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Let $V$ be a vector space with inner product $(\phi,\psi)$ antilinear in the second argument - not necessarily a Hilbert space. Let $\Phi$ be an antilinear functional on $V$.

What are the precise (necessary and sufficient) conditions that must be imposed on $\Phi$ that will guarantee the existence of a sequence or net of vectors $\phi_k\in V$ such that $$\Phi(\psi)=\lim_{k} (\phi_k,\psi) ~~~\forall~ \psi\in V?$$ The answer is given by the Riesz representation theorem when $V$ is a Hilbert space, but I am interested in the general case.

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    $\begingroup$ Do you want that limit to just hold pointwise, or do you want something stronger? If you just want it pointwise, the Riesz representation theorem has nothing to do with this--you can find such a net for any $\Phi$ (because you can find a $\phi$ that works for any finite set of $\psi$s). On the other hand, if you restrict to sequences, $\Phi$ does have to be bounded if $V$ is complete, but this is not obvious (it follows from the Banach-Steinhaus theorem). $\endgroup$ – Eric Wofsey Aug 24 '15 at 15:18
  • $\begingroup$ @EricWofsey: The limit is a limit of numbers, hence there is no weaker or stronger version. How to construct the net? $\endgroup$ – Arnold Neumaier Aug 24 '15 at 16:08
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    $\begingroup$ I think Eric is asking whether you wanted some kind of uniformity with respect to $\psi$ in that limit. (It can be interpreted as a limit of functions of $\psi$ rather than just a limit of numbers.) $\endgroup$ – Qiaochu Yuan Aug 24 '15 at 22:56
  • $\begingroup$ @QiaochuYuan: I see. Although asking for more seems interesting, the pointwise construction is enough for me. $\endgroup$ – Arnold Neumaier Aug 25 '15 at 9:14
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Such a net exists for any $\Phi$ (in fact, there is a canonical such net). First, note that if $F\subseteq V$ is a finite-dimensional subspace, then there is a unique $\phi_F\in F$ such that $\Phi(\psi)=(\phi_F,\psi)$ for all $\psi\in F$. The collection of such $F$ form a directed set under inclusion, and the net $(\phi_F)$ will have the desired property.

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