Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, 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

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
thanks! So the sum I often see comes from the scalar product then, I guess. So for example if we look at the space $\ell^1$ the sum comes from the scalar product, hence Riesz? – Johan Dec 5 '12 at 16:09
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. – Martin Argerami Dec 5 '12 at 16:25

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.

share|cite|improve this answer
Just to add, the theorem is also known as Riesz-Markov. Don't know about the history of the naming, though. – Martin Argerami Dec 5 '12 at 16:26
Interesting! Lax seems to be the only book (that I have, anyway) that doesn't credit Markov. – icurays1 Dec 5 '12 at 16:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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