A sequence of nilpotent operators on H which converges with respect to the norm topology on B(H) to an operator which is not topologically nilpotent.

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This example is due to Kakutani (cf. [Ric, p. 282]). Let H be a separable Hilbert space with orthonormal basis (f_{m})_{m∈N}. Define α_{m} = e^{−k} for m = 2^{k}(2l +1), k, l = 0, 1, ⋯ and also the operator T by T f_{m} = α_{m}f_{m+1}, m ∈ N. Then ∥ T ∥= sup_{m∈N} |α_{m}|, T^{n}f_{m} = α_{m}α_{m+1} ⋯ α_{m+n−1}f_{m+n} and so ∥ T^{n} ∥= sup_{m∈N} (α_{m}α_{m+1} ⋯ α_{m+n−1}). Moreover, by the definition of the α_{m}, we have α_{1}α_{2} ⋯ α_{2^{t}−1} = ∏_{j=1}^{t−1}exp(−j2^{t−j−1}).

Therefore (α_{1}α_{2} ⋯ α_{2^{t}−1})\frac{1}{2^{t−1}} > (∏_{j=1}^{t−1}exp[\frac{−j}{2^{j+1}} ])^{2} and if σ = ∑_{j=1}^{∞} \frac{j}{2^{j+1}} , then

e^{−2σ} ≤ lim_{n} ∥ T^{n} ∥\frac{1}{n}. So T is not topologically nilpotent. Next define the operator T_{k} by

T_{k}f_{m} = \{ \begin{matrix}0 & m = 2^{k}(2l + 1), l = 0, 1, ⋯ \\ α_{m}f_{m+1} & otherwise \end{matrix}

Then T_{k} is nilpotent. But

(T − T_{k})f_{m} = \{ \begin{matrix}e^{−k}f_{m+1} & m = 2^{k}(2l + 1), l = 0, 1, ⋯ \\ 0 & otherwise \end{matrix}

Thus ∥ T_{k} − T ∥= e^{−k}, hence lim_{k} T_{k} = T in the norm topology on B(H). Ref.

[Ric] C.E. Rickart, General theory of Banach algebras, Princeton, Van Nastrand, 1960.