Complex Analysis and Operator Theory

, Volume 13, Issue 3, pp 737–751

Fredholm Equation in Smooth Banach Spaces and Its Applications

• Paweł Wójcik
Open Access
Article

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

The terminology “Fredholm operator” recognizes the pioneering work of Erik Fredholm. In 1903 he published a paper that, in modern language, dealt with equations of the form
\begin{aligned} f(t)-\int \limits _a^b k(t,u)f(u)du=h(t), \end{aligned}
(1.1)
where k(tu) is in $$L^2([a,b]\times \![a,b])$$ and fh 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
\begin{aligned} x-Ax=y_o, \end{aligned}
with a single unknown vector $$x\in X$$. We recall the celebrated Fredholm Alternative. This great result has a number of applications in the theory of integral equations.

Theorem 1.1

(The Fredholm Alternative) Let 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
\begin{aligned} \lambda x-Ax=y_o \end{aligned}
(1.2)
if and only if the only vector 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) $$\forall _{x\in X}:\quad \Vert A_*(x)\Vert \le \Vert A\Vert \cdot \Vert x\Vert$$.

For a vector x in a normed space X, we consider the set
\begin{aligned} J(x):=\{\varphi \in X^*: \ \varphi (x)=\Vert x\Vert ,\ \Vert \varphi \Vert =1\}. \end{aligned}
(1.3)
By the Hahn-Banach Extension Theorem we get $$J(x)\ne \emptyset$$. In this paper, for a normed space 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

[5] Let X be a reflexive Banach spaces over the field $$\mathbb {K}$$. The following conditions are equivalent:
1. (a)

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

2. (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 pointx 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

[7] Suppose that $$X=\left( \bigoplus _{n=1}^\infty X_n\right) _{l_p}$$ with $$p\in (1,\infty )$$, $$\dim X_n<\infty$$ for some $$1< p<\infty$$. Then T is a smooth point of the unit sphere of $$\mathcal {L}(X)$$ if and only if
1. (s1)

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

2. (s2)

there is exactly one $$x_o$$ in the unit sphere of X (up to multiplication with scalars of modulus 1) for which $$\Vert T(x_o)\Vert =1$$;

3. (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 AB instead of only one. We are ready to prove the main result of this section.

Theorem 2.1

Assume that 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:
1. (a)

$$I-AB$$ is invertible;

2. (b)

$$I-AB$$ is surjective;

3. (c)

$$I-AB$$ is injective;

4. (d)

$$I-AB$$ has dense range;

5. (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.

Now, we prove (c)$$\Rightarrow$$(e). Suppose that $$I-AB$$ is injective. Assume, for a contradiction, that $$\Vert I+AB\Vert \ge 2$$. From the inequalities $$2\le \Vert I+AB\Vert \le 1+\Vert AB\Vert \le 1+\Vert A\Vert \cdot \Vert B\Vert \le 2$$ we get
\begin{aligned} \Vert AB\Vert =\Vert A\Vert \cdot \Vert B\Vert \ \ \text{ and }\ \ \Vert I+AB\Vert = 1+\Vert AB\Vert . \end{aligned}
(2.1)
Now, we prove that $$J(I+AB)\subset J(I)\cap J(AB)$$. Let $$\varphi \in J(I+AB)$$. This gives $$\varphi \in \mathcal {L}(X)^*$$, $$\varphi (I+AB)=\Vert I+AB\Vert$$ and $$\Vert \varphi \Vert =1$$. We will prove that $$\varphi (I)=1$$ and $$\varphi (AB)=\Vert AB\Vert$$. It follows that
\begin{aligned} 1+\Vert AB\Vert \overset{(2.1)}{=}\Vert I+AB\Vert =\varphi (I+AB)=\varphi (I)+\varphi (AB), \end{aligned}
and $$|\varphi (I)|\le 1$$ and $$|\varphi (AB)|\le \Vert AB\Vert$$. This clearly forces $$\varphi (I)= 1$$ and $$\varphi (AB)= \Vert AB\Vert$$ and, in consequences, the inclusion
\begin{aligned} J(I+AB)\subset J(I)\cap J(AB) \end{aligned}
(2.2)
is true. The set $$J(I+AB)$$ is convex. Furthermore, the set $$J(I+AB)$$ is a nonempty weak*-closed subset of weak*-compact unit ball $$\mathbf {B}(\mathcal {L}(X)^*)$$. From this it follows that $$J(I+AB)$$ is weak*-compact. Applying the Krein-Milman Theorem we see that there exists $$\mu \in \text {ext}J(I+AB)$$. An easy computation shows that the set $$J(I+AB)$$ is an extremal subset of $$\mathbf B (\mathcal {L}(X)^*)$$. Therfore
\begin{aligned} \text {ext}J(I+AB)\subset \text {ext}\mathbf {B}(\mathcal {L}(X)^*), \end{aligned}
and, in consequences, we get $$\mu \in \text {ext}\mathbf {B}(\mathcal {L}(X)^*)$$.
Now we prove that $$\mu \in \text {ext}\mathbf {B}(\mathcal {K}(X)^*)$$. Recall that $$\mathcal {L}(X)^*=\mathcal {K}(X)^*\oplus _1 \mathcal {K}(X)^\bot$$. From this it may be concluded that
\begin{aligned} \text {ext}\mathbf {B}(\mathcal {L}(X)^*)=\text {ext}\mathbf {B}(\mathcal {K}(X)^*)\oplus _1\{0\} \ \cup \ \{0\}\oplus _1\text {ext}\mathbf {B}(\mathcal {K}(X)^\bot ). \end{aligned}
Suppose that $$\mu = \mu _1+\mu _2$$ is the suitable decay of $$\mu$$. Namely, we have $$\mu _1\in \mathcal {K}(X)^*$$, $$\mu _2\in \mathcal {K}(X)^\bot$$ and $$1=\Vert \mu \Vert = \Vert \mu _1\Vert +\Vert \mu _2\Vert$$. This implies that
\begin{aligned} \mu =\mu _1\in \text {ext}\mathbf {B}(\mathcal {K}(X)^*) \ or \ \mu =\mu _2\in \text {ext}\mathbf {B}(\mathcal {K}(X)^\bot ), \end{aligned}
So it suffices to show that $$\mu _2=0$$. Assume, contrary to our claim, that $$\mu =\mu _2\in \text {ext}\mathbf {B}(\mathcal {K}(X)^\bot )$$. Hence $$\Vert \mu _2\Vert =1$$. From the assumption, we have that $$\Vert AB-C\Vert < \Vert AB\Vert$$ and $$C\in \mathcal {K}(X)$$. It follows that
\begin{aligned} \Vert AB\Vert&=\mu (AB)=\mu _2(AB)=\mu _2(AB)-0=\mu _2(AB)-\mu (C)=\\&=\mu _2(AB-C)\le \Vert AB-C\Vert <\Vert AB\Vert \end{aligned}
and we obtain a contradiction. Thus we must have $$\mu =\mu _1\in \text {ext}\mathbf {B}(\mathcal {K}(X)^*)$$.
Therefore it follows from (2.2) and Theorem 1.2 that
\begin{aligned} \mu (I+AB)=y^*(x+ABx) \ and \ \mu (I)=y^*(Ix)=y^*(x) \ and \ \mu (AB)=y^*(ABx) \end{aligned}
for some $$y^{*}\in \text {ext}\mathbf {B}(X^{*})$$ and $$x\in \text {ext}\mathbf {B}(X)$$.
This is summarized as follows:
\begin{aligned} y^*(x+ABx)=\Vert I+AB\Vert \ \ \text{ and } \ \ y^*(Ix)=1 \ \ \text{ and } \ \ y^*(ABx)=\Vert AB\Vert . \end{aligned}
Then we have $$\Vert I+AB\Vert =y^*(x+ABx)\le \Vert x+ABx\Vert \le \Vert I+AB\Vert$$ and $$\Vert AB\Vert =y^*(ABx)\le \Vert ABx\Vert \le \Vert AB\Vert$$. Thus we obtain the equalities:
\begin{aligned} \Vert x+ABx\Vert =\Vert I+AB\Vert \ \ \text{ and } \ \ \Vert ABx\Vert =\Vert AB\Vert , \end{aligned}
(2.3)
whence $$\mathcal {M}(AB)\ne \emptyset$$.

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 $$\frac{Bx}{\Vert Bx\Vert }\in \mathcal {M}(A)$$, $$\mathcal {M}(A)\ne \emptyset$$.

Combining (2.1) and (2.3), we immediately get
\begin{aligned} \Vert x+ABx\Vert =\Vert x\Vert +\Vert ABx\Vert . \end{aligned}
(2.4)
Using the properties of s.i.p., we obtain
\begin{aligned} \Vert A\Vert ^2\cdot \Vert B\Vert ^2&=\Vert AB\Vert ^2=\Vert ABx\Vert ^2\overset{\small {\text{(sip4) }}}{=}\left[ ABx|ABx\right] =\\&=\left[ Bx|A_*(ABx)\right] \overset{\small {\text{(sip3) }}}{\le }\Vert Bx\Vert \cdot \Vert A_*(ABx)\Vert \overset{\small {\text{(sip7) }}}{\le }\\&\le \Vert Bx\Vert \cdot \Vert A\Vert \cdot \Vert ABx\Vert \le \Vert Bx\Vert \cdot \Vert A\Vert \cdot \Vert A\Vert \cdot \Vert Bx\Vert \le \Vert A\Vert ^2\cdot \Vert B\Vert ^2. \end{aligned}
It follows from the above inequalities that
\begin{aligned} \Vert A_*(ABx)\Vert =\Vert A\Vert \cdot \Vert A\Vert \cdot \Vert Bx\Vert \end{aligned}
(2.5)
and $$\left[ Bx|A_*(ABx)\right] =\Vert Bx\Vert \cdot \Vert A_*(ABx)\Vert$$, whence $$\left[ Bx|\frac{A_*(ABx)}{\Vert A_*(ABx)\Vert }\right] =\Vert Bx\Vert$$.  Therefore
\begin{aligned} \left[ \ \cdot \ |\frac{A_*(ABx)}{\Vert A_*(ABx)\Vert }\right] \in J\left( Bx\right) . \end{aligned}
On the other hand it is easy to verify that
\begin{aligned} \left[ \ \cdot \ |\frac{Bx}{\Vert Bx\Vert }\right] \in J\left( Bx\right) . \end{aligned}
The Banach space X is smooth. Thus we get $$card J(Bx)=1$$, whence
\begin{aligned} \left[ \ \cdot \ |\frac{A_*(ABx)}{\Vert A_*(ABx)\Vert }\right] =\left[ \ \cdot \ |\frac{Bx}{\Vert Bx\Vert }\right] . \end{aligned}
(2.6)
Using again the properties of s.i.p., we obtain
\begin{aligned} \Vert x+ABx\Vert ^2&=\left[ x+ABx|x+ABx\right] =\left[ x|x+ABx\right] +\left[ ABx|x+ABx\right] \overset{\small {\text{(sip3) }}}{\le }\\&\le \Vert x\Vert \cdot \Vert x+ABx\Vert +\Vert ABx\Vert \cdot \Vert x+ABx\Vert =\\&= \left( \Vert x\Vert +\Vert ABx\Vert \right) \cdot \Vert x+ABx\Vert \overset{(2.4)}{=}\\&=\Vert x+ABx\Vert \cdot \Vert x+ABx\Vert =\Vert x+ABx\Vert ^2\\ \end{aligned}
From the above, we have
\begin{aligned} \left[ x|x+ABx\right] =\Vert x\Vert \cdot \Vert x+ABx\Vert and \left[ ABx|x+ABx\right] =\Vert ABx\Vert \cdot \Vert x+ABx\Vert , \end{aligned}
whence
\begin{aligned} \left[ x|\frac{x+ABx}{\Vert x+ABx\Vert }\right] =\Vert x\Vert and \left[ ABx|\frac{x+ABx}{\Vert x+ABx\Vert }\right] =\Vert ABx\Vert . \end{aligned}
Therefore,
\begin{aligned} \left[ \ \cdot \ |\frac{x+ABx}{\Vert x+ABx\Vert }\right] \in J(x)\quad and \quad \left[ \ \cdot \ |\frac{x+ABx}{\Vert x+ABx\Vert }\right] \in J(ABx). \end{aligned}
(2.7)
On the other hand,
\begin{aligned} \left[ \ \cdot \ |x\right] \in J(x)\quad and \quad \left[ \ \cdot \ |\frac{ABx}{\Vert ABx\Vert }\right] \in J(ABx). \end{aligned}
(2.8)
Smoothness of X yields that $$card J(x)=1$$ and $$card J(ABx)=1$$. So, combining (2.7) and (2.8), we immediately get
\begin{aligned} \left[ \ \cdot \ |\frac{x+ABx}{\Vert x+ABx\Vert }\right] =\left[ \cdot |x\right] \quad and \quad \left[ \ \cdot \ |\frac{x+ABx}{\Vert x+ABx\Vert }\right] =\left[ \cdot |\frac{ABx}{\Vert ABx\Vert }\right] . \end{aligned}
It follows from the above equalities that
\begin{aligned} \left[ \ \cdot \ |x\right] =\left[ \ \cdot \ |\frac{ABx}{\Vert ABx\Vert }\right] . \end{aligned}
(2.9)
Fix $$y\in X$$. Finally, we deduce
\begin{aligned} \Vert B\Vert ^2\cdot \left[ Ay|x\right]&\overset{(2.9)}{=}\Vert B\Vert ^2\cdot \left[ Ay|\frac{ABx}{\Vert ABx\Vert }\right] \overset{(2.3)}{=}\Vert B\Vert ^2\cdot \left[ Ay|\frac{1}{\Vert AB\Vert }ABx\right] =\\&\overset{(2.1)}{=}\Vert B\Vert ^2\cdot \left[ Ay|\frac{1}{\Vert A\Vert \cdot \Vert B\Vert }ABx\right] \overset{\small {\text{(sip2) }}}{=}\frac{\Vert B\Vert }{\Vert A\Vert }\cdot \left[ Ay|ABx\right] =\\&=\frac{\Vert B\Vert }{\Vert A\Vert }\cdot \left[ y|A_*(ABx)\right] \overset{{(2.6)},(2.5)}{=}\frac{\Vert B\Vert }{\Vert A\Vert }\cdot \left[ y|\ \Vert A\Vert ^2\cdot Bx\right] \end{aligned}
and by (sip2) we have $$\Vert B\Vert \cdot \left[ Ay|x\right] =\Vert A\Vert \cdot \left[ y|Bx\right]$$. So we conclude that
\begin{aligned} \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] . \end{aligned}
(2.10)
Putting Bx in place of y in the above equality we get
\begin{aligned} \left[ ABx|x\right] \overset{(2.10)}{=}\frac{\Vert A\Vert }{\Vert B\Vert }\cdot \left[ Bx|Bx\right] =\frac{\Vert A\Vert }{\Vert B\Vert }\cdot \Vert Bx\Vert ^2=\frac{\Vert A\Vert }{\Vert B\Vert }\cdot \Vert B\Vert ^2\cdot \Vert x\Vert ^2=\left[ x|x\right] =1 \end{aligned}
From the above, we have $$\left[ ABx|x\right] =\left[ x|x\right]$$. Since $$\Vert AB\Vert =\Vert A\Vert \cdot \Vert B\Vert =1$$, we have $$\left[ AB(\cdot )|x\right] ,\left[ \cdot |x\right] \in J(x)$$. Smoothness of X yields that $$card J(x)=1$$. So, we immediately get
\begin{aligned} \left[ AB(\cdot )|x\right] =\left[ \cdot |x\right] . \end{aligned}
(2.11)
Define $$Y:=\{y\in X: \left[ y|x\right] =1\}\cap \mathbf {B}(X)$$. By (2.11) we obtain $$AB(Y)\subset Y$$. It is known that $$AB:(X, weak)\rightarrow (X, weak)$$ is continuous. It is easy to check that 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

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

$$\Vert I+AB\Vert =2$$;

2. (ii)

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

3. (iii)

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

Proof

In a similar way as in the proof of Theorem 2.1 (see the proof of (c)$$\Rightarrow$$(e)) we obtain the implications $$\Vert I+AB\Vert =2$$$$\Rightarrow$$(2.10)$$\Rightarrow$$(2.11) and we may consider (i)$$\Rightarrow$$(ii)$$\Rightarrow$$(iii) as shown. Finally, if (iii) holds, then
\begin{aligned} 2&=\left[ x|x\right] +\left[ x|x\right] \overset{\text{(iii) }}{=}\left| \left[ x|x\right] +\left[ ABx|x\right] \right| =\left| \left[ x+ABx|x\right] \right| \le \\&\le \Vert x+ABx\Vert \cdot \Vert x\Vert \le \Vert I+AB\Vert \le 1+\Vert AB\Vert \le 1+\Vert A\Vert \cdot \Vert B\Vert =2 \end{aligned}
which yields $$\Vert I+AB\Vert =1+\Vert A\Vert \cdot \Vert B\Vert$$. Thus we get (iii)$$\Rightarrow$$(i). $$\square$$

We can add another proposition to our list.

Proposition 2.3

Let 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:
1. (i)

$$\Vert I+E\Vert =2$$;

2. (ii)

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

3. (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

Assume that 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:
1. (i)

for every $$y_o$$ in X there is an unique u in X such that $$u-ABu=y_o$$;

2. (ii)

for every $$y_o$$ in X there is a u in X such that $$u-ABu=y_o$$;

3. (iii)

the only vector x such that $$x-ABx=0$$ is $$x=0$$;

4. (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

Let 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:
1. (i)

for every $$y_o$$ in X there is an unique u in X such that $$u-Du=y_o$$;

2. (ii)

for every $$y_o$$ in X there is a w in X such that $$w-Dw=y_o$$;

3. (iii)

the only vector x such that $$x-Dx=0$$ is $$x=0$$.

Proof

The implications (i)$$\Rightarrow$$(ii), (i)$$\Rightarrow$$(iii) are trivial, so we look first at (iii)$$\Rightarrow$$(i). Assume that $$I-D$$ is injective. Since $$I-AB$$ is injective, we conclude that $$I-AB$$ is invertible (see Theorem 2.1). Suppose, for a contradiction, that $$I-D$$ is not invertible. Writing
\begin{aligned} I-D=(I-AB)+(AB-D), \end{aligned}
we can assert that
\begin{aligned} I-D=(I-AB)\left( I+(I-AB)^{-1}(AB-D)\right) . \end{aligned}
(2.12)
The operator $$I-AB$$ is invertible and $$I-D$$ is not invertible. Therefore, it follows from (2.12) that $$\left( I+(I-AB)^{-1}(AB-D)\right)$$ is also not invertible. The operator $$(I-AB)^{-1}(AB-D)$$ is compact. By Theorem 1.1, the operator $$I+(I-AB)^{-1}(AB-D)$$ is not injective, and thus there is a vector $$c\in X\setminus \{0\}$$ such that $$c+(I-AB)^{-1}(AB-D)c=0$$. So, it is not difficult to check that $$c-Dc=0$$. Hence $$\ker (I-D)\ne \{0\}$$, a contradiction.

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

Let XABC be as in Theorem 2.1. Then
\begin{aligned} \mathbb {T}\cap \sigma (AB)=\mathbb {T}\cap \sigma _p(AB). \end{aligned}

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

Let 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
\begin{aligned} \mathbb {T}\cap (\sigma (A+K)\setminus \sigma _p(A+K))\subset \mathbb {T}\cap \sigma _p(A), \end{aligned}
for each compact operator 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

Let XA be as in Theorem 3.3. Then
\begin{aligned} \mathbb {T}\ \ \cap \bigcup \limits _{K\in \mathcal {K}(X)}(\sigma (A+K)\setminus \sigma _p(A+K))\subset \mathbb {T}\cap \sigma _p(A). \end{aligned}

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

Suppose that $$X=\left( \bigoplus _{n=1}^\infty X_n\right) _{l_p}$$ with $$p\in (1,\infty )$$, $$\dim X_n<\infty$$. Suppose that the spaces $$X_n$$ are smooth. Let $$A_1,A_2,B\in \mathcal {L}(X)$$, $$\Vert A_1\Vert =\Vert A_2\Vert =\Vert B\Vert =1$$, $$\Vert A_1B+A_2B\Vert =2$$. Assume that B is a smooth point of the unit sphere of $$\mathcal {L}(X)$$. Suppose that $$A_1,A_2$$ are linearly independent. Then
\begin{aligned} \mathbb {T}\cap \sigma (A_1B)\cap \sigma (A_2B)\subset \mathbb {T}\cap \sigma _p\left( \alpha A_1B+(1-\alpha ) A_2B\right) \end{aligned}
for any $$\alpha \in [0,1]$$.

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 AC 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.

Since $$\Vert A_1\Vert =\Vert A_2\Vert =\Vert B\Vert =1$$ and $$\left\| \frac{1}{2}A_1B+\frac{1}{2}A_2B\right\| =1$$ we see that $$\Vert A_1B\Vert =\Vert A_2B\Vert =1$$. From this we conclude that $$\Vert \alpha A_1B+(1-\alpha )A_2B\Vert =1$$. Hence $$\Vert AB\Vert =1$$. We have
\begin{aligned} \Vert AB-C\Vert&=\Vert \alpha A_1B+(1-\alpha ) A_2B-\alpha A_1K-(1-\alpha ) A_2K \Vert \le \\&\le \Vert \alpha A_1+(1-\alpha ) A_2\Vert \cdot \Vert B-K\Vert <\Vert \alpha A_1+(1-\alpha ) A_2\Vert \cdot 1\le \\&\le \alpha \Vert A_1\Vert +(1-\alpha ) \Vert A_2\Vert =1=\Vert AB\Vert \end{aligned}
so that $$\Vert AB-C\Vert <\Vert AB\Vert$$. Since $$\lambda \in \mathbb {T}\cap \sigma (A_1B)\cap \sigma (A_2B)$$, it means that $$I-\frac{1}{\lambda }A_1B$$, $$I-\frac{1}{\lambda }A_1B$$ are not invertible.
We define $$C_1,C_2$$ as follows: $$C_1:=A_1K$$ and $$C_2:=A_2K$$. Thus we have $$C_1,C_2\in \mathcal {K}(X)$$. It follows easily that $$\Vert A_1B-C_1\Vert <1=\Vert A_1B\Vert$$ and $$\Vert A_2B-C_2\Vert <1=\Vert A_2B\Vert$$. Now we may apply Theorem 2.1 to conclude that
\begin{aligned} \Vert I+\frac{1}{\lambda }A_1B\Vert =2,\quad \Vert I+\frac{1}{\lambda }A_2B\Vert =2. \end{aligned}
(3.1)
Applying Theorem 1.3, we can find the unique vector $$x_o$$ in the unit sphere of 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
\begin{aligned} \forall _{y\in X}:\ \left[ \frac{1}{\lambda }A_1B(y)|x_o\right] =\left[ y|x_o\right] , \quad \left[ \frac{1}{\lambda }A_2B(y)|x_o\right] =\left[ y|x_o\right] . \end{aligned}
(3.2)
As an immediate consequence we see that
\begin{aligned} \forall _{y\in X}:\ \left[ \frac{1}{\lambda }(\alpha A_1B+(1-\alpha )A_2B)(y)|x_o\right] =\left[ y|x_o\right] . \end{aligned}
(3.3)
Summarizing, we have proved $$\frac{1}{\lambda }C\in \mathcal {K}(X)$$ and
\begin{aligned} \left\| \frac{1}{\lambda }AB-\frac{1}{\lambda }C\right\| <\left\| \frac{1}{\lambda }AB\right\| ,\quad \forall _{y\in X}:\ \left[ \frac{1}{\lambda }AB(y)|x_o\right] =\left[ y|x_o\right] . \end{aligned}
(3.4)
Combining (3.4) and Proposition 2.2 we see that $$\left\| I+\frac{1}{\lambda }AB\right\| =2$$. It follows from Theorem 2.1 ((c)$$\Leftrightarrow$$(e)) that $$I-\frac{1}{\lambda }AB$$ is not injective. This means $$\lambda \in \sigma _p(AB)=\sigma _p(\alpha A_1B+(1-\alpha ) A_2B)$$, and the proof is complete. $$\square$$

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

Let X be as in Theorem 2.1. Let $$A\in \mathcal {L}(X)$$, $$\Vert A\Vert =1$$. Suppose that
\begin{aligned} \mathbb {T}\cap \left( \sigma (A)\setminus \sigma _p(A)\right) \ne \emptyset . \end{aligned}
Then $$dist (A,\mathcal {K}(X))=1$$. In particular, $$\Vert A\Vert \le \Vert A+K\Vert$$, for each compact operator $$K\in \mathcal {L}(X)$$.

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 $$\frac{1}{|\alpha |}\le \Vert A_{\alpha ,\beta }\Vert$$.

Let x be an arbitrary unit vector in H. Thus we have
\begin{aligned} 1=&\ \Vert x\Vert ^2=\Vert A_{\alpha ,\beta }^{-1}(A_{\alpha ,\beta }x)\Vert ^2=\Vert (\alpha I+\beta T)(A_{\alpha ,\beta }x)\Vert ^2=\\ =&\ \Vert \alpha A_{\alpha ,\beta }x+\beta T(A_{\alpha ,\beta }x)\Vert ^2=\left\langle \alpha A_{\alpha ,\beta }x+\beta T(A_{\alpha ,\beta }x)|\alpha A_{\alpha ,\beta }x+\beta T(A_{\alpha ,\beta }x)\right\rangle =\\ =&\ |\alpha |^2\cdot \Vert A_{\alpha ,\beta }x\Vert ^2+\overline{\alpha }\beta \left\langle T(A_{\alpha ,\beta }x)|A_{\alpha ,\beta }x\right\rangle +\alpha \overline{\beta }\left\langle A_{\alpha ,\beta }x|T(A_{\alpha ,\beta }x)\right\rangle +\\&+\Vert \beta T(A_{\alpha ,\beta }x)\Vert ^2=^{(\alpha \overline{\beta }\in (0,\infty ),\ so\ \alpha \overline{\beta }=\overline{\alpha }\beta )}\\ =&\ |\alpha |^2\cdot \Vert A_{\alpha ,\beta }x\Vert ^2+\alpha \overline{\beta }\left\langle T(A_{\alpha ,\beta }x)|A_{\alpha ,\beta }x\right\rangle +\alpha \overline{\beta }\left\langle T^*(A_{\alpha ,\beta }x)|A_{\alpha ,\beta }x\right\rangle +\\&+\Vert \beta T(A_{\alpha ,\beta }x)\Vert ^2\ge \\ \ge&\ |\alpha |^2\cdot \Vert A_{\alpha ,\beta }x\Vert ^2+\alpha \overline{\beta }\left\langle (T+T^*)(A_{\alpha ,\beta }x)|A_{\alpha ,\beta }x\right\rangle \overset{(T+T^*\ge 0)}{\ge }|\alpha |^2\cdot \Vert A_{\alpha ,\beta }x\Vert ^2. \end{aligned}
We obtain $$\Vert A_{\alpha ,\beta }x\Vert \le \frac{1}{|\alpha |}$$. Passing to the supremum over $$\Vert x\Vert =1$$ we get $$\Vert A_{\alpha ,\beta }\Vert \le \frac{1}{|\alpha |}$$. Therefore $$\Vert A_{\alpha ,\beta }\Vert =\frac{1}{|\alpha |}$$.

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 $$1\in \sigma (\alpha A_{\alpha ,\beta })$$.

We want to show that $$1\notin \sigma _p(\alpha A_{\alpha ,\beta })$$. Assume, for a contradiction, that $$\alpha A_{\alpha ,\beta }w=w$$ for some $$w\in H\setminus \{0\}$$. Thus
\begin{aligned} \alpha w=A_{\alpha ,\beta }^{-1}w=(\alpha I+\beta T)w=\alpha w\,+\,\beta Tw. \end{aligned}
It follows that $$Tw=0$$. As 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

Suppose that $$X=\left( \bigoplus _{n=1}^\infty X_n\right) _{l_p}$$ with $$p\in (1,\infty )$$, $$\dim X_n<\infty$$. Suppose that the spaces $$X_n$$ are smooth. Let $$A\in \mathcal {L}(X)$$, $$K\in \mathcal {K}(X)$$ be nonvanishing operators with $$\Vert A\Vert ^p+\Vert K\Vert ^p\le 1$$, $$\Vert A\Vert <\Vert K\Vert$$. Let us consider the system of linear equations:
\begin{aligned} \left\{ \begin{array}{ccc}x_1-Ax_2=y_1,\\ Kx_1-x_2=y_2.\end{array}\right. \end{aligned}
(5.1)
The following three statements are equivalent:
1. (a)

for every $$y_1,y_2$$ in X there is an unique solution $$x_1,x_2\in X$$ of (5.1);

2. (b)

for every $$y_1,y_2$$ in X there is a solution $$x_1,x_2\in X$$ of (5.1);

3. (c)

the only vectors $$x_1,x_2$$ such that $$\left\{ \begin{array}{ccc}x_1-Ax_2=0\\ Kx_1-x_2=0\end{array}\right.$$ are $$x_1=0=x_2$$.

Proof

We define $$A\oplus _p K\in \mathcal {L}(X\oplus _p X)$$ by the formula $$(A\oplus _p K) (u_1,u_2):=(A(u_1),K(u_2))$$. It is not difficult to show that $$\Vert A\oplus _p K\Vert =\Vert K\Vert \le 1$$ and $$0\oplus _p K\in \mathcal {K}(X\oplus _p X)$$ and
\begin{aligned} \Vert A\oplus _p K - 0\oplus _p K\Vert =\Vert A\oplus _p 0\Vert =\Vert A\Vert <\Vert K\Vert =\Vert A\oplus _p K\Vert . \end{aligned}
Thus we have $$\Vert A\oplus _p K - 0 \oplus _p K\Vert <\Vert A\oplus _p K\Vert$$. Moreover, $$X\oplus _p X=\left( \bigoplus _{n=1}^\infty Y_n\right) _{l_p}$$, $$Y_n\in \{X_k:k\in \mathbb {N}\}$$ and $$\mathcal {L}(X\oplus _p X)^*=\mathcal {K}(X\oplus _p X)\oplus _1\mathcal {K}(X\oplus _p X)^\bot$$. Applying Theorem 2.4 again, this time to $$A\oplus _p K$$, we obtain (a)$$\Leftrightarrow$$(b)$$\Leftrightarrow$$(c). $$\square$$

6 Remarks: Invertibility

Let $$E\in \mathcal {L}(X)$$, where 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
\begin{aligned} \Vert I-T\Vert =\left\| I-\frac{1}{2}I+\frac{1}{2}AB \right\| =\frac{1}{2}\left\| I+AB\right\| \le \frac{1}{2}(1+\Vert A\Vert \cdot \Vert B\Vert )< 1, \end{aligned}
which means that 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

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

$$\Vert I-T\Vert <1$$;

2. (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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