Metric completion of polynomial function space

Take the vector space of all polynomial functions from $\Bbb R$ to $\Bbb R$ with an inner product $$\langle f,g \rangle = \int_{-\infty}^{\infty} f(x)g(x)\sigma(x)dx$$ where $\sigma$ is a positive, smooth function such that for all natural numbers $n$ $$\lim_{|x| \to \infty} \sigma(x)x^n=0.$$ (For example, $\sigma (x)$ could be $e^{-x^2}$.)

Producing a metric from this inner product in the usual way turns the space of polynomial functions into a metric space. I think I can show that this metric space is not complete. For instance, I believe the exponential function is a limit that does not exist in the space. What I am not certain of is how to go about finding the completion of this space.

So my question is: what is the completion of this space?

I have a feeling that it is something like the space of all functions which are square-integrable, do not go to infinity faster than $\sigma$ goes to zero, and are equal to their Taylor series on all of $\Bbb R$... but I don't know how to show that.

Thanks!

• Are there any other conditions on $\sigma$, such as $\sigma > 0$? Otherwise, you could take $\sigma = 0$ and not even get an inner product. – Paul Sinclair Oct 19 '15 at 22:39
• Ah, thanks. You are correct, $\sigma$ must be everywhere-positive. – Matt Dickau Oct 19 '15 at 22:41
• It's pretty clear that any Taylor series with infinite radius of convergence is in the closure, But I think the closure is likely to be much bigger. It seems likely to me, though I haven't proved it, that the closure is all of $L^2(\Bbb R)$. – Paul Sinclair Oct 20 '15 at 2:17
• @PaulSinclair : If $\sigma(x)=e^{-|x|}$, then $e^{2x}$ is not in closure, even though $e^{2x}$ has an infinite radius of convergence. – DisintegratingByParts Oct 20 '15 at 20:14
• @TrialAndError - true. I misspoke, and the $L^2(\Bbb R)$ in my previous comment should be $L^2(\Bbb R, \sigma)$ as well. – Paul Sinclair Oct 20 '15 at 20:35

In the case of $e^{-x^{2}}$, you have an answer, but a general answer is non-trivial. In the case of $\sigma(x)=e^{-x^{2}}$, Hermite polynomials give you the answer: $$H_n(x) = (-1)^{n}C_ne^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}$$ You have $$\int_{-\infty}^{\infty}H_{n}(x)H_{m}(x)e^{-x^{2}}dx =0,\;\;\; n \ne m.$$ By choosing positive constants $C_n$ appropriately, $\{ H_n \}_{n=0}^{\infty}$ becomes an orthonormal set with respect to the weighted inner product. For any $f \in L^{2}(\mathbb{R})$, it is known that $\{ h_n=e^{-x^{2}/2}H_{n}\}$ is an orthonormal basis of $L^{2}(\mathbb{R})$. Therefore, if $f \in L^{2}(\mathbb{R})$, then $$\int_{-\infty}^{\infty}|f-\sum_{n=0}^{N}(f,h_n)h_n|^{2}dx =\int_{-\infty}^{\infty}|fe^{x^{2}/2}-\sum_{n}^{N}(f,h_n)H_n|^{2}e^{-x^{2}}dx\rightarrow 0$$ as $N\rightarrow \infty$. So the completion of the space with weight $\sigma(x)=e^{-x^{2}}$ is $e^{x^{2}/2}L^{2}(\mathbb{R})$, which is as nice as you can reasonably expect. The completion $X$ of your space is isometrically isomorphic to $L^{2}$ under the map $U : L^{2}(\mathbb{R})\rightarrow X$ defined by $Uf=e^{x^{2}/2}f$. Proving the completeness of the Hermite functions $\{ h_n \}$ is not a simple matter, and, for an arbitrary weight $\sigma$, I don't think things turn out so nice. I wish I had a reference for you, but I don't. You might try searching "orthogonal polynomials" for references.