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


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.

  • $\begingroup$ 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

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 $n>0$.


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