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Let $H$ be a separable infinite dimensional Hilbert space, with orthonormal basis $(e_n)_{n=1}^\infty$. Consider the operators $U_n$ on $H$ such that $U_ne_k = e_1$ if $n=k$ and $U_ne_k=0$ if $n\neq k$. Note that $\|U_n\|=1$.

My question is inspired by the following results, which are parts of exercises in section 4.6 of Gert Pedersen's book Analysis Now.

Fact 1: $0$ is in the closure of the set $A=\{\sqrt{n}U_n:n\geq 1\}$ in the strong operator topology.

Fact 2: No sequence in $A$ converges weakly or strongly to $0$. (Consequently, these topologies are not metrizable.)

The proof of Fact 2 goes by showing that if a sequence in $A$ weakly/strongly converges to $0$, then it is pointwise bounded, so an application of the Principle of Uniform Boundedness yields a contradiction, since such a sequence cannot be uniformly bounded in norm unless it is finite, in which case it cannot converge to $0$.

Question: Why doesn't this argument hold for nets? By Fact 1, there is a net in $A$ which converges strongly to $0$, and it seems like the same argument would imply that the net is pointwise bounded, hence uniformly bounded in norm.

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As opposed to a sequence, convergence of a net does not imply boundedness.

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  • $\begingroup$ Is the issue here that if $T_\alpha$ is a net of operators with $\|T_\alpha x\|\to 0$ for each $x$, then though $\|T_\alpha x\|$ is bounded for all $\alpha\geq \alpha_x$ for some $\alpha_x$, this depends on $x$, and one cannot just assume that all of the $\|T_\alpha x\|$ start being bounded at the same time? $\endgroup$ – Iian Smythe Feb 7 '14 at 2:23
  • $\begingroup$ Exactly. ${\ }$ $\endgroup$ – Martin Argerami Feb 7 '14 at 2:24

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