=\{1,2,3,\dots \}$. Since $s_j\to 0$, for each pair $(k,n)$ the set of integers $m>n$ with $s_m<2^{-(k+1)}$ is nonempty; [F12] gives its least element, so [F11] defines recursively $n_0=0<n_1<n_2<\cdots$ and $s_{n_k}<2^{-k}$ for every $k\ge 1$. Set $y_N={\sum }_{k=1}^N{s}_{n_k}{u}_{n_k}$. For $M>N$, [F6] gives $\mathrm{\parallel }{y}_M-y_N{\mathrm{\parallel }}^2={\sum }_{k=N+1}^M{s}_{n_k}^2\le {\sum }_{k=N+1}^M{4}^{-k}\le 4^{-N}\mathrm{/}3$, so $(y_N)$ is Cauchy and converges to some $y\in Y$ by [F8]. Each $y_N$ belongs to $R(K)$ because $s_{n_k}{u}_{n_k}=K{v}_{n_k}$, hence $y\in \overline{R(K)}$. If $y=Kx$, then for every $k\ge 1$ and every $N\ge k$, orthonormality gives $⟨y_N,u_{n_k}⟩=s_{n_k}$; by [F9] we may pass to the limit to get $⟨y,u_{n_k}⟩=s_{n_k}$. The SVD and adjoint identity [F3]–[F4] then give $s_{n_k}=⟨Kx,u_{n_k}⟩=s_{n_k}⟨x,v_{n_k}⟩$, so $⟨x,v_{n_k}⟩=1$ for every $k\ge 1$. Applying [F5] to any first $N$ of these orthonormal vectors gives $N\le \mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $N\ge 1$. Taking $\epsilon =1\mathrm{/}(\mathrm{\parallel }x{\mathrm{\parallel }}^2+1)$ in [F10] yields an integer $N\ge 1$ with $1\mathrm{/}N<1\mathrm{/}(\mathrm{\parallel }x{\mathrm{\parallel }}^2+1)$, hence $N>\mathrm{\parallel }x{\mathrm{\parallel }}^2$, a contradiction. Thus $y\mathrm{\notin }R(K)$ and the range is not closed.

For each $k\ge 1$, $K{v}_{n_k}=s_{n_k}{u}_{n_k}\in D(K^{†})$ and $K^{†}K{v}_{n_k}=v_{n_k}$ by [F2], since $v_{n_k}\in N(K{)}^{\perp }$. Therefore $\mathrm{\parallel }{K}^{†}K{v}_{n_k}\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }K{v}_{n_k}\mathrm{\parallel }=1\mathrm{/}{s}_{n_k}>2^k$, so $K^{†}$ is unbounded when the positive singular system is infinite.

By [F3], every compact $K$ has either finite range dimension, covered by step 1.1, or an infinite positive singular system, covered by steps 1.2 and 2.1. Thus $R(K)$ is closed exactly in the finite-dimensional case, and in exactly that case $K^{†}$ is bounded. This proves both stated equivalences.

Suppose $y=Kx$. Then $y\perp N(K^{\ast })$ by [F3]. For each $j\in J$, the adjoint identity and singular relation give $⟨y,u_j{⟩}_Y=⟨Kx,u_j{⟩}_Y=⟨x,K^{\ast }{u}_j{⟩}_X=s_j⟨x,v_j{⟩}_X$. Applying [F5] to finite subsets of the orthonormal family $(v_j)$ shows $P_K(y)={\sum }_{j\in J}\mathrm{\mid }⟨x,v_j{⟩}_X{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_X^2<\mathrm{\infty }$. This proves necessity in claim 1.

Now let $y\perp N(K^{\ast })$ and $P_K(y)<\mathrm{\infty }$. By [F3], $y\in \overline{R(K)}$. For finite $J$ take the finite sum; for infinite $J$ let $x_N={\sum }_{j=1}^N{s}_j^{-1}⟨y,u_j{⟩}_Y{v}_j$. For $M>N$, [F6] gives $\mathrm{\parallel }{x}_M-x_N{\mathrm{\parallel }}_X^2={\sum }_{j=N+1}^M{s}_j^{-2}\mathrm{\mid }⟨y,u_j{⟩}_Y{\mathrm{\mid }}^2$, whose tails tend to zero because $P_K(y)<\mathrm{\infty }$. Thus $(x_N)$ is Cauchy and converges to $x\in X$ by [F9]. The complete right singular family places $x$ in $N(K{)}^{\perp }$.

By [F11], applying $K$ to the finite sums and passing to the limit gives $Kx={\lim }_{N\to \mathrm{\infty }}K{x}_N={\lim }_{N\to \mathrm{\infty }}{\sum }_{j=1}^N⟨y,u_j{⟩}_Y{u}_j=y$, where the last equality is [F7] on $\overline{R(K)}$. The finite-$J$ case is the same equality with a finite sum. Since $x\in N(K{)}^{\perp }$ and $Kx=y$, [F10] identifies $x=K^{†}y$. This proves sufficiency in claim 1 and the formula there.

For arbitrary $y\in Y$, apply [F8] to the closed subspace $M=\overline{R(K)}$ and write $y=p+q$ with $p\in M$ and $q\in M^{\perp }$. Since $R(K)\subseteq M$, $q\in R(K{)}^{\perp }$. The Fourier coefficients of $p$ are $⟨p,u_j{⟩}_Y=⟨y,u_j{⟩}_Y$, and [F7] expands $p$ in the complete left singular family. If $P_K(y)<\mathrm{\infty }$, the construction in steps 1.2–2.1 applied to $p\in M=(N(K^{\ast }){)}^{\perp }$ gives $p\in R(K)$ and $K^{†}p={\sum }_j{s}_j^{-1}⟨y,u_j{⟩}_Y{v}_j$; hence $y=p+q\in R(K)\oplus R(K{)}^{\perp }=D(K^{†})$ and [F10] gives the displayed formula.

Conversely, let $y=r+q\in D(K^{†})$ with $r\in R(K)$ and $q\in R(K{)}^{\perp }$. By [F3], $q\in N(K^{\ast })$ and each $u_j\in N(K^{\ast }{)}^{\perp }$, so $⟨q,u_j{⟩}_Y=0$. Step 1.1 applied to $r$ shows $P_K(y)=P_K(r)<\mathrm{\infty }$; [F10] gives $K^{†}y=K^{†}r$ and the same series formula. Thus the domain is exactly the set where the Picard sum is finite, completing claim 2.