4
$\begingroup$

I have always seen the linear functionals in $R^n$ expressed at $\ell(x) = \sum_{i=0}^n a_ix_i$ And in an countable metric space $\ell(x) = \sum_{i=0}^{\infty} a_ix_i$. I guess that this follows directly from http://en.wikipedia.org/wiki/Riesz_representation_theorem, for Hilbert spaces. But what if we are not in an Hilbert space or if the space is uncountable. If X was a 1 dimensional space I would get $f(x) = f(1)x$ by continuity and linearity (by derivation and integration) and by partial derivation it would look like $f(x) = \sum_{i=0}^n f(1)x_i$ for the n dimensional case

2 Answers 2

3

In general, linear functionals can be very different than that.

The easiest example of a functional not of the form you claim, is probably there case where you take $X=C[0,1]$ and consider functionals like $\ell(f)=f(0)$ (or any other point, for that matter).

Note also that when you consider infinite-dimensional spaces, they most often come with a topology, and one considers bounded (i.e. continuous) linear functionals. Unbounded linear functionals exist but are a lot tricker to deal with.

  • 0
    Bounded functionals of $\ell^1(\mathbb N)$ are of the form you want, i.e. every bounded functional is given by an element of $\ell^\infty(\mathbb N)$ (which we say is the *dual* of $\ell^1(\mathbb N)$). But the proof is ad-hoc, you cannot apply the Riesz-Representation Theorem since the $\ell^1$ norm is not given by an inner product.2012-12-05
1

I can think of a nice representation theorem that holds in a non-Hilbert space. It goes by the name Riesz-Kakutani-Markov:

Let $X$ be a compact Hausdorff space and $(C(X),\|\cdot\|_\infty)$ the space of continuous real valued functions on $X$ endowed with the maximum norm. Then, every bounded linear functional $F$ on $C(X)$ can be written as an integral against a signed, finite Borel measure $\mu$ on $X$:

$ F(f)=\int_X fd\mu $

with norm

$ \|F\|=\int_X\vert d\mu\vert $ where $\vert d\mu\vert$ is the absolute variation of $\mu$.

A good resource for this theorem is Lax: Functional Analysis. Granted, this is more sophisticated than the Riesz representation theorem on Hilbert spaces, but that's to be expected.

  • 0
    Interesting! Lax seems to be the only book (that I have, anyway) that doesn't credit Markov.2012-12-05