Spatial Numerical Range of Operators on Weighted Hardy Spaces
Abstract
We consider the spatial numerical range of operators on weighted Hardy spaces and give conditions for closedness of numerical range of compact operators. We also prove that the spatial numerical range of finite rank operators on weighted Hardy spaces is star shaped; though, in general, it does not need to be convex.
1. Introduction
In [2], Gaur and Husain have studied the spatial numerical range of elements of Banach algebras without identity. Specifically, the relationship between spatial numerical ranges, numerical ranges, and spectra has been investigated. Among other results, it has been shown that the closure of the spatial numerical range of an element of a Banach algebra without identity but with regular norm is exactly its numerical range as an element of the unitized algebra.
A complete survey on numerical ranges of operators can be found in the books by Bonsall and Duncan [3, 4]; we refer the reader to these books for general information and background.
In Section 2, after giving some background material, we give useful formula for the spatial numerical range of operators on weighted Hardy space. In Section 3, we show that the spatial numerical range of an operator needs not to be convex, and we also prove that the spatial numerical range of finite rank operators is star shaped. Finally, in Section 4, we give conditions for closedness of numerical range of compact operators.
2. Preliminaries
- (i)
[x + y, z] = [x, z]+[y, z] for all x, y, z ∈ X,
- (ii)
[λx, y] = λ[x, y] for all x, y ∈ X and λ ∈ ℂ,
- (iii)
[x, x] ≥ 0 for all x ∈ X,
- (iv)
|[x, y]|2 ≤ [x, x][y, y] for all x, y ∈ X and λ ∈ ℂ.
The unit ball of X is called smooth if for all x, with ∥x∥ = 1, there is a unique x* ∈ X*, such that ∥x*∥ = 1 and 〈x, x*〉 = 1. In this case, there is a unique semi-inner products on X satisfying (2.1), and then V(T) coincides with numerical range W(T) corresponding to the unique semi-inner product satisfying (2.1).
- (i)
W(T)⊆V(T),
- (ii)
,
- (iii)
sup {|λ | : λ ∈ W(T)} = sup {|λ | : λ ∈ V(T)}.
It is of course trivial that every eigenvalue of T is actually in V(T).
- (a)
if α ≥ 0 and f ∈ Hp(β), then (αf) * = αp−1f*,
- (b)
if f ∈ Hp(β), .
Lemma 2.1. If T is a bounded linear operator on Hp(β), then
Proof. Suppose that f ∈ Hp(β), g ∈ (Hp(β)) *, ∥f∥ = ∥g∥ = 1, and 〈f, g〉 = 1. Then
3. Shape of the Spatial Numerical Range
The usual numerical range of a bounded linear operator on a Hilbert space is convex, and for every bounded linear operator T on a normed space X, we know that V(L(X), T) is convex. Although V(T) needs not to be convex (see [3]), B. E. Cain and H. Schneider proved that it is connected. Also in [9], Kuliyev proved that the spatial numerical range of a given operator on a separable Banach space is pathwise connected.
Recall that V(T) is star shaped with respect to zero if tz ∈ V(T) for 0 ≤ t ≤ 1 and z ∈ V(T).
In Theorem 3.1, we give a necessary and sufficient condition for the numerical range of a bounded linear operator to be star shaped. In Example 3.2, we show that the spatial numerical range of linear operator T on Hp(β) needs not to be convex, even if T is compact (see also [3]). We also determine the shape of V(T), when T is a finite rank operator. Finally, in Theorem 3.3, we prove that there is an operator T on Hp(β) that may not be star shaped.
Theorem 3.1. Let T be a bounded linear operator on Hp(β). Then
- (a)
V(T) is star shaped with respect to zero if and only if
(3.1) - (b)
if T is finite rank on Hp(β) and 0 ∈ V(T), then V(T) is star shaped with respect to zero.
Proof. The proof is trivial, as 〈T(kf), (kf) *〉 = kp〈Tf, f*〉, for each nonnegative real number k and f ∈ Hp(β).
Example 3.2. Let β(1) = 1 and T be the linear operator on Hp(β) given by

Theorem 3.3. There is an operator T on Hp(β) with 0 ∈ V(T), such that V(T) is not star shaped.
Proof. We proof this theorem by contradiction. Suppose that the spatial numerical range of each linear operator that allowed origin is star shaped. If a, z ∈ V(T), then 0 ∈ V(T − a), and so t(z − a) ∈ V(T − a) for 0 ≤ t ≤ 1. Hence, tz + (1 − t)a ∈ V(T), and it follows that V(T) is convex which is a contradiction to the previous example.
4. Compact Operators
Since H2(β) is a Hilbert space, the numerical range of a compact operator on H2(β) is closed if and only if it contains the origin. Also the numerical range of a compact operator on Hp(β) contains all nonzero extreme points of its closure, and since Hp(β) is infinite dimensional, there is a compact operator T on Hp(β), such that V(T) is not closed (see [10] and page 103-109 of [11]). So, in general, the spatial numerical range of a compact operator needs not to be closed. In the following theorem, we give a closedness condition of such operators.
Theorem 4.1. Let T be a compact operator on Hp(β). If V(T) is star shaped with respect to zero, then it is closed.
Proof. Since V(T) is star shaped with respect to zero, then by Theorem 3.1,
For given , there is a sequence hn with ∥hn∥p = 1 and . By reflexivity of Hp(β) and Alaogul′s Theorem, there is a sequence , such that in weak topology and in weak* topology for some h ∈ ball(Hp(β)) and g ∈ ball((Hp(β)) *).
Now, let m ∈ ℕ. Define the bounded linear functionals x, x* by
On the other hand,
Since in infinite dimensional spaces 0 is allowed in spectrum of any compact operator, then we have the following corollary.
Corollary 4.2. Let T be a compact operator on Hp(β), such that V(T) is convex. Then V(T) is closed if and only if 0 ∈ V(T).
Acknowledgments
The authors would like to thank Professor Alexander Rosa, the Editor-in-Chief of the International Journal of Mathematics and Mathematical Sciences, and the referee for useful and helpful comments and suggestions.