An operator U on a Hilbert space, other than I, such that sp(U) = {1} and ∥ U ∥= 1.

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Suppose that H = L^{2}(0, 1) with respect to the Lebesgue measure and (T f)(x) = ∫_{0}^{x} f(t)dt. It follows from BA15.DVI, sp(T) = 0, so that sp(I + T) = {1}. Hence U = (I + T)^{−1} ≠ I is well-defined, moreover sp(U) = {λ^{−1}; λ ∈ sp(I + T)} = {1}. Therefore

1 = r(U) ≤∥ U ∥ .

But ∥ U ∥≤ 1, since

∥ U^{−1}ξ ∥^{2}=∥ (I + T)ξ ∥^{2}=∥ f ∥^{2} + < (T + T^{∗})ξ, ξ > + ∥ T ξ ∥^{2}≥∥ f ∥^{2} .

(Note that T + T^{∗} is a projection onto the space of constant functions, since (T^{∗}f)(t) = ∫_{t}^{1} f(t)dt.) Thus ∥ U ∥= 1.