6
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – GSofer May 31 '18 at 5:36
  • $\begingroup$ @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 $\overline{c_{00}} = \ell^2$. $\endgroup$ – mechanodroid May 31 '18 at 8:01
1
$\begingroup$

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 $n>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.