Given a compact subset K of C such that \overline{K^{0}} = 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 = L^{2}(K) in which K is equipped with the Lebesgue measure m on R^{2}. Define T on H as the following:

(T f)(μ) = μf(μ); μ ∈ K, f ∈ H.

If λ ∉ K, then sup{|λ − μ|^{−1}; μ ∈ K} < ∞ and so we can define an operator S on H by (Sf)(μ) = (λ − μ)^{−1}f(μ); f ∈ H, μ ∈ K. Hence S(T − λI) = (T − λI)S = I so that λ ∉ sp(T). If λ ∈ K, (λI − T)^{−1} ∈ B(H) and f denotes the characteristic function of {μ; |λ−μ| < ϵ} multiplied by m({μ; |λ−μ| < ϵ})^{−1/2} , then

1 = ∥ f ∥_{2}≤∥ (λI − T)^{−1} ∥∥ (λI − T)f ∥_{2}
= ∥ (λI − T)^{−1} ∥ ∫_{K}(λ − μ)f(μ)dm(μ) ≤∥ (λI − T)^{−1} ∥ ϵ,

a contradiction. Hence (λI − T) is not invertible. So λ ∈ sp(T). It follows that sp(T) = K. In addition, if T f = αf for some α ∈ C, then for all μ ∈ K, μf(μ) = αf(μ). So f = 0 almost every where. Thus T has no eigenvalue.