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I was wondering if anyone could give me some interesting "counter examples" to the Riesz representation theorem about functionals over Hilbert spaces. When I say counter examples, I'm obviously talking about examples where some of the basic assumptions of the theorem aren't met, so the theorem doesn't hold. In other words - could you show me some non-trivial examples of functionals over inner-product spaces that cannot be expressed as an inner-product with some vector in the vector space? I already have an example from $C [0, 1]$ based on the standard $L^2$ integral inner-product, but I was wondering if anyone could enlighten me with a more interesting example. I don't have much background, but I'm very interested to hear about this topic, and I'd appreciate it if you could give full explanations so I could understand. Thanks in advance

Martin Sleziak
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GSofer
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2 Answers2

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Consider $c_{00}$, the space of all finitely supported sequences, equipped with the $\|\cdot\|_2$ norm.

The functional $$(x_n)_n \mapsto \sum_{n=1}^\infty \frac{x_n}{n}$$

is bounded on $c_{00}$ but it is not represented by any vector from $c_{00}$.

Namely, it is represented by $\left(\frac1n\right)_n \in \ell^2$, which is in the completion of $c_{00}$.

All examples will be of this form, i.e. the functional will be representable by some vector from the completion of your incomplete inner product space.

mechanodroid
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  • Thank you for the example. How do you prove that it can only be represented by 1/n? Why can't there be another representation? – GSofer May 31 '18 at 05:36
  • @GSofer The Riesz representation is always unique: if $\langle x, y\rangle = \langle x, z\rangle, \forall x \in c_{00}$ then $$0 = \langle x, y-z\rangle, \forall x \in c_{00}$$ so $y - z \perp c_{00}$ and then $y = z$ because $\overline{c_{00}} = \ell^2$. – mechanodroid May 31 '18 at 08:01
  • @mechanodroid How can we say "All examples will be of this form"? – Meet Patel Feb 09 '24 at 12:54
  • @MeetPatel Suppose that $f$ is a bounded functional on an incomplete inner product space $X$. Then $f$ by density extends uniquely to a bounded functional $\overline{f}$ on the completion $\overline{X}$ of $X$. Since $\overline{X}$ is a Hilbert space, by the Riesz representation theorem there exists a unique $y \in \overline{X}$ such that $\overline{f} = \langle \cdot, y\rangle$. The question is only whether $y \in X$. If yes, then $f = \langle \cdot, y\rangle$ is representable in $X$. If not, $f$ is certainly not representable as $f = \langle \cdot, x\rangle$ for some $x \in X$. – mechanodroid Feb 09 '24 at 14:34
  • @mechanodroid Can we use Hahn Banach theorem to extend $f$.? Because in this case we may have multiple extension – Meet Patel Feb 09 '24 at 15:02
  • @MeetPatel Hahn-Banach extension on a Hilbert space is always unique, see here: https://math.stackexchange.com/q/4724346/144766 – mechanodroid Feb 10 '24 at 12:52
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If $V$ is the space of trigonometric polynomials endowed with the restriction of the inner product of $L^2[-\pi,\pi]$, then the functional $\phi\in V^*$ such that $\phi(f)=\int_{-\pi}^\pi tf(t)\,dt$ for all $f$ is not representable as $\langle\bullet,p\rangle$ for any trigonometric polynomial $p$. That is because the sequence $a_n=\langle \sin(nx), p\rangle$ must be eventually zero for any $p$, while $\phi(\sin(nx))\ne 0$ for all $n>0$.