Fredholm Equation in Smooth Banach Spaces and Its Applications

The object of this paper is to consider the Fredholm equation (i.e., $$x-ABx=y_o$$x-ABx=yo) 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.


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  [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 ∈ K(X ), y o ∈ X . From our modern perspective we can think of the Fredholm equation as with a single unknown vector x ∈ 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 ∈ K(X ), λ ∈ K and λ = 0, then for every y o in X there is an x in X such that if and only if the only vector u such that λu − 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 K(X ) (see Theorem 4.2).

Semi-inner Product
Let (X, · ) be a normed space over K ∈{R, C}. Lumer [6] and Giles [2] proved that in a space X there always exists a mapping [·|·] : X × X → K satisfying the following properties: Such a mapping is called a semi-inner product (s.i.p.) in X (generating the norm · ). 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. [·|·] : X × X → K. If A is a bounded linear operator from X to itself, then f z (·) := [A(·)|z] 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 [Ax|z] = [x|A * (z)] for all x in X . Of course, in a Hilbert space we have A * = A * . In general case the mapping A * : X → X is not linear but it still has some good properties: By the Hahn-Banach Extension Theorem we get J (x) = ∅. In this paper, for a normed space X , we denote by B(X ) the closed unit ball in X . By extD we will denote the set of all extremal points of a set D.
Let L(X ) denote the space of all bounded linear operators on a space X , and I the identity operator. We write K(X ) for the space of all compact operators on X . For

Geometry of Space 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 K(X ). Theorem 1.2 [5] Let X be a reflexive Banach spaces over the field K. The following conditions are equivalent: (a) f ∈ extB (K(X ) * ), (b) there exist y * ∈ extB(X * ) and x ∈ extB(X ) such that f (T ) = y * (T x) for every T ∈ K(X ).
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 * ∈ B(X * ), i.e., cardJ (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 L(X ) (cf. [7]). Theorem 1.3 [7] Suppose that X = ∞ n=1 X n l p with p ∈ (1, ∞), dim X n < ∞ for some 1 < p < ∞. Then T is a smooth point of the unit sphere of L(X ) if and only if (s1) inf{ T + K : K ∈ 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 T (x o ) = 1; (s3) the point T x o is smooth.

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 Assume that X is a reflexive smooth Banach space over
The following five conditions are equivalent: Proof The implications (a)⇒(b)⇒(d) and (a)⇒(c) are immediate. First we prove (e)⇒(a). We get I − 1 2 (I − AB) = 1 2 I + AB < 1, which means that 1 . This gives ϕ ∈ L(X ) * , ϕ(I + AB) = I + AB and ϕ = 1. We will prove that ϕ(I ) = 1 and ϕ(AB) = AB . It follows that and |ϕ(I )| ≤ 1 and |ϕ(AB)| ≤ AB . This clearly forces ϕ(I ) = 1 and ϕ(AB) = AB and, in consequences, the inclusion 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 B(L(X ) * ). From this it follows that J (I + AB) is weak*-compact. Applying the Krein-Milman Theorem we see that there exists μ ∈ extJ (I + AB). An easy computation shows that the set J (I + AB) is an extremal subset of B(L(X ) * ). Therfore and, in consequences, we get μ ∈ extB(L(X ) * ).
Suppose that μ = μ 1 + μ 2 is the suitable decay of μ. Namely, we have So it suffices to show that μ 2 = 0. Assume, contrary to our claim, that μ = μ 2 ∈ extB(K(X ) ⊥ ). Hence μ 2 = 1. From the assumption, we have that AB − C < AB and C ∈ K(X ). It follows that and we obtain a contradiction. Thus we must have μ = μ 1 ∈ extB(K(X ) * ). Therefore it follows from (2.2) and Theorem 1.2 that for some y * ∈ extB(X * ) and x ∈ extB(X ). This is summarized as follows: Thus we obtain the equalities: Combining (2.1) and (2.3), we immediately get Using the properties of s.i.p., we obtain It follows from the above inequalities that On the other hand it is easy to verify that The Banach space X is smooth. Thus we get cardJ (Bx) = 1, whence Using again the properties of s.i.p., we obtain From the above, we have Therefore, On the other hand, Smoothness of X yields that cardJ (x) = 1 and cardJ (ABx) = 1. So, combining (2.7) and (2.8), we immediately get It follows from the above equalities that Fix y ∈ X . Finally, we deduce Putting Bx in place of y in the above equality we get [ABx|x] (2.10)  ∈ M(B), i.e., (2.11). Thus we have [(I − AB)y|x] = 0 for all y ∈ X . Since I − AB has dense range, x = 0 and we obtain a contradiction.
Careful reading of the proof of Theorem 2.1 (more precisely, the proof of (c)⇒(e)) shows that we can get the following.

Proposition 2.2 Let X, A, B, C be as in Theorem
Then the following three statements are equivalent: Proof In a similar way as in the proof of Theorem 2.1 (see the proof of (c)⇒(e)) we obtain the implications I + AB = 2 ⇒(2.10)⇒(2.11) and we may consider (i)⇒(ii)⇒(iii) as shown. Finally, if (iii) holds, then which yields I + AB = 1 + A · B . Thus we get (iii)⇒(i).
We can add another proposition to our list.

Proposition 2.3 Let X be as in Theorem
Suppose that there is an operator C ∈ K(X ) such that E − C < E . Then the following three statements are equivalent: Proof Putting I E in place of AB in the above proposition we get (i)⇔(ii). Putting E I in place of AB in the above proposition we get (i)⇔(ii).
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: L * = K * ⊕ 1 K ⊥ , A · B ≤ 1.

Theorem 2.4
Assume that X is a reflexive smooth Banach space. Suppose that L(X ) * = K(X ) * ⊕ 1 K(X ) ⊥ . Let A, B ∈ L(X ) and A · B ≤ 1. Suppose that there is an operator C ∈ K(X ) such that AB − C < AB . Then the following four statements are equivalent: The proof of (ii)⇒(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)⇒(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 +

An Application: Spectrum
As an application of the results in the previous section, we consider the notion of a spectrum of an operator. Let T := {λ ∈ K : |λ| = 1}. We next explore some consequences of Theorem 2.1.
The following theorem was discovered by Weyl [8].

Theorem 3.2 [8] Suppose that H is a complex Hilbert space. If A ∈ L(H ) and K is a compact operator, then σ (
It makes sense to replace the operator AB by AI . Some part of Theorem 3.2 can be strengthen as follows.
for each compact operator K .
Proof Fix a number λ ∈ T. Assume that λ ∈ σ (A + K ) \ σ p (A + K ). Suppose, for a contradiction, that λ / ∈ σ p (A). Then I − 1 λ A is injective. Define an operator D by D : We have our desired contradiction.
As a consequence of this result, we get a corollary.

Corollary 3.4 Let 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
Suppose that X = ∞ n=1 X n l p with p ∈ (1, ∞), dim X n < ∞. Suppose that the spaces X n are smooth. Let A 1 , A 2 Assume that B is a smooth point of the unit sphere of L(X ). Suppose that A 1 , A 2 are linearly independent. Then for any α ∈ [0, 1].
Proof By assumption, X is smooth. It follows from Theorem 1.3 (condition (s1)) that, with some K ∈ K(X ), are not invertible. We define C 1 , C 2 as follows: C 1 := A 1 K and C 2 := A 2 K . Thus we have C 1 , C 2 ∈ K(X ). It follows easily that A 1 B − C 1 < 1 = A 1 B and A 2 B − C 2 < 1 = A 2 B . Now we may apply Theorem 2.1 to conclude that As an immediate consequence we see that Summarizing, we have proved 1 λ C ∈ K(X ) and Combining (3.4) and Proposition 2.2 we see that I + 1 λ AB = 2. It follows from Theorem 2.1 ((c)⇔(e)) that I − 1 λ AB is not injective. This means λ ∈ σ p (AB) = σ p (α A 1 B + (1 − α)A 2 B), and the proof is complete.

Distance from Operator to Compact Operators
The problem of best approximation of bounded linear operators L(X ) on a Banach space X by compact operators K(X ) has been of great interest in the twentieth century. This section is a small sample of it.
Let L(H ) be the algebra of all bounded operators on infinite-dimensional complex 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 ∈ L(H ) is hyponormal and has no isolated eigenvalues of finite multiplicity, then A ≤ A + K , for each compact operator K .
As an immediate consequence of the above result, we deduce that if U ∈ L(l 2 ) is the unilateral shift, then dist(U, 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
Then dist(A, K(X )) = 1. In particular, A ≤ A + K , for each compact operator K ∈ L(X ).
If H is a complex Hilbert space and T ∈ L(H ) such that T x|x = 0 for all x in H , then T = 0. This is not true for real Hilbert spaces. For example, let U ∈ L(R 2 ) be the linear operator given by U (x, y) := (−y, x). Then we we have U x|x = 0 for all x in H However, U = 0. So it makes sense to consider the following result.
which means that T is invertible. So the assumption A · B ≤ 1 makes the invertibility problem more interesting. In particular, the implication (b)⇒(a) (in Theorem 6.1) seem to be amazing.