Strong limits of nilpotent operators

Question

Let $H$ be an infinite-dimensional Hilbert space.

Is it possible that the Identity $H\to H$ is a strong limit of nilpotent compact operators?

Answer 1

Yes, on a separable infinite-dimensional Hilbert space $H$ there exists a sequence of compact and nilpotent linear operators $T_n$ that converges strongly to the identity.

To construct such a sequence $(T_n)$ we may assume that $H = L^2(0,1)$ (it doesn't matter whether the scalar field is real or complex).

For each integer $n \ge 1$ let $S_n$ be the right shift on $L^2(0,1)$ with shift length $\frac{1}{n}$; everything that leaves $[0,1]$ on the right is dropped and on the left only $0$ enters. Clearly each operators $S_n$ is nilpotent and the sequence $(S_n)$ converges strongly to the identity operator.

Also for each integer $n \ge 1$, let $C_n: L^2(0,1) \to L^2(0,1)$ denote the convolution with $2n \, 1_{[\frac{1}{2n}, \frac{1}{n}]}$. Each operator $C_n$ is compact and the sequence $(C_n)$ also converges strongly to the identity operator.

Finally, set $T_n := C_n S_n$ for each $n \ge 1$. Then each $T_n$ is compact and the sequence $(T_n)$ also converges strongly to the identity. Moreover, for every $a \in [0,1]$ each $C_n$ leaves the subspace of those functions that live on $[a,1]$ invariant. Hence, each $T_n$ is nilpotent.