then N e f(5(E)) if and only if there is a bounded set E
c E such that a(N) c
f(Ex) . If f is analytic and completely non-constant on the open set G , then a
normal operator N belongs to f(5(G)) if and only if a(N) e f(G) .
Proof. If N = f(A) for some A in f(5(E)) , the n the Spectral Mapping
Theorem implies that the set E
may be taken to be a (A) . Conversely, if
such a bounded set E
exists, then there is a Borel function g: a(N) - Ex
such that g{X) e E
n f
!!) for all X in a(N) . (This last statement follows by
any of many measurable selection theorems, or, using the analyticity of f, the
reader can give a direct proof.) So g is bounded and if A = g(N) , then N =
f(A) .
In the second statement of the proposition, one implication is, again,
an immediate consequence of the Spectral Mapping Theorem. If a(N) c
f(G) , then the assumptions about f imply that there is a compact set K
contained in G such that f(K) = a(N) . To see this note that G = k J
1 ^ ,
where each Kn is compact and Kn c int K
n + 1
. Since f is completely non-
constant on G , f(int K J is open for each n . Thus { f(int Kn) } is an open
cover of a(N) and there is an integer n such that a(N) c fCK^) . The result
now follows by the first part of the proposition.
The stated condition on f and E in the preceding proposition is not
always satisfied. For example, let E = {0} u {
+ 2?ini : n e M } and let f(z)
. So f(E) = { 1, e
1 / 2
, e
1 / 3
, . . . } . If N is the diagonal operator with
entries 1, e
1 / 2
, e
1 / 3
, . . . , then N e f(5(E)) . It is not too difficult to see that
N E cl[ f(5(E)) ] . See Corollary 2.7 for a characterization of the normal
operators belonging to cl[ f(5(E)) ] .
1.5 Proposition (a) If X and Y are operators such that there is no non-zero
[ V Q ]
operator S with XS = SY , then for any operator Q the spectrum of T =
is G(X) U a(Y) . Moreover, if T e f(5(E)) , then X and Y belong to f(5(E)) .
(b) If X and Y are operators with Gj(X) n ar(Y) = 0 and X 0 Y belongs
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