Riesz representation theorem - yet another "counter example"

Question

this is a follow up on a previous question I asked.

I was looking for examples of when the Riesz representation theorem doesn't hold because not all conditions are met. Meaning, I was looking for examples of functionals over inner-product spaces that cannot be represented as an inner-product with some specific vector.

I came up with an examples of my own, but I'm not entirely sure it is correct, and I'd like to hear your opinion on if it's true. If it is true, I'd appreciate it if you'd help me prove it formally, because I'm having some trouble in this area:

I'm looking at the space of all real sequences who's sum absolutely converges, with the inner product $\langle a, b\rangle=\Sigma_{k=1}^{\infty} a_{k}b_{k}$ . On this space, I define the linear functional that maps any sequence to its sum.

It's pretty easy to see that the functional could hypothetically by represented as an inner product with the sequence: $(b_{n})=(1, 1, 1,...)$. But because this sequence is not a part of the vector space, the inner product is not defined on it.

Do you think this example is correct? If it is, how would you prove that there can't be some other vector that can represent this functional? I'm completely new to this world of content, so any feedback would be appreciated (namely, if I made some mistake, please explain where it is and if there's a way to get around it somehow by changing parts of the example).

Thanks in advance!

Answer 1

Yes, it's correct. To prove this functional can't be represented by any other vector, consider what the functional does to $e_j$ which has $1$ in position $j$ and $0$ everywhere else.

Answer 2

It is true that $f : \ell^1 \to \mathbb{R}$ given by $f(x_i)_i = \sum_{i=1}^\infty x_i$ cannot be represented by scalar multiplication with a vector $\ell^1$, and that the only candidate is $(1, 1, \ldots )$.

However, this example isn't as interesting because $f$ isn't a bounded linear functional on $\ell^1$ in the first place (remember, you're looking at $\ell^1$ equipped with the $\ell^2$ norm!).

Consider $x_n = \left(1, \frac12, \ldots, \frac1n, 0,0, \ldots\right) \in \ell^1$. We have $\|x_n\|_2 = \sum_{i=1}^n \frac1{i^2} \le \frac{\pi^2}6$ but $|f(x_n)| = \sum_{i=1}^n \frac1i$ which is unbounded as $n \to \infty$.

The Riesz representation theorem claims nothing about unbounded functionals.