# Fredholm Equation in Smooth Banach Spaces and Its Applications

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## Abstract

The object of this paper is to consider the Fredholm equation (i.e., \(x-ABx=y_o\)) for smooth Banach spaces. In particular, we will prove that the classical assumption of compactness of *AB* is redundant in some circumstances. In this paper, we show that Coburn’s theorem holds for another classes of generally of nonnormal operators. Moreover, as a corollary, we find the distance from some operator to compact operators.

## Keywords

Semi-inner product Smooth Banach space Fredholm equation Spectrum Compact operators## Mathematics Subject Classification

Primary 46B20 45B05 46C50 Secondary 47A50 47A10 47A53## 1 Preliminaries

*k*(

*t*,

*u*) is in \(L^2([a,b]\times \![a,b])\) and

*f*,

*h*are in \(L^2[a,b]\). The Fredholm equation and the Fredholm operator have been intensively studied in connection with integral equations, as well as operator theory. Let

*X*be a real or complex Banach space. Assume that \(A\in \mathcal {K}(X)\), \(y_o\in X\). From our modern perspective we can think of the Fredholm equation as

### Theorem 1.1

*X*be a Banach space. If \(A\in \mathcal {K}(X)\), \(\lambda \in \mathbb {K}\) and \(\lambda \ne 0\), then for every \(y_o\) in

*X*there is an

*x*in

*X*such that

*u*such that \(\lambda u-Au=0\) is \(u=0\). If this condition is satisfied, then the solution to (1.2) is unique.

In this paper in Sect. 2 we consider a generalized Fredholm equation. Our main result (Theorems 2.4) is an extension of the Fredholm Alternative to the setting of possibly noncompact operators. The most important theorems of this paper will be contained in Sects. 2 and 4. In Sect. 2 we will give a characterization for the general idea of the Fredholm equation (i.e., \(x-ABx=y_o\)) under certain assumptions. In particular, we will prove that the assumption of compactness of *AB* is not needed (in some circumstances). The second part of this work (i.e., Sects. 3, 4, 5) is devoted to applications of Theorems 2.4 and 2.5. Our main goal in Sect. 3 is to get some information on the spectrum of a noncompact operator on smooth space. Moreover, in Sect. 4 we will compute the distance from some operator to the subspace \(\mathcal {K}(X)\) (see Theorem 4.2).

### 1.1 Semi-inner Product

Let \((X, \Vert \cdot \!\Vert )\) be a normed space over \(\mathbb {K}\in \{\mathbb {R},\mathbb {C}\}\). Lumer [6] and Giles [2] proved that in a space *X* there always exists a mapping \(\left[ \cdot |\cdot \right] : X\times X\rightarrow \mathbb {K}\) satisfying the following properties:

(sip1) \(\forall _{x,y,z\in X}\ \forall _{\alpha ,\beta \in \mathbb {K}}:\quad \left[ \alpha x+\beta y|z\right] =\alpha \left[ x|z\right] +{\beta }\left[ y|z\right] \);

(sip2) \(\forall _{x,y\in X}\ \forall _{\alpha \in \mathbb {K}}:\quad \left[ x|\alpha y\right] =\overline{\alpha }\left[ x|y\right] \);

(sip3) \(\forall _{x,y\in X}:\quad |\left[ x|y\right] |\le \Vert x\Vert \cdot \Vert y\Vert \);

(sip4) \(\forall _{x\in X}:\quad \left[ x|x\right] =\Vert x\Vert ^2\).

Such a mapping is called a *semi-inner product* (s.i.p.) in *X* (generating the norm \(\Vert \cdot \Vert \)). There may exist infinitely many different semi-inner products in *X*. There is a unique one if and only if *X* is *smooth* (i.e., there is a unique supporting hyperplane at each point of the unit sphere). If *X* is an inner product space, the only s.i.p. on *X* is the inner-product itself.

We quote some additional result concerning s.i.p.. Let *X* be a smooth, reflexive Banach space. Then there exists a unique s.i.p. \(\left[ \cdot |\cdot \right] :X\times X\rightarrow \mathbb {K}\). If *A* is a bounded linear operator from *X* to itself, then \(f_z(\cdot ):=\left[ A(\cdot )|z\right] \) is a continuous linear functional, and from the generalized Riesz-Fischer representation theorem it follows that there is a unique vector \(A_*(z)\) such that \(\left[ Ax|z\right] =\left[ x|A_*(z)\right] \) for all *x* in *X*. Of course, in a Hilbert space we have \(A_*=A^*\). In general case the mapping \(A_*:X\rightarrow X\) is not linear but it still has some good properties:

(sip5) \((AB)_*=B_*A_*\),

(sip6) \((\alpha A)_*=\overline{\alpha } A_*\),

(sip7) \(\forall _{x\in X}:\quad \Vert A_*(x)\Vert \le \Vert A\Vert \cdot \Vert x\Vert \).

*x*in a normed space

*X*, we consider the set

*X*, we denote by \(\mathbf {B}(X)\) the closed unit ball in

*X*. By \(\text {ext}D\) we will denote the set of all extremal points of a set

*D*.

Let \(\mathcal {L}(X)\) denote the space of all bounded linear operators on a space *X*, and *I* the identity operator. We write \(\mathcal {K}(X)\) for the space of all compact operators on *X*. For \(A\in \mathcal {L}(X)\), we denote the set \(\mathcal {M}(A):=\{x\in \mathbf {B}(X): \Vert Ax\Vert =\Vert A\Vert , \Vert x\Vert =1 \}\).

### 1.2 Geometry of Space \(\mathcal {L}(X)\)

The main tool in our approach in the next section is a theorem due to Lima and Olsen [5] which characterizes the extremal points of the closed unit ball in \(\mathcal {K}(X)\).

### Theorem 1.2

*X*be a reflexive Banach spaces over the field \(\mathbb {K}\). The following conditions are equivalent:

- (a)
\(f\in \mathrm{ext}\mathbf {B}\left( \mathcal {K}(X)^*\right) \),

- (b)
there exist \(y^*\in \mathrm{ext}\mathbf {B}(X^*)\) and \(x\in \mathrm{ext}\mathbf {B}(X)\) such that \(f(T)=y^*(Tx)\) for every \(T\in \mathcal {K}(X)\).

It is possible that there is space *X* such that \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \) where \(\mathcal {K}(X)^\bot := \left\{ f\in \mathcal {L}(X)^* : \mathcal {K}(X)\subset \ker f\right\} \). In this case, if \(\varphi =\varphi _1+\varphi _2\) is the unique decay of \(\varphi \) in \(\mathcal {L}(X)^*\), then \(\Vert \varphi \Vert =\Vert \varphi _1\Vert +\Vert \varphi _2\Vert \). This fact is referred to [4, p. 28] (see also [3, Theorem 4]), where it was proved that \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \) when \(X=\left( \bigoplus _{n=1}^\infty X_n\right) _{l_p}\) or \(X=\left( \bigoplus _{n=1}^\infty X_n\right) _{c_0}\), with \(p\in (1,\infty )\), \(\dim X_n<\infty \). In particular, \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \) if \(X=l_p\) or \(X=c_0\), with \(p\in (1,\infty )\). Moreover, Lima has obtained the equality \(\mathcal {L}(L_2(\mu ))^*=\mathcal {K}(L_2(\mu ))^*\oplus _1 \mathcal {K}(L_2(\mu ))^\bot \) for some measure \(\mu \) (see [3]). On the other hand, if \(p\ne 2\) and \(\mu \) is not purely atomic, then it is known that \(\mathcal {L}(L_p(\mu ))^*\ne \mathcal {K}(L_p(\mu ))^*\oplus _1 \mathcal {K}(L_p(\mu ))^\bot \) (see [3, Theorem 11]).

If \(\dim X<\infty \), then the desired equality \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \) holds trivially, because in such a case, \(\mathcal {L}(X)=\mathcal {K}(X)\) and \(\mathcal {K}(X)^\bot =\{0\}\).

Recall that a *smooth point**x* of the unit sphere of a Banach space *X* is defined by the requirement that \(x^*(x) = 1\) for a uniquely determined \(x^*\in B(X^*)\), i.e., \(\text{ card } J(x)=1\). In this case the norm of *X* is Gâteaux differentiable at *x* with derivative \(x^*\). We will need a characterization of points of smoothness in the space of operators \(\mathcal {L}(X)\) (cf. [7]).

### Theorem 1.3

*T*is a smooth point of the unit sphere of \(\mathcal {L}(X)\) if and only if

- (s1)
\(\inf \{\Vert T+K\Vert : K\in \mathcal {K}(X)\}<1\);

- (s2)
there is exactly one \(x_o\) in the unit sphere of

*X*(up to multiplication with scalars of modulus 1) for which \(\Vert T(x_o)\Vert =1\); - (s3)
the point \(Tx_o\) is smooth.

## 2 Fredholm Equation

The following theorem can be considered as an extension of the Fredholm Alternative (i.e., Theorem 1.1). We want to show that the assumption of compactness of operators may be redundant (in some circumstances). It is more convenient to consider two operator *A*, *B* instead of only one. We are ready to prove the main result of this section.

### Theorem 2.1

*X*is a reflexive smooth Banach space over \(\mathbb {K}\). Suppose that \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \). Let \(A,B\in \mathcal {L}(X)\), \(\Vert A\Vert \cdot \Vert B\Vert \le 1\). Suppose that there is an operator \(C\in \mathcal {K}(X)\) such that \(\Vert AB-C\Vert <\Vert AB\Vert \). The following five conditions are equivalent:

- (a)
\(I-AB\) is invertible;

- (b)
\(I-AB\) is surjective;

- (c)
\(I-AB\) is injective;

- (d)
\(I-AB\) has dense range;

- (e)
\(\Vert I+AB\Vert <2\).

### Proof

The implications (a)\(\Rightarrow \)(b)\(\Rightarrow \)(d) and (a)\(\Rightarrow \)(c) are immediate. First we prove (e)\(\Rightarrow \)(a). We get \(\left\| I-\frac{1}{2}\left( I-AB\right) \right\| =\frac{1}{2}\left\| I+AB\right\| < 1\), which means that \(\frac{1}{2}\left( I-AB\right) \) is invertible. Therefore \(I-AB\) is invertible.

From the above, we have \(\Vert A\Vert \cdot \Vert B\Vert =\Vert AB\Vert =\Vert ABx\Vert \le \Vert A\Vert \cdot \Vert Bx\Vert \le \Vert A\Vert \cdot \Vert B\Vert \), whence \(\Vert Bx\Vert =\Vert B\Vert \), and consequently \(x\in \mathcal {M}(B)\), \(\mathcal {M}(B)\ne \emptyset \). Furthermore, we obtain \(\left\| A\left( \frac{Bx}{\Vert Bx\Vert }\right) \right\| =\frac{1}{\Vert Bx\Vert }\left\| ABx\right\| =\frac{1}{\Vert B\Vert }\left\| AB\right\| \overset{(2.1)}{=}\frac{1}{\Vert B\Vert }\Vert A\Vert \cdot \Vert B\Vert =\Vert A\Vert \), whence \(\frac{Bx}{\Vert Bx\Vert }\in \mathcal {M}(A)\), \(\mathcal {M}(A)\ne \emptyset \).

*X*is smooth. Thus we get \(card J(Bx)=1\), whence

*X*yields that \(card J(x)=1\) and \(card J(ABx)=1\). So, combining (2.7) and (2.8), we immediately get

*Bx*in place of

*y*in the above equality we get

*X*yields that \(card J(x)=1\). So, we immediately get

*Y*is a closed convex subset of

*X*. By a theorem of James,

*Y*is a weak-compact subset. Let us consider a function \(AB|_{Y}:Y\rightarrow Y\). By the Schauder Fixed-Point Theorem, there is a vector \(x_1\) in

*Y*such that \(ABx_1=x_1\). That means \(0=(I-AB)x_1\) which implies that also \(\ker (I-AB)\ne \{0\}\). Since \(I-AB\) was assumed injective, \(\ker (I-AB)=\{0\}\), a contradiction.

We prove (d)\(\Rightarrow \)(e). Suppose that \(I-AB\) has dense range. Assume, for a contradiction, that \(\Vert I+AB\Vert \ge 2\). Using similar elementary techniques, one may prove \(\left[ AB(\cdot )|x\right] =\left[ \cdot |x\right] \) for some \(x\in \mathcal {M}(B)\), i.e., (2.11). Thus we have \(\left[ (I-AB)y|x\right] =0\) for all \(y\in X\). Since \(I-AB\) has dense range, \(x=0\) and we obtain a contradiction. \(\square \)

Careful reading of the proof of Theorem 2.1 (more precisely, the proof of (c)\(\Rightarrow \)(e)) shows that we can get the following.

### Proposition 2.2

*X*,

*A*,

*B*,

*C*be as in Theorem 2.1. Suppose that \(\Vert A\Vert \cdot \Vert B\Vert =1\). Then the following three statements are equivalent:

- (i)
\(\Vert I+AB\Vert =2\);

- (ii)
\(\exists _{x\in \mathcal {M}(B)}\ \forall _{y\in X}:\ \Vert B\Vert \cdot \left[ Ay|x\right] =\Vert A\Vert \cdot \left[ y|Bx\right] \);

- (iii)
\(\exists _{x\in \mathcal {M}(B)}\ \forall _{y\in X}:\ \left[ AB(y)|x\right] =\left[ y|x\right] \).

### Proof

We can add another proposition to our list.

### Proposition 2.3

*X*be as in Theorem 2.1. Let \(E\in \mathcal {L}(X)\), \(\Vert E\Vert \le 1\). Suppose that there is an operator \(C\in \mathcal {K}(X)\) such that \(\Vert E-C\Vert <\Vert E\Vert \). Then the following three statements are equivalent:

- (i)
\(\Vert I+E\Vert =2\);

- (ii)
\(\exists _{x\in \mathcal {M}(E)}\ \forall _{y\in X}:\ \left[ y|x\right] =\left[ y|Ex\right] \);

- (iii)
\(\exists _{x\in \mathbf {B}(X), \Vert x\Vert =1}\ \forall _{y\in X}:\ \left[ Ey|x\right] =\left[ y|x\right] \).

### Proof

Putting *IE* in place of *AB* in the above proposition we get (i)\(\Leftrightarrow \)(ii). Putting *EI* in place of *AB* in the above proposition we get (i)\(\Leftrightarrow \)(ii). \(\square \)

As an immediate consequence of Theorem 2.1, we have the following. Actually, it is not necessary to assume that *AB* is a compact operator. In fact, it suffices to assume reflexivity, smoothness and the two properties: \(\mathcal {L}^*=\mathcal {K}^*\oplus _1 \mathcal {K}^\bot \), \(\Vert A\Vert \cdot \Vert B\Vert \le 1\).

### Theorem 2.4

*X*is a reflexive smooth Banach space. Suppose that \(\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot \). Let \(A,B\in \mathcal {L}(X)\) and \(\Vert A\Vert \cdot \Vert B\Vert \le 1\). Suppose that there is an operator \(C\in \mathcal {K}(X)\) such that \(\Vert AB-C\Vert <\Vert AB\Vert \). Then the following four statements are equivalent:

- (i)
for every \(y_o\) in

*X*there is an unique*u*in*X*such that \(u-ABu=y_o\); - (ii)
for every \(y_o\) in

*X*there is a*u*in*X*such that \(u-ABu=y_o\); - (iii)
the only vector

*x*such that \(x-ABx=0\) is \(x=0\); - (iv)
\(\Vert I+AB\Vert <2\);

### Proof

It is clear that \(I-AB\) is invertible, [surjective, (injective)] if and only if (i) holds , [(ii) holds, ((iii) holds)]. Now by applying Theorem 2.1 we arrive at the desired assertion. \(\square \)

The Fredholm Alternative, together with Theorem 2.1, allows us to give an extension of Theorem 2.4.

### Theorem 2.5

*X*be as in Theorem 2.1. Let \(A,B,D\in \mathcal {L}(X)\). Suppose that \(1\notin \sigma _p(AB)\) and \(\Vert A\Vert \cdot \Vert B\Vert \le 1\). Suppose that there is an operator \(C\in \mathcal {K}(X)\) such that \(\Vert AB-C\Vert <\Vert AB\Vert \). Suppose that \(AB-D\in \mathcal {K}(X)\) Then the following three statements are equivalent:

- (i)
for every \(y_o\) in

*X*there is an unique*u*in*X*such that \(u-Du=y_o\); - (ii)
for every \(y_o\) in

*X*there is a*w*in*X*such that \(w-Dw=y_o\); - (iii)
the only vector

*x*such that \(x-Dx=0\) is \(x=0\).

### Proof

The proof of (ii)\(\Rightarrow \)(i) runs similarly, but it is presented here for the convenience. Now suppose that \(I-D\) is surjective. Since \(I-AB\) is injective, whence (applying again Theorem 2.1) \(I-AB\) is a bijection. Suppose, for a contradiction, that \(I-D\) is not invertible. In a similar way as in the proof of (ii)\(\Rightarrow \)(i) we obtain (2.12). The operator \(I-AB\) is invertible and \(I-D\) is not invertible. Thus, it follows from (2.12) that \(I+(I-AB)^{-1}(AB-D)\) is also not invertible. Since the operator \((I-AB)^{-1}(AB-D)\) is compact, this tells us that \(\left( I+(I-AB)^{-1}(AB-D)\right) \) is not surjective (see Theorem 1.1). Hence there exists a vector \(e\in X\) so that \(x+(I-AB)^{-1}(AB-D)x\ne e\) for all \(x\in X\). Now, it is easy to check that \(x-Dx\ne e-ABe\) (for all \(x\in X\)). Now we have a contradiction: \(I-D\) is not surjective. The proof is complete. \(\square \)

## 3 An Application: Spectrum

As an application of the results in the previous section, we consider the notion of a spectrum of an operator. Let \(\mathbb {T}:=\{\lambda \in \mathbb {K}: |\lambda |=1\}\). We next explore some consequences of Theorem 2.1.

### Theorem 3.1

*X*,

*A*,

*B*,

*C*be as in Theorem 2.1. Then

### Proof

Clearly, \(\sigma _p(AB)\subset \sigma (AB)\). Let us fix \(\lambda \) in \(\mathbb {T}\cap \sigma (AB)\). Then \(I-\frac{1}{\lambda }AB\) is not invertible. According to Theorem 2.1 the operator \(I-\frac{1}{\lambda }AB\) is not injective. Thus \(\lambda \in \sigma _p(AB)\) and so \(\lambda \in \mathbb {T}\cap \sigma _p(AB)\). \(\square \)

The following theorem was discovered by Weyl [8].

### Theorem 3.2

[8] Suppose that *H* is a complex Hilbert space. If \(A\in \mathcal {L}(H)\) and *K* is a compact operator, then \(\sigma (A+K)\setminus \sigma _p(A+K)\subset \sigma (A)\).

It makes sense to replace the operator *AB* by *AI*. Some part of Theorem 3.2 can be strengthen as follows.

### Theorem 3.3

*X*be as in Theorem 2.1. Let \(A\in \mathcal {L}(X)\), \(\Vert A\Vert =1\). Suppose that there is an operator \(C\in \mathcal {K}(X)\) such that \(\Vert A-C\Vert <1\). Then

*K*.

### Proof

Fix a number \(\lambda \in \mathbb {T}\). Assume that \(\lambda \in \sigma (A+K)\setminus \sigma _p(A+K)\). Suppose, for a contradiction, that \(\lambda \notin \sigma _p(A)\). Then \(I-\frac{1}{\lambda }A\) is injective. Define an operator *D* by \(D:=\frac{1}{\lambda }(A+K)\). Since \(\lambda \notin \sigma _p(A+K)\), it follows that \(I-\frac{1}{\lambda }(A+K)\) is injective; this means that \(I-D\) is injective.

The inequality \(\Vert A-C\Vert <1\) implies \(\left\| \frac{1}{\lambda }A-\frac{1}{\lambda }C\right\| <\left\| \frac{1}{\lambda }A \right\| \). It is easy to see that \(\frac{1}{\lambda }A-D\in \mathcal {K}(X)\). Moreover, we have \(1\notin \sigma _p\left( \frac{1}{\lambda }A\right) \). Using Theorem 2.5 we see that \(I-D\) is also invertible, so \(I-\frac{1}{\lambda }(A+K)\) is invertible. Hence \(\lambda \notin \sigma (A+K)\). We have our desired contradiction. \(\square \)

As a consequence of this result, we get a corollary.

### Corollary 3.4

*X*,

*A*be as in Theorem 3.3. Then

In the following we will show how our knowledge on the smooth operators can be used in order to characterize some points in the spectrum of bounded operator. Note that the following applies to \(l^p\)-spaces and, most particularly, to Hilbert spaces.

### Proposition 3.5

*B*is a smooth point of the unit sphere of \(\mathcal {L}(X)\). Suppose that \(A_1,A_2\) are linearly independent. Then

### Proof

By assumption, *X* is smooth. It follows from Theorem 1.3 (condition (s1)) that, with some \(K\in \mathcal {K}(X)\), \(\Vert B-K\Vert <1=\Vert B\Vert \). Fix \(\alpha \in [0,1]\). Fix \(\lambda \in \mathbb {T}\) such that \(\lambda \in \sigma (A_1B)\cap \sigma (A_2B)\). We define *A*, *C* as follows: \(A:=\alpha A_1+(1-\alpha ) A_2\) and \(C:=\alpha A_1K+(1-\alpha ) A_2K\). Thus we have \(C\in \mathcal {K}(X)\). We have \(A\ne 0\), because \(A_1,A_2\) are linearly independent.

*X*(up to multiplication with scalars of modulus 1) for which \(\Vert Bx_o\Vert =1\). In other words, \(\mathcal {M}(B)=\{\beta x_o: \beta \in \mathbb {K},\, |\beta |=1\}\). Combining (3.1) with to Proposition 2.2 gives that

## 4 Distance from Operator to Compact Operators

The problem of best approximation of bounded linear operators \(\mathcal {L}(X)\) on a Banach space *X* by compact operators \(\mathcal {K}(X)\) has been of great interest in the twentieth century. This section is a small sample of it.

Let \(\mathcal {L}(H)\) be the algebra of all bounded operators on infinite-dimensional complex Hilbert space *H*. An operator \(A\in \mathcal {L}(H)\) is *hyponormal* if \(A^*A\ge AA^*\). Clearly every hyponormal operator is normal. The interesting result on hyponormal operators may be found in [1, Corollary 3.2].

### Theorem 4.1

[1] If \(A\in \mathcal {L}(H)\) is hyponormal and has no isolated eigenvalues of finite multiplicity, then \(\Vert A\Vert \le \Vert A+K\Vert \), for each compact operator *K*.

As an immediate consequence of the above result, we deduce that if \(U\in \mathcal {L}(l^2)\) is the unilateral shift, then \(dist (U,\mathcal {K}(l^2))=1\). But, this fact can be obtained (or even more generally) by using our next result.

We prove some modifications of Theorem 4.1, where the Hilbert space *H* is replaced by a smooth Banach space *X*, and where the condition is satisfied for unit operator only.

### Theorem 4.2

*X*be as in Theorem 2.1. Let \(A\in \mathcal {L}(X)\), \(\Vert A\Vert =1\). Suppose that

### Proof

Fix a number \(\lambda \in \mathbb {T}\cap \left( \sigma (A)\setminus \sigma _p(A)\right) \). Assume, for a contradiction, that \(dist (A,\mathcal {K}(X))\ne 1\). Since \(\lambda \notin \sigma _p(A)\), we see that \(I-\frac{1}{\lambda }A\) is injective. Since \(dist (A,\mathcal {K}(X))\le \Vert A-0\Vert =\Vert A\Vert =1\), we obtain \(dist (A,\mathcal {K}(X))< \Vert A\Vert \). It follows that \(dist \left( \frac{1}{\lambda }A,\mathcal {K}(X)\right) < \left\| \frac{1}{\lambda }A\right\| \). Hence, there is \(C\in \mathcal {K}(X)\) such that \(\left\| \frac{1}{\lambda }A-C\right\| <\left\| \frac{1}{\lambda }A\right\| \). Applying Theorem 2.1 (more precisely, (a)\(\Leftrightarrow \)(c)) we see that \(I-\frac{1}{\lambda }A\) is invertible. This is a contradiction, since \(\lambda \in \sigma (A)\). \(\square \)

Here a small application is given.

### Corollary 4.3

If \(p\in (1,\infty )\), define \(U\in \mathcal {L}(l^p)\) by \(U(x_1,x_2,\ldots ):=(0,x_1,x_2,\ldots )\). \(U(x_1,x_2,\ldots ):=(0,x_1,x_2,\ldots )\). Then \(dist (U,\mathcal {K}(l^p))=1\).

### Proof

We know that \(\sigma (U)=\{\lambda \in \mathbb {K}:|\lambda |\le 1\}\). It is easy to check that \(\sigma _p(U)=\emptyset \). From this it follows that \(\mathbb {T}\cap \left( \sigma (U)\setminus \sigma _p(U)\right) =\mathbb {T}\). By Theorem 4.2 we get \(dist (U,\mathcal {K}(l^p))=1\). \(\square \)

### Theorem 4.4

Let *H* be a (real or complex) Hilbert space. Suppose that \(T\in \mathcal {L}(H)\) is injective but not invertible. Let \(\alpha \overline{\beta }\in (0,+\infty )\) and \(-\frac{\alpha }{\beta }\notin \sigma (T)\). Let *T* satisfy \(T+T^*\ge 0\). If \(A_{\alpha ,\beta }:=(\alpha I+\beta T)^{-1}\), then \(dist (A_{\alpha ,\beta }, \mathcal {K}(H))=\frac{1}{|\alpha |}\). In particular, \(A_{\alpha ,\beta }\) is a well-defined operator.

### Proof

First it must be shown that \(A_{\alpha ,\beta }\) is well defined. By assumption, \(-\alpha \notin \beta \sigma (T)\). This clearly forces \(0\notin \alpha +\sigma (\beta T)=\sigma (\alpha I+\beta T)\). From this we deduce that \(\alpha I+\beta T\) is invertible. So \(A_{\alpha ,\beta }\) is a well-defined operator.

We want to show that \(\Vert A\Vert =\frac{1}{|\alpha |}\). By assumption, \(\beta T\) is not invertible, and hence \(0\in \sigma (\beta T)\). From this it follows that \(\alpha \in \alpha +\sigma (\beta T)\), and thus \(\alpha \in \sigma (\alpha I+\beta T)=\sigma (A_{\alpha ,\beta }^{-1})\). So we conclude that \(\frac{1}{\alpha }\in \sigma (A_{\alpha ,\beta })\) and hence \(\frac{1}{|\alpha |}\le \Vert A_{\alpha ,\beta }\Vert \).

*x*be an arbitrary unit vector in

*H*. Thus we have

We next show that \(dist (A_{\alpha ,\beta }, \mathcal {K}(H))=\frac{1}{|\alpha |}\). It suffices to show that \(dist (\alpha A_{\alpha ,\beta }, \mathcal {K}(H))=1\). Recalling that \(\frac{1}{\alpha }\in \sigma (A_{\alpha ,\beta })\), we see that \(1\in \sigma (\alpha A_{\alpha ,\beta })\).

*T*is injective we have \(w=0\). This is a contradiction. Thus \(1\notin \sigma _p(\alpha A_{\alpha ,\beta })\). Finally, we deduce that \(1\in \sigma (\alpha A_{\alpha ,\beta })\setminus \sigma _p(\alpha A_{\alpha ,\beta })\).

This gives \(\mathbb {T}\cap (\sigma (\alpha A_{\alpha ,\beta })\setminus \sigma _p(\alpha A_{\alpha ,\beta }))\ne \emptyset \). Applying Theorem 4.2, we see that \(dist (\alpha A_{\alpha ,\beta }, \mathcal {K}(H))=1\). \(\square \)

If *H* is a complex Hilbert space and \(T\in \mathcal {L}(H)\) such that \(\left\langle Tx|x\right\rangle =0\) for all *x* in *H*, then \(T=0\). This is not true for real Hilbert spaces. For example, let \(U\in \mathcal {L}(\mathbb {R}^2)\) be the linear operator given by \(U(x,y):=(-y,x)\). Then we we have \(\left\langle Ux|x\right\rangle =0\) for all *x* in *H* However, \(U\ne 0\). So it makes sense to consider the following result.

### Corollary 4.5

Let *H* be a real Hilbert space. Suppose that \(T\in \mathcal {L}(H)\) is injective but not invertible. Let \(\alpha \beta \in (0,+\infty )\) and \(-\frac{\alpha }{\beta }\notin \sigma (T)\). Let *T* satisfy \(\left\langle Tx|x\right\rangle =0\) for all *x* in *H*. If \(A_{\alpha ,\beta }:=(\alpha I+\beta T)^{-1}\), then \(dist (A_{\alpha ,\beta }, \mathcal {K}(H))=\frac{1}{|\alpha |}\).

### Proof

Fix \(x\in H\). It follows that \(\left\langle (T+T^*)x|x\right\rangle =\left\langle Tx|x\right\rangle +\left\langle x|T^*x\right\rangle =\left\langle Tx|x\right\rangle +\left\langle Tx|x\right\rangle =0\ge 0\). Theorem 4.4 now leads to \(dist (A_{\alpha ,\beta }, \mathcal {K}(H))=\frac{1}{|\alpha |}\). \(\square \)

## 5 An Application: System of Linear Equations

Systems of linear equations (such as (5.1)) arise in a number of physical problems. In this section we will discuss solutions of such system by relating the system to a certain noncompact operator.

### Theorem 5.1

### Proof

## 6 Remarks: Invertibility

*X*is a Banach space. It is standard that if \(\Vert I-E\Vert <1\), then

*E*is invertible. The converse does not hold in general. We prove that under certain conditions it is true. We end this paper with a simple result. However, a few words motivating the proof are appropriate. Let us consider the operator \(T:=\frac{1}{2}(I- AB)\). If \(\Vert A\Vert \cdot \Vert B\Vert <1\), then

*T*is invertible. So the assumption \(\Vert A\Vert \cdot \Vert B\Vert \le 1\) makes the invertibility problem more interesting. In particular, the implication (b)\(\Rightarrow \)(a) (in Theorem 6.1) seem to be amazing.

### Proposition 6.1

*X*,

*A*,

*B*,

*C*be as in Theorem 2.1. If \(T:=\frac{1}{2}(I-AB)\), then the following three statements are equivalent:

- (a)
\(\Vert I-T\Vert <1\);

- (b)
*T*is invertible.

### Proof

It is easy to check that \(\Vert I+AB\Vert <2\) if and only if \(\Vert I-T\Vert <1\). It is clear that \(I-AB\) is invertible if and only if *T* is invertible. Applying Theorem 2.1 we at the desired conclusion. \(\square \)

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