Given a compact subset Given a compact subset K of C such that [`(K0)] = K, there exists an operator T acting on a Hilbert space H such that sp(T) = K and T has no eigenvalue.

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Let H = L2(K) in which K is equipped with the Lebesgue measure m on Â2. Define T on H as the following:
(Tf)(m) = mf(m);   m Î K , f Î H.
If l Ï K, then sup{|l- m|-1 ; m Î K} < ¥ and so we can define an operator S on H by (Sf)(m) = ( l- m)-1f(m) ; f Î H, m Î K. Hence S(T - lI) = (T - lI)S = I so that l Ï sp(T). If l Î K, (lI - T)-1 Î B(H) and f denotes the characteristic function of {m; |l- m| < e} multiplied by m({m; |l- m| < e})-1/2 , then
1
=
||f ||2 £ ||(lI - T)-1||||(lI - T)f ||2
=
||(lI - T)-1|| ó
õ

K 
 (l- m)f(m)dm(m) £ ||(lI - T)-1||e,
a contradiction. Hence (lI - T) is not invertible. So l Î sp(T). It follows that sp(T) = K. In addition, if Tf = af for some a Î C, then for all m Î K,  mf(m) = af(m). So f = 0 almost every where. Thus T has no eigenvalue.

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On 22 Feb 2001, 00:21.