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Here's my question:

Let $H$ be a (separable) Hilbert space. Let $W$ be a dense subspace of $H$ and $P: H \to H$ be a linear operator satisfying $\langle Pw, w\rangle \ge 0$ for each $w\in W$. Then $P$ is a positive operator on $H$.

My goal is to show that $\langle Pf, f\rangle \ge 0$ for each $f \in H$. To this end, let $f \in H$ and let $(w_n)$ be a sequence in $W$ such that $(w_n)$ converges to $f$. If $\langle Pf , f\rangle = \lim \langle Pw_n , w_n \rangle$ then we would be done but I do not see how this would work.

If we fix $m \in \mathbb N$ then we can show that $\langle Pw_n , w_m \rangle$ converges to $\langle Pf, w_m \rangle$ as $n \to \infty$. But I do not think interchanging the limit would work here.

Hints are appreciated.

ashK
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  • Inner product is jointly continuous in its two operands, so your first limit already works. – David Gao Dec 10 '23 at 07:46
  • Is $P$ bounded, or is it an unbounded operator that happens to be defined everywhere? If $P$ is bounded, then the bilinear/sesquilinear map $(\varphi,\psi) \mapsto \langle P(\varphi), \psi\rangle$ is continuous. – Bruno B Dec 10 '23 at 07:48
  • @BrunoB It is a bounded operator. – ashK Dec 10 '23 at 07:51
  • @DavidGao I was unaware of that fact. Thanks. – ashK Dec 10 '23 at 07:51

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