Theorem 6
Let
\(A\in L(H)\), there exists a compact operator
\(K\in K(H)\)
such that
$$ \rho_{e}(A)=M_{e}(A+K). $$
Correction of the operator A
First correction
This correction involves the construction of a compact perturbation \(K_{1}\) such that if \(A\in L(H)\) and \(B=(A+K_{1})\in L(H)\) then
$$ \rho_{e}(A)\cap \operatorname{reg}(A+K_{1})\subseteq M_{e}(A+K_{1}). $$
Lemma 2
Let
\(A\in L(H)\)
such that
A
is invertible on the left or on the right, \(K_{1}\)
an operator of
\(K(H)\)
and
\(B=A+K_{1}\). If
\(\Vert K_{1}\Vert < c(A)\)
then

(1)
\(\delta(N(B),N(A))\leq\frac{\Vert K_{1}\Vert }{c(A)}\),

(2)
B
is invertible on the left or on the right.
Proof
We have by definition \(\delta(N(B),N(A))=\Vert (IP_{N(B)})P_{N(A)}\Vert \). Assume that \(u\in H\) and let \(v=(IP_{N(A)})P_{N(B)}u\). Then v is orthogonal to \(N(A)\) and we have
$$ \Vert Av\Vert \geq c(A)\Vert v\Vert , $$
or still
$$\begin{aligned}& \Vert v\Vert \leq\frac{\Vert Av\Vert }{c(A)}, \\& \frac{\Vert Av\Vert }{c(A)} = \frac{\Vert A((IP_{N(A)})P_{N(B)}u)\Vert }{c(A)} \\& \hphantom{\frac{\Vert Av\Vert }{c(A)}} = \frac{\Vert A(P_{N(B)}u)\Vert }{c(A)} \\& \hphantom{\frac{\Vert Av\Vert }{c(A)}} = \frac{\Vert (AB)(P_{N(B)}u)\Vert }{c(A)} \\& \hphantom{\frac{\Vert Av\Vert }{c(A)}} \leq \frac{\Vert (AB)\Vert \Vert P_{N(B)}u \Vert }{c(A)}. \end{aligned}$$
Using Proposition 2, we have
$$ \delta\bigl(N(B),N(A)\bigr)< 1\quad \Rightarrow\quad \dim N(B)\leq\dim N(A). $$
If A is onetoone, then \(\dim N(A)=0\Rightarrow\dim N(B)=0\). Therefore \(N(B)=\{0\}\), hence B is onetoone. As \(R(B)\) is closed (compact perturbation of semiFredholm operator), we deduce that B is left invertible.
If \(N(A)\neq \{ 0 \} \), then A is not left invertible. Hence A is right invertible. Then \(N(A^{\ast})= \{ 0 \} \). \(A,B\in L(H)\Rightarrow A^{\ast},B^{\ast}\in L(H)\), \(A^{\ast}B^{\ast}=AB\), and \(c(A)=c(A^{\ast})\).
By a similar reasoning we will have
$$ \delta\bigl(N\bigl(B^{\ast}\bigr),N\bigl(A^{\ast}\bigr)\bigr)\leq \frac{\Vert A^{\ast }B^{\ast }\Vert }{c(A^{\ast})}< 1\quad \Rightarrow\quad \dim N\bigl(B^{\ast}\bigr)=0. $$
Then \(N(B^{\ast})=\{0\}\) and \(B^{\ast}\) is onetoone. \(B=(A+K_{1})\in SF(H)\), in particular \(R(B)\) is a closed subspace of H, \(R(B)=N(B^{\ast })^{\perp}=\{0\}^{\perp}=H\), then B is onto. Thus it is right invertible. □
Lemma 3
Let
M, N
be two closed subspace of
H
such that
\(\dim N\geq\dim M>0\)
and
\(\dim M<\infty\). Then there is
U, a partial isometry of
H
such that
$$ R(U)\subseteq N\quad \textit{and}\quad R\bigl(U^{\ast}\bigr)=M. $$
Proof
Let \(e_{1},e_{2},\ldots,e_{m}\) be an orthonormal bases of M and \(f_{1},f_{2},\ldots,f_{m}\) an orthonormal family of vectors of N. Let P the orthogonal projection of H onto M and Q the orthogonal projection onto the space spanned by \(f_{1},f_{2},\ldots,f_{m}\). If \(u\in H\), there are m complex numbers \(a_{1},a_{2},\ldots,a_{m}\) such that \(Pu= \sum_{i=1}^{m}a_{i}e_{i}\). We put \(Uu=\sum_{i=1}^{m}a_{i}f_{i}\). Then \(R(U)\subseteq N\) and if \(v\in H\), there are m complex numbers \(b_{1},b_{2},\ldots,b_{m}\) such that \(Qv=\sum_{i=1}^{m}b_{i}f_{i}\). Then \(\langle Uu,v \rangle =\sum_{i=1}^{m}a_{i}\overline{b}_{i}= \langle u, \sum_{i=1}^{m}b_{i}e_{i} \rangle = \langle u,U^{\ast }v \rangle\), hence \(U^{\ast}v=\sum_{i=1}^{m}b_{i}e_{i}\). It is easy to see that \(R(U\ast)=M\), U is a partial isometry of H, and
$$ U^{\ast}U=P, \qquad UU^{\ast}=Q. $$
□
Lemma 4
Let
\(A\in L(H)\), \(\beta\in\rho_{e}(A)\), and
\(\alpha\in\mathbb {R}_{+}^{\ast }\). Then there is
\(T(A,\beta)\in L(H)\)
such that

(1)
\(T(A,\beta)\)
is of finite rank,

(2)
\(\Vert T(A,\beta)\Vert \leq1\),

(3)
\(\beta\in M_{e}(A+\alpha T(A,\beta))\).
Proof
We suppose at first \(\operatorname{ind}(A\beta I)\geq0\). Hence \(\dim N(A^{\ast }\beta I)<\infty\). By Lemma 3, there is U such that \(R(U)\subseteq N(A\beta I)\), \(R(U^{\ast})=N(A^{\ast}\beta I)\). We take \(U^{\ast}=T(A,\beta)\). Then \(T(A,\beta)\) is of finite rank, thus compact, and we have \(\beta \in\rho _{e}(A+\alpha T(A,\beta))\). Furthermore \(A+\alpha T(A,\beta)\beta I\) is onto. Indeed \(R(A\beta I)\) is a closed subspace of H. Hence \(H=R(A\beta I)\oplus N(A^{\ast}\beta I)\). Let \(u\in H\), then \(u=u_{1}+u_{2}\) with \(u_{1}\in R(A\beta I)\) and \(u_{2}\in N(A\ast\beta I)\). Let Q be a projection onto \(N(A\beta I)\). There is \(v_{1}\) and \(v_{2}\) such that \((A\beta I)v_{1}=u_{1}\) where \(Qv_{1}=0\) and \(u_{2}=\alpha Tv_{2}\), where \(v_{2}=Qv_{2}\), that is, \(v_{2}\in N(A\beta I)\). If we take \(v=v_{1}+v_{2}\) it follows that
$$\begin{aligned} \bigl(A+\alpha T(A,\beta)\beta I\bigr)v =&(A\beta I) (IQ)v_{1}+ \alpha T(A,\beta )v_{2} \\ =&(A\beta I)v_{1}+\alpha T(A,\beta)v_{2}=u_{1}+u_{2}=u. \end{aligned}$$
Hence the operator \((A+\alpha T(A,\beta)\beta I)\) is onto and \(\beta \in M_{e}(A+\alpha(T,\beta))\). If \(\operatorname{ind}(A\beta I)<0\) and \(\dim N(A\beta I)>0\), we proceed as above by replacing \(A\beta I\) by \(A^{\ast}\beta I\) and T by \(T^{\ast}\) as \(\operatorname{ind}(A\beta I)=\operatorname{ind}(A^{\ast}\beta I)\). The operator \((A+\alpha T(A,\beta)\beta I)\) is onetoone, it follows that
$$ \beta\in M_{e}\bigl(A+\alpha T(A,\beta)\bigr). $$
□
Remark 3
If \(\operatorname{ind}(A\beta I)=0\), then \(A+\alpha T(A,\beta)\beta I\) is onto and onetoone.
Remark 4
If \(\min \operatorname{ind}(A\beta I)=0\), we take \(T=0\). Indeed in this case \(A+\alpha T(A,\beta)\beta I=A\beta I\) is invertible on the left or right.
Lemma 5
Let
\(A\in L(H)\). Then for all
\(\varepsilon>0\), there is
\(K_{1}\in K(H)\)
such that

(1)
\(\Vert K_{1}\Vert \leq\varepsilon\),

(2)
\(\rho_{e}(A+K_{1})\cap \operatorname{reg}(A+K_{1})\subseteq M_{e}(A+K_{1})\).
Proof
Suppose that \(\rho_{e}(A)=\rho_{e}(B)=\bigcup_{1\leq i\leq n}C_{i}\) where \(C_{i}\), \(i=1,2,3,\ldots,n\), is a connected component. For each i, let \(\beta_{i}\in C_{i}\). We inductively define a sequence of positive real numbers and a sequence of operators in \(L(H)\) as follows:

(1)
\(\alpha_{1}=\alpha\), \(A_{0}=A\),

(2)
\(A_{i}=A_{i1}+\alpha_{i}T_{i}\), \(i=1,2,3,\ldots,n\), where \(T_{i}=T(A_{i1},\beta_{i})\),

(2)
\(\alpha_{i+1}=\frac{1}{3}\min \{ \alpha_{i},c(A_{i}\beta _{i}I), i=1,2,3,\ldots,n1 \} \).
Note that (1) and (3) give \(\alpha_{i+1}\leq\frac{1}{3} \alpha _{i}\Rightarrow\alpha_{i+1}\leq\frac{\alpha}{3^{i}}\). We set \(S_{i}=\sum_{1\leq j\leq i}\alpha_{j}T_{j}\); then
$$ \Vert S_{i}\Vert \leq\sum_{1\leq j\leq i} \alpha_{j} \Vert T_{j}\Vert \leq\sum _{1\leq j\leq i}\alpha_{j}\leq\sum _{1\leq j\leq \infty}\alpha_{j}\leq \alpha\sum _{1\leq j\leq\infty}\frac {1}{3^{j1}}=\frac{3}{2}\alpha. $$
Therefore \(\Vert S_{i}\Vert \leq\frac{3}{2}\alpha\). For each i, \(A_{i}=A+S_{i}\) where \(S_{i}\) is a finite rank operator and therefore compact. We must have \(\rho_{e}(A)=\rho_{e}(A_{i})\) and \(c(A_{i}\beta _{i}I)>0\). This shows that all \(\alpha_{i}\) are strictly positive. In addition, by the previous lemma \(A_{i}\beta_{i}I\) is left or right invertible, from which we deduce that \(C_{i}\cap \operatorname{reg}(A_{i})\subseteq M_{e}(A_{i})\). Finally
$$ \Vert S_{n}S_{i}\Vert \leq\sum _{i+1\leq j\leq n}\alpha _{j}\leq \frac{3}{2} \alpha_{i+1}\leq\frac{1}{2}c(A_{i}\beta _{i}I)< c(A_{i}\beta _{i}I). $$
Hence \(A_{n}\beta_{i}I(A_{i}\beta_{i}I)=S_{n}S_{i}\); using Lemma 2, we deduce that for \(i=1,2,3,\ldots,n\), \(A_{n}\beta_{i}I\) is left or right invertible and therefore \(\rho_{e}(A+S_{n})\cap \operatorname{reg}(A+S_{n})\subseteq M_{e}(A+S_{n})\). Accordingly we get \(K_{1}=S_{n}\) and \(\alpha=\frac {2}{3}\varepsilon\), and the theorem is proved. If \(\rho_{e}(A)\) has a countable infinite numbers of connected components, the only other possible case, we observe that the sequence \((S_{n})\) converges normally, and taking \(K_{1}=\lim_{n\rightarrow\infty}S_{n}\), we proceed exactly as in the previous case. As, for every n, \(S_{n}\) is an operator of finite rank, \(K_{1}\) is a compact operator. We set \(B=A+K_{1}\). □
Second correction
Let \(B\in L(H)\) such that \(\rho_{e}(B)\cap \operatorname{reg}(B)\subseteq M_{e}(B)\). We will build a second compact perturbation \(K_{2}\) such that \(\rho _{e}(B)\subseteq M_{e}(B+K_{2})\). If \(B\in L(H)\) such that \(\rho _{e}(B)\cap \operatorname{reg}(B)\subseteq M_{e}(B)\), then \(\rho_{e}(B)\backslash M_{e}(B)\) is a finite or countable set denoted \(\mu_{j}\). Note that \(\sigma _{e}(B)=\emptyset\), because in [6], there exist \(n\in\mathbb{N}\) and \(\mu_{1},\mu_{2},\mu_{3},\ldots,\mu_{n}\in\mathbb{C}\) such that \(\sigma (B)=\{\mu_{1},\mu_{2},\mu_{3},\ldots,\mu_{n}\}\). Hence \(\dim H<\infty\), this is contrary to our hypotheses. As \(\sigma_{e}(B)\) is a closed set of \(\sigma(B)\) (which is compact), \(\sigma_{e}(B)\) is also compact. Since the distance function is continuous on \(\sigma_{e}(B)\), there is a sequence of points \((\mu_{l}^{\prime})_{j\in\mathbb{N}^{\ast}}\in \sigma _{e}(B)\) such that \(d(\mu_{j},\sigma_{e}(B))=\mu_{j}\mu _{j}^{\prime }\). Finally if the sequence \((\mu_{n})\) is infinite, then \(\lim_{j\rightarrow\infty}\mu_{j}\mu_{j}^{\prime}=0\), because otherwise \(\mu_{n}\) would have an accumulation point in \(\rho_{e}(B)\), which is contradictory. For every \(\mu_{j}\), let \(M_{j}\), \(N_{j}\) be the two subspaces of H corresponding to the decomposition of Kato of the operator \(B\mu_{j}I\) and let \(P_{j}\) be the associated projection of the kernel \(M_{j}\) and the range \(N_{j}\).
Construction of subspaces \(M_{j}^{\prime}\) and \(N_{j}^{\prime}\)
For \(j\in\mathbb{N}\), let \(M _{0}^{\prime}=H\), \(M_{j+1}^{\prime }=M_{j}^{\prime}\cap M_{j+1}\), and \(N_{1}^{\prime}=M_{1}^{\perp}\), \(N_{j+1}^{\prime}=M_{j}^{\prime}\cap M_{j+1}^{\prime\perp}\). Finally, we denote by \(P_{j}^{\prime}\) the orthogonal projection onto \(N_{j}^{\prime}\).
Lemma 6

(1)
If
\(1\leq i\leq n\), \(1\leq j\leq n\), and
\(i\neq j\)
then
\(N_{i}^{\prime }\perp N_{j}^{\prime}\)
if and only if
\(P_{i}^{\prime }P_{j}^{\prime}=0\).

(2)
If
\(1\leq j\leq n\), then
\(N_{j}^{\prime}\oplus M_{j}^{\prime}=N_{j} \oplus M_{j}^{\prime}=M_{j1}^{\prime}\)
and
\(\dim N_{j}^{\prime}=\dim N_{j}<\infty\).

(3)
If
\(1\leq j\leq n\), then
\(\bigcap_{1\leq i\leq j}N_{i}^{\prime}=M_{j}^{\prime}\)
and
\(\operatorname{co}\dim M_{j}^{\prime}=\sum_{1\leq i\leq j}\dim N_{i}\).
Proof
For any \(i>j\), we have \(N_{i}^{\prime}\perp N_{j}^{\prime}\). Indeed \(N_{i}^{\prime}\subseteq M _{i1}^{\prime}\subseteq M _{j}^{\prime}\); and \(N _{j}^{\prime}\subseteq M_{j}^{\prime\perp}\), then \(M_{j}^{\prime}\subseteq N_{j}^{\prime\perp}\) and we deduce \(N_{i}^{\prime}\subseteq N_{j}^{\prime\perp}\). Suppose \(P_{i}^{\prime }P_{j}^{\prime}=0\) and let \(x\in N_{i}^{\prime}\). Then \(\langle x,y \rangle= \langle P_{i}^{\prime}x,P_{j}^{\prime }y \rangle= \langle x,P_{i}^{\prime}P_{j}^{\prime}y \rangle=0\). \(N_{i}^{\prime}\) is orthogonal to \(N_{j}^{\prime}\). Reciprocally if \(N_{i}^{\prime}\) is orthogonal to \(N_{j}^{\prime}\) and \(u\in H\), then \(P_{j}^{\prime}u \in N_{j}^{\prime}\) and \(P_{j}^{\prime}u\in N_{j}^{\prime\perp}\) then \(P_{i}^{\prime }P_{j}^{\prime}u=0\) and \(P_{i}^{\prime}P_{j}^{\prime}=0\).
\(N_{j}^{\prime}\cap M_{j}^{\prime}\subseteq M_{j}^{\prime\perp}\cap M_{j}^{\prime}=\{0\}\) and \(N_{j}^{\prime}\cap M_{j}^{\prime}\subseteq N_{j}^{\prime}\cap M_{j}^{\prime}=\{0\}\). \(M_{j}^{\prime}+N_{j}^{\prime }=M_{j1}^{\prime}\) (because \(N_{j}^{\prime}=M_{j1}^{\prime}\cap M_{j}^{\prime\perp}\)).
For any \(i>j\), \(N_{i}^{\prime}\subseteq M_{j1}^{\prime}=N_{j}^{\prime }\oplus M_{j}^{\prime}\), then \(M_{j}^{\prime}\oplus N_{j}^{\prime }\subseteq M_{j}^{\prime}\oplus N_{j}^{\prime}\); \(N_{j}^{\prime }\subseteq M_{j1}^{\prime}=M_{j}^{\prime}\oplus N_{j}\Rightarrow M_{j}^{\prime}\oplus N_{j}^{\prime}\subseteq M_{j}^{\prime}\oplus N_{j}\).
If \(i=j\), the result is obvious. Suppose that it is true for j. Then \(\operatorname{co}\dim M_{j}^{\prime}<\infty\) and therefore \(M_{j}^{\prime\perp }\oplus M_{j+1}^{\prime}\) is closed. Hence \(M_{j}^{\prime\perp}\oplus M_{j+1}^{\prime}=N_{j+1}^{\prime\perp}\) and we have
$$ \bigcap_{1\leq i\leq j+1}N_{i}^{\prime\perp}=M_{j}^{\prime} \cap N_{j+1}^{\prime\perp}=M_{j}^{\prime}\cap \bigl(M_{j}^{\prime\perp }\oplus M_{j+1}^{\prime} \bigr)=M_{j+1}^{\prime}. $$
It follows that for any \(j\in\mathbb{N}\), \(1\leq j\leq n\); \(M_{j+1}^{\prime \perp}=M_{j}^{\prime\perp}\oplus N_{j}^{\prime}\) It is easy to see that \(\operatorname{co}\dim M_{j}^{\prime}=\sum_{1\leq i\leq j}\dim N_{i}^{\prime}<\infty\). □
Remark 5
If \((P_{i})\) is a sequence of mutually orthogonal projections in H, then \(\sum_{1\leq i\leq n}P_{i}\) converges to P, which is an orthogonal projection (see [13]).
Suppose first that the sequence \((\mu_{n})\) is finite, \(1\leq j\leq n\). We put \(H_{1}=\bigcap_{1\leq j\leq n}M_{j}\) and \(K_{2}=\sum_{1\leq j\leq n}(\mu _{j}^{\prime}\mu_{j})P_{j}\). It is obvious that \(K_{2}\) is normal.
Lemma 7
If
\(\mu\in\rho_{e}(B)\), then
\(N(B+K_{2}\mu I)\subseteq N(B\mu I)\cap H_{1}\).
Proof
First show by induction that if \(u\in N(B+K_{2}\mu I)\) then \(u\in H_{1}\). Let \(u\in N(B+K_{2}\mu I)\), then \(P_{1}^{\prime}=0\). Indeed
$$ (B+K_{2}\mu I) \bigl(IP_{1}^{\prime} \bigr)u+(B+K_{2}\mu I)P_{1}^{\prime }u=(B+K_{2} \mu I)=0. $$
But
$$ K_{2}P_{1}^{\prime}u=\biggl[\sum _{1\leq k\leq n}\bigl(\mu_{k}^{\prime}\mu _{k}\bigr)P_{k}^{\prime}\biggr]P_{1}^{\prime}u= \bigl(\mu_{1}^{\prime}\mu _{1}\bigr)P_{1}^{\prime}u. $$
Hence
$$\begin{aligned} 0 =&(B+K_{2}\mu I) \bigl(IP_{1}^{\prime} \bigr)u+B\mu_{1}I+\bigl(\mu_{1}^{\prime } \mu_{1}\bigr)IP_{1}^{\prime}u \\ =&(B+K_{2}\mu I) \bigl(IP_{1}^{\prime}\bigr)u+B \mu_{1}I+\bigl(\mu_{1}^{\prime }\mu _{1} \bigr) (IP_{1})P_{1}^{\prime}u \\ &{}+B\mu_{1}I+\bigl(\mu_{1}^{\prime} \mu_{1}\bigr)IP_{1}P_{1}^{\prime}u. \end{aligned}$$
Now we account for the last two terms of the amount in \(M_{1}\) (because \(K_{2}(IP_{1}^{\prime})=(IP_{1}^{\prime})K_{2}\) is invariant under B) and the last part of \(N_{1}\) which is invariant under B. So as \(H=M_{1}\oplus N_{1}\) then \(B\mu_{1}I+(\mu_{1}^{\prime}\mu )IP_{1}P_{1}^{\prime}u=0\). The operator \((B\mu_{1}I)N_{1}\) is nilpotent; and as \(\mu_{1}= \mu\), then \(B\mu_{1}I+(\mu_{1}^{\prime} \mu)I\) is onetoone. It follows that \(P_{1}P_{1}^{\prime}u=0\). Hence \(P_{1}^{\prime}u=0\). Suppose that always \(u\in N(B+K_{2} \mu I)\); \(P_{j}^{\prime}=0\) with \(j=1,2,3,\ldots,m\). According to Lemma 5, \(u\in M_{m}^{\prime}u\), we show \(P_{m+1}^{\prime}u=0\). Indeed
$$ (B+K_{2}\mu I) \bigl(IP_{m+1}^{\prime} \bigr)u+(B+K_{2}\mu I)P_{m+1}^{\prime }u=(B+K_{2} \mu I)=0. $$
But \(K_{2}P_{m+1}^{\prime}u=( \mu_{m+1}^{\prime} \mu_{m+1})P_{m+1}^{\prime}u\), and \((IP_{m+1}^{\prime})\in M_{m+1}^{\prime }\subseteq M_{m+1}\). Hence
$$ (B+K_{2}\mu I) \bigl(IP_{m+1}^{\prime}\bigr)u+B \mu_{m+1}I+\bigl(\mu_{m+1}^{\prime}\mu \bigr)IP_{m+1}^{\prime}u=0. $$
(2.1)
But
$$(B+K_{2} \mu I) \bigl(IP_{m+1}^{\prime}\bigr)u=(B \mu I) \bigl(IP_{m+1}^{\prime}\bigr)u+K_{2} \bigl(IP_{m+1}^{\prime}u\bigr)\in M_{m+1}. $$
Because \(M_{m+1}\) is invariant under \(B \mu I\) and \(M_{m+1}^{\prime}\) is invariant under \(K_{2}\),
$$\begin{aligned}& B\mu_{m+1}I+\bigl( \mu_{m+1}^{\prime} \mu \bigr)IP_{m+1}^{\prime}u \\& \quad =B \mu_{m+1}I+\bigl( \mu_{m+1}^{\prime} \mu \bigr)I(IP_{m+1})P_{m+1}u \\& \qquad {}+B \mu_{m+1}I+\bigl( \mu_{m+1}^{\prime} \mu \bigr)IP_{m+1}P_{m+1}^{\prime}u. \end{aligned}$$
The first term is in \(M_{m+1}\) while the second is in \(N_{m+1}\) (same reasons as before). So as \(H=M_{m+1}\oplus N_{m+1}\), and as above, we deduce from (2.1) that \(B \mu_{m+1}I+( \mu_{m+1}^{\prime} \mu)IP_{m+1}P_{m+1}^{\prime}u=0\) then \(P_{m+1}P_{m+1}^{\prime}u=0\). Hence \(P_{m+1}u\in M_{m+1}\). So by Lemma 6 and using the induction hypothesis \(u\in M_{m+1}\) and therefore \(P_{m+1}^{\prime}u=0\), which states that \(u\in H_{1}\) and \(K_{2}u=0\). If the sequence \(\mu_{n}\) is infinite, we set \(H_{1}=\bigcap_{1\leq j\leq\infty} M_{j}\) and \(K_{2}\sum_{1\leq j\leq \infty}(\mu _{j}^{\prime} \mu_{j})P_{j}^{\prime}\), the previous proposal remains valid because the series defining \(K_{2}\) is normally convergent. Indeed \(\lim_{j\rightarrow \infty}( \mu_{j}^{\prime} \mu_{j})=0\) ⇔ (\(\forall\varepsilon>0\)) (\(\exists N(\varepsilon )>0\)) (\(\forall j\geq N(\varepsilon)\Rightarrow \mu_{j}^{\prime}\mu_{j}<\varepsilon\)). Let \(u\in H\) and \(n\geq N(\varepsilon)\) then
$$\begin{aligned} \biggl\Vert \sum_{k\geq n}\bigl( \mu _{k}^{\prime} \mu _{k}\bigr)P_{k}^{\prime}u \biggr\Vert ^{2} =& \biggl\langle \sum_{k\geq n} \bigl( \mu _{k}^{\prime} \mu _{k} \bigr)P_{k}^{\prime}u,\sum_{j\geq n} \bigl( \mu _{j}^{\prime} \mu _{j} \bigr)P_{j}^{\prime}u \biggr\rangle \\ =& \biggl\langle \sum_{k\geq n}\sum _{j\geq n}\bigl( \mu _{k}^{\prime} \mu _{k}\bigr)\overline{\bigl( \mu _{j}^{\prime} \mu _{j}\bigr)}P_{j}^{\prime}P_{k}^{\prime}u,u \biggr\rangle \\ =&\sum_{k\geq n}\bigl\vert \mu _{k}^{\prime} \mu _{k}\bigr\vert ^{2} \bigl\langle P_{k}^{\prime}u,P_{k}^{\prime }u \bigr\rangle \leq\varepsilon^{2}\biggl\Vert \sum _{k\geq n}P_{k}^{\prime }u\biggr\Vert ^{2} \\ \leq&\varepsilon^{2}\bigl\Vert P^{\prime}u\bigr\Vert ^{2}\leq \varepsilon^{2}\Vert u\Vert ^{2}. \end{aligned}$$
Therefore \(\Vert \sum_{k\geq n}( \mu _{k}^{\prime} \mu_{k})P_{k}^{\prime}u \Vert \leq\varepsilon\), then \(\sum_{1\leq j\leq \infty}( \mu _{j}^{\prime} \mu _{j})P_{j}^{\prime}\) is normally convergent.
The rest of the proof is similar to the case of finite \(\mu_{j}\). □
Lemma 8
If
\(\mu\in\rho_{e}(B)\), then
\(H_{1}^{\perp}\subseteq R(B+K_{2}\mu I)+H_{1}\).
Proof
(A) First prove that for any \(j\in\mathbb{N}\), if \(\mu_{j}\in N_{j}^{\prime}\) then there exists one and only one \(w_{j}\in N_{j}^{\prime}\) such that
$$ P_{j}^{\prime}(B+K_{2} \mu I)w_{j}=u_{j}. $$
(2.2)
Uniqueness. Suppose that \(w_{j}\) and \(w_{j}^{\prime}\) satisfy the requirement. Then \(w_{j}w_{j}^{\prime}\in N(B+K_{2} \mu I)\subseteq H_{1}\) (see Lemma 6), hence \(w_{j}w_{j}^{\prime}\in N_{j}^{\prime}\cap H_{1}= \{ 0 \} \).
Existence. If \(u_{j}\in N_{j}^{\prime}\) then \(u_{j}=x_{j}+y_{j}\) with \(x_{j}\in N_{j}\), \(y_{j}\in M_{j}^{\prime}\) and \(u_{j}=P_{j}^{\prime}x_{j}\). We have \(B\mu I+(\mu_{j}^{\prime}\mu _{j})I=B \mu_{j}I+( \mu_{j}^{\prime} \mu)I\) and as \(\mu_{j}^{\prime}\neq \mu\), there is \(v_{j}\in N_{j}\) such that \(x_{j}=[B \mu I+(\mu_{j}^{\prime} \mu_{j})I]v_{j}\) then
$$\begin{aligned} u_{j} =&P_{j}^{\prime}\bigl[B \mu I+\bigl( \mu_{j}^{\prime} \mu_{j}\bigr)I\bigr] \bigl[P_{j}^{\prime}v_{j}+\bigl(IP_{j}^{\prime} \bigr)v_{j}\bigr] \\ =&P_{j}^{\prime}[B \mu I+K_{2}]P_{j}^{\prime}v_{j}+P_{j}^{\prime} \bigl[B \mu I+\bigl(\mu_{j}^{\prime} \mu_{j}\bigr)I \bigr]\bigl(IP_{j}^{\prime}\bigr)v_{j}. \end{aligned}$$
But \((IP_{j}^{\prime})vj\in M_{j}^{\prime}\). Hence \([B \mu I+( \mu_{j}^{\prime} \mu_{j})I](IP_{j}^{\prime})v_{j}\in M_{j}^{\prime}\) and therefore \(P_{j}^{\prime}[B \mu I+( \mu_{j}^{\prime} \mu_{j})I](IP_{j}^{\prime})v_{j}=0\). We take \(w_{j}=P_{j}^{\prime }v_{j}\in N_{j}^{\prime}\).
(B) Let \(Q=\sum_{j}P_{j}^{\prime}\), \(u_{j}=P_{j}^{\prime}u\), and \(u\in H_{1}^{\perp}\). We show that there is \(w\in H_{1}^{\perp}\) such that
$$ u=Q(B+K_{2} \mu I)w. $$
(2.3)
It follows from the existence of \(w_{1}\in N_{1}^{\prime}\) that \(P_{1}^{\prime}(B+K_{2} \mu I)w_{1}=u_{1}\). Suppose we know \(w_{1},w_{2},w_{3},\ldots,w_{n}\), \(u_{n+1}\sum_{1\leq j\leq n}P_{n+1}^{\prime}(B+K_{2} \mu I)w_{j}\in N_{n+1}^{\prime}\), then there is \(w_{n+1}\in N_{n+1}^{\prime}\) such that
$$ P_{n+1}^{\prime}(B+K_{2} \mu I)w_{n+1}=u_{n+1} \sum_{1\leq j\leq n}P_{n+1}^{\prime}(B+K_{2}\mu I)w_{j}, $$
and therefore
$$ u_{n+1}=\sum_{1\leq j\leq n}P_{n+1}^{\prime}(B+K_{2}\mu I)w_{j}. $$
(2.4)
Let \(w=\sum_{j}w_{j}\). Then
$$\begin{aligned} Q(B+K_{2} \mu I)w =&\sum_{j}P_{j}^{\prime}(B+K_{2} \mu I)\sum_{i}w_{i} \\ =&\sum_{j}\sum_{i}P_{j}^{\prime}(B+K_{2} \mu I)w_{i} \\ =&\sum_{j}\sum_{i}P_{j}^{\prime} \bigl[B \mu I+\bigl(\mu _{i}^{\prime} \mu_{i}\bigr)I \bigr]w_{i}. \end{aligned}$$
But \(w_{i}\in M_{i1}^{\prime}\), \([B\mu I+(\mu _{i}^{\prime} \mu_{i})I]w_{i}\in M_{i1}^{\prime}\) and if \(j< i\) we have \(P_{j}^{\prime }(M _{i1}^{\prime})=0\). Then
$$ Q(B+K_{2} \mu I)w=\sum_{j}\sum _{1\leq i\leq j}P_{j}^{\prime}(B+K_{2} \mu I)w_{i}. $$
Using this we find \(Q(B+K_{2} \mu I)w=\sum_{j}u_{j}=u\); then \(u=(B+K_{2} \mu I)w(IQ)(B+K_{2} \mu I)w\). Hence \(u\in R(B+K_{2} \mu I)+H_{1}\). This completes the proof of the lemma. □
Lemma 9
Let
\(B\in L(H)\)
such that
\(\rho_{e}(B)\cap \operatorname{reg}(B)\subseteq M_{e}(B)\). Then there is a compact operator
\(K_{2}\in K(H)\)
such that
$$ \rho_{e}(B)=\rho_{e}(B+K_{2})\subseteq M_{e}(B+K_{2}). $$
Proof
We show that
$$ \rho e(B)\subseteq \operatorname{reg}(B+K_{2}). $$
(2.5)
Let \(\mu \in\rho_{e}(B)\) and \(u\in N(B+K_{2} \mu I)\). Then \(u\in N(B \mu I)\cap H_{1}\). But \((B \mu I)H_{1}\) is regular because for any \(j\in\mathbb{N}\), \((B \mu I)M_{j}\) is also regular. Hence for any \(j\in\mathbb{N}\), \(u\in R[(B\mu I)^{j}H_{1}]\subseteq R[(B+K_{2} \mu I)^{j}]\). As further \(B+K_{2} \mu I\) is a compact perturbation of the semiFredholm operator \(B\mu I\), it is itself semiFredholm and hence \(R(B+K_{2} \mu I)\) is a closed subspace, which completes the demonstration We now show that \(\operatorname{reg}(B+K_{2})\subseteq M_{e}(B+K_{2})\). Let \(\rho e(B)=\bigcup_{k}C_{k}=\operatorname{reg}(B+K_{2})\) where \(C_{k}\) is the nth kconnected component of \(\rho_{e}(B)\).
For any \(k\in \mathbb{N}^{\ast}\), \(C_{k}\cap \operatorname{reg}(B)=\emptyset\). Let \(\mu \in C_{k}\cap \operatorname{reg}(B)=C_{k}\cap M_{e}(B)\).
Two cases are possible.
First case. \(B \mu I\) is onetoone and therefore \(N(B+K_{2} \mu I)\subseteq N(B \mu I)=0\Rightarrow B+K_{2}\mu I\) is onetoone, hence \(C_{k}\subseteq M_{e}(B+K_{2})\).
Second case. \(B \mu I\) is onto and therefore \(H_{1}\subseteq R(B \mu I)\). For any \(u\in H_{1}\), there is v such that \((B \mu I)v=u\);
$$ (B \mu I)Qv=(B \mu I)v(B \mu I) (IQ)v, $$
\((B  \mu I)v\in H_{1}\), \((IQ)v\in H_{1}\Rightarrow(B \mu I)(IQ)v\in H_{1}\), \(Qv=\sum_{j}P_{j}^{\prime}v=v_{j}\) where \(v_{j}=\sum_{j}P_{j}^{\prime}v\). We have
$$(B \mu I)Qv=\sum_{j}(B \mu I)v_{j}= \sum_{j}(B \mu I)P_{j}v_{j}+ \sum_{j}(B \mu I) (IP_{j})v_{j}, $$
where
$$\sum_{j}(B \mu I)P_{j}v_{j} \in N_{j} \quad \mbox{and}\quad \sum_{j}(B \mu I) (IP_{j})v_{j}\in M_{j}^{\prime} $$
because \(v_{j}\in N_{j}^{\prime }\subseteq M_{j1}^{\prime}\Rightarrow(IP_{j})v_{j}\in M_{j}^{\prime }\). We have
$$(B \mu I)P_{1}v_{1}=(B \mu I)Qv\sum _{j>1}(B \mu I)P_{j}v_{j}\sum _{j}(B \mu I) (IP_{j})v_{j}\in M_{1}. $$
Hence \((B \mu I)P_{1}v_{1}=0\) because
$$(B \mu I)P_{1}v_{1}\in M_{1}\cap N_{1}=0. $$
As \(\mu \in \operatorname{reg}(B)\Rightarrow \mu _{1}= \mu\) and \(B \mu I=B \mu _{1}+( \mu _{1} \mu)I\). We can see that \(P_{1}v_{1}=0\), hence \(v_{1}\in M_{1}\cap N_{1}=0\Rightarrow v_{1}= 0\). Suppose we have shown that \(v_{1}=v_{2}=v_{3}=\cdots=v_{n}=0\) and we demonstrate that \(v_{n+1}=0\). As previously \((B \mu I)P_{n+1}v_{n+1}\in M_{n+1}^{\prime}\) hence \((B \mu I)P_{n+1}v_{n+1}=0\), and as \(\mu\in \operatorname{reg}(B)\Rightarrow \mu = \mu _{n+1}\Rightarrow P_{n+1}v_{n+1}=0\), then \(v_{n+1}\in M_{n+1}\cap N_{n+1}^{\prime}=0\Rightarrow v_{n+1}=0\). It follows that \((B \mu I)Qv=0\) hence \(u=(B \mu I)w\) with \(w=(IQ)v\in H_{1}\); therefore \(u=(B +K_{2} \mu I)w\Rightarrow u\in R(B+K_{2} \mu I)\). Hence \(H_{1}\subseteq R(B+K_{2} \mu I)\) and (by Lemma 7) we have \(H_{1}^{\perp}\subseteq R(B+K_{2} \mu I)\); hence \(H\subseteq R(B+K_{2} \mu I)\), which implies that \(B+K_{2} \mu I\) is onto and \(C_{k}\subseteq M_{e}(B+K)_{2}\) and therefore \(\rho _{e}(B)=\sum_{k}C_{k}\subseteq M_{e}(B+K_{2})\), which is equivalent to \(\rho_{e}(B)=M_{e}(B+K_{2})\). □
Theorem 7
Let
\(A\in L(H)\). There is a compact operator
\(K\in K(H)\)
such that
$$ \rho_{e}(A)=M_{e}(A+K). $$
Proof
This is an immediate consequence of Lemma 5 and Lemma 9; simply take \(K=K_{1}+K_{2}\). □
Corollary 1
Let
A
be a Riesz operator. Then there exists a normal compact operator
K
such that
\((AK)\in QN(H)\).
Proof
A satisfies the hypothesis of Lemma 9 and also \(\rho_{e}(A)=\mathbb{C}\backslash0\) and \(\operatorname{reg}(A)=\mathbb{C}\backslash\sigma(A)=\rho(A)\).
\(\rho_{e}(A)\cap \operatorname{reg}(A)=\rho(A)=M_{e}(A)\). Hence there is a compact operator K such that \(\rho(AK)=M_{e}(AK)=\rho_{e}(A)=\mathbb{C}\backslash0\Rightarrow\sigma(AK)=0\Rightarrow(AK)\in QN(H)\). We find the result of [14]. □
Remark 6
The result of Stampfli is a special case of Theorem 7 because if \(\mu\in\Phi_{0}(A)\), then \(\mu \in\rho_{e}(A)\subseteq M_{e}(A+K)\), hence \(A+K \mu I\) is onetoone or onto. If \(A+K \mu I\) is onetoone (resp. onto), then \(N(A+K \mu I)=0\) (resp. \(N(A^{\ast}+K^{\ast} \mu I)=0\)) and \(\operatorname{ind}(A+K \mu I)=\operatorname{ind}(A \mu I)=0N(A^{\ast}+K^{\ast} \mu I)=0\) (resp. \(N(A+K \mu I)=0\)) and therefore \(A+K \mu I\) is onetoone and onto. Hence \(\mu \in\rho(A+K)\Rightarrow\Phi_{0}(A)=\rho(A+K)\).
Remark 7
In 1966, West has shown in [14] that for any Riesz operator \(A\in L(H)\) such that \(\rho_{e}(A)\subseteq\mathbb{C}\backslash0\) there is a compact operator \(K\in K(H)\) and \(Q\in QN(H)\) such that \(A=K+Q\). We find this result as a special case of our result (see Corollary 1).