Give an example of a Hilbert space, H, and a sequence of compact operators, $(S_n)_{n=1}^\infty$ on H such that

i.) $\|S_n\| \leq 1$ for $n = 1,2,...$

ii.) The operators $V_N=\sum_{n=1}^N \frac{1}{n}S_n$ converge strongly as $N \rightarrow \infty$

iii.) The strong limit of the operators $V_N$ is not compact.

The closest I've come is:

$S_n(x): \ell_2 \rightarrow \ell_2, S_n(x) = (0,0,...,0,x_n,0,...)$,

but its limit is compact.


Let's turn it around a bit: you want a sequence $V_n$ of compact operators such that $\|V_n - V_{n-1}\| \le 1/n$ and $V_n$ converges strongly to a non-compact operator. Then you can take $S_n = n (V_n - V_{n-1})$. The orthogonal projections $P_m: x \to (x_1, \ldots, x_m, 0, \ldots)$ on $\ell_2$ converge strongly to the identity, but $\|P_m - P_{m-1}\| = 1$. So instead of going directly from $P_m$ to $P_{m+1}$ in one step, go in many steps. Actually, to make things a bit simpler, I'll go from $P_m/2$ to $P_{m+1}/2$. Thus for $2^{m} \le n < 2^{m+1}$ take $$ V_n = \dfrac{2^{m+1}-n}{2^{m+1}} P_m + \dfrac{n - 2^m}{2^{m+1}} P_{m+1}$$


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