MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

As someone who mostly does symbolic computation, I've always been puzzled by the fascination mathematicians seem to have with Lp(R) (for p<∞)? To be more precise, there are no non-trivial polynomials in that space and, to me, polynomials are not only the simplest functions, they are the building blocks of most everything which can be (easily) manipulated algorithmically. And restricting to a compact support is really a non-answer, since one of the great things about polynomials is that they are global, analytic functions.

To ask a more precise question: are there some spaces of (total, real-valued) functions which are both nice from a functional analysis point of view, and contain all the polynomials?

share|cite|improve this question
By any measure (ha!), characteristic functions of intervals have to be considered simpler than polynomials. – Qiaochu Yuan Feb 14 '10 at 20:49
up vote 13 down vote accepted

You are referring to $L^p(\mathbb{R}, \mathcal{B}, \mu)$ in the case that $\mathbb{R}$ is endowed with Lebesgue measure $\mu$. Consider instead the measure $\nu$ given by $d\nu = f d\mu$, where $f$ is in the Schwartz space and $f$ does not take the value zero. Because the product of a polynomial with $f$ is also in the Schwartz space, and the Schwartz space is contained in $L^p(\mathbb{R}, \mathcal{B}, \mu)$, it follows that polynomials are in $L^p(\mathbb{R}, \mathcal{B}, \nu)$.

share|cite|improve this answer
Wonderful answer. I guess the only drawback is that such measures are no longer translation-invariant? – Jacques Carette Feb 14 '10 at 16:45
They are not, but you can do lots of nice things with (e.g.) Gaussian measures: – Steve Huntsman Feb 14 '10 at 16:47

If one wants to do algebra, or symbolic computation, then polynomials are indeed the simplest type of function. But if one wants to do analysis, or numerical computation, then actually the best functions are the bump functions - they are infinitely smooth, but also completely localised. (Gaussians are perhaps the best compromise, being extremely well behaved in both algebraic and analytic senses.)

That said, I'm not sure what your question is really after. If you want a function space that contains the polynomials, you could just take ${\bf R}[x]$. Of course, this space does not come equipped with a special norm, but polynomials, being algebraic objects rather than analytic ones, are not naturally equipped with any canonical notion of size. Due to their growth at infinity, any such notion of size would have to be mostly localised, as is the case with the weighted spaces and distribution spaces given in other answers.

share|cite|improve this answer
The question is really about trying to find links between the algebraic world where we have lots of exact algorithms and functional analysis. As you well know, lot of successful mathematics is about finding ways to transport theorems from one setting to another. But, before I encountered tempered distributions and the Schwartz space, I could not find links between the two that were 'natural'. And yes, I want a norm (or a metric), since that's a fundamental tool of functional analysis. – Jacques Carette Feb 17 '10 at 3:38

Distributions (or tempered distributions).

share|cite|improve this answer
Distributions are total functions? I always thought of them as equivalence classes, thus being essentially impossible to evaluate pointwise. Thanks for the pointer to tempered distributions, I had not encountered them before. – Jacques Carette Feb 14 '10 at 16:35
Hm. Well, I suppose you could look at distributions that are locally in some Sobolev space or something like that. – Akhil Mathew Feb 14 '10 at 16:55
Heh, tempered distributions are somewhat "dual" to the schwartz space mentioned in the answer above, Jacques. – Harry Gindi Feb 14 '10 at 22:09

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.