On smooth local resolvents

We exhibit an example of a bounded linear operator on a Banach space which admits an everywhere defined local resolvent with continuous derivatives of all orders. Let T be a bounded linear operator acting on a complex Banach space X. It is well known that the resolvent z 7→ (T − z)−1 defined on the complement of the spectrum σ(T ) is unbounded. More precisely, ‖(T − z)−1‖ → ∞ whenever z approaches the spectrum σ(T ). Let x ∈ X be a nonzero vector. By a local resolvent of T at x we mean a function f : U → X defined on a set U ⊃ C \ σ(T ) such that (T − z)f(z) = x (z ∈ U). Clearly the local resolvent is uniquely determined for all z ∈ C \ σ(T ) and f(z) = (T − z)−1x. Thus any local resolvent is analytic on the complement of the spectrum. It was observed in [G] that a local resolvent can be bounded. Bounded local resolvents were further studied in [BG], [N], [BM] and it was shown that they are rather frequent. In [BM], an example of a continuous everywhere defined local resolvent was given (such a local resolvent is necessarily bounded since each local resolvent vanishes for z →∞). The aim of this note is to exhibit an example of an everywhere defined C∞ local resolvent. Note that by a basic result of local spectral theory there are no analytic everywhere defined local resolvents. Denote by C,R and Z+ the sets of all complex numbers, real numbers and nonnegative integers, respectively. Let X be a complex Banach space and f : C → X a function. As usually we identify C with R and consider f to be a function of two real variables x and y. We say that f is a C∞-function if it has continuous derivatives ∂ f ∂xk∂yl of all orders. Theorem 1. There exist a Banach space X, an operator T ∈ B(X), a nonzero vector x ∈ X and a C∞-function f : C→ X such that (T − z)f(z) = x (z ∈ C). Proof. Let φ : C→ 〈0, 1〉 be a C∞-function such that φ(z) = { 1 (|z| < 1/3), 0 (|z| > 2/3). Mathematics Subject Classification: 47A10, 47A11.