Suppose that H = L²(0,1) with respect to the Lebesgue measure and (Tf)(x) = ò₀^x f(t)dt. It follows from BA15.DVI, sp(T) = 0, so that sp(I+T) = {1}. Hence U = (I + T)^-1 ¹ I is well-defined, moreover sp(U) = {l^-1 ; l Î sp(I + T)} = {1}. Therefore

1 = r(U) £ ||U||.

But ||U|| £ 1, since

||U-1x||2 = ||(I + T)x||2 = ||f||2 + < (T+T*)x, x > + ||T x||2 ³ ||f ||2.

(Note that T+T^* is a projection onto the space of constant functions, since (T^*f)(t) = ò_t¹ f(t)dt.)
Thus ||U|| = 1.