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Consider the Hilbert Space of weighted-square-integrable functions f(x):

$$ \begin{equation} \int_{-\infty}^{\infty}\frac{f(x)^2}{e^{x}+e^{-x}}dx<\infty. \end{equation} $$

Note this integral is taken over the real line in both directions, and the weighting is roughly exponential. Hilbert Polynomials are complete using a different weighting, while Laguerre polynomials use a similar exponential weighting, but only on the positive real line.

What is a simple proof or reference that polynomials are complete in this space?

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  • $\begingroup$ Do you suppose that finding some $g$ such that $g'(u) = \frac{1}{e^{g(u)}+e^{-g(u)}}$ and doing a u-substitution to get $\int_{-\infty}^{\infty}f(g(u))^2 du$ in the integral be a useful approach? $\endgroup$ – Milo Brandt Sep 28 '14 at 17:16
  • $\begingroup$ Do you want to know that/if/why the set $\Bbb{R}[x]$ of polynomials is dense in the given $L^2$ space, or do you want to have an explicit orthonormal basis consisting of polynomials? $\endgroup$ – PhoemueX Sep 28 '14 at 18:03
  • $\begingroup$ @phoemueX: I want to know that polynomials are dense in the space. Then I can easily construct an orthonormal polynomial basis using the Gram-Schmidt procedure. $\endgroup$ – Steven Heston Sep 28 '14 at 20:10
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Disclaimer: The following proof is not completely elementary and uses methods from complex analysis as well as the Fourier transform, but at least it does the job:

Assume that the polynomials are not dense in your $L^{2}$-space. This implies the existence of $f\neq0$ in your $L^{2}$space for which $$ \int_{\mathbb{R}}\frac{f\left(x\right)\cdot x^{n}}{e^{x}+e^{-x}}\, dx=0\qquad\forall n\in\mathbb{N}_{0}. $$ Now consider the function $$ \Gamma:\left\{ s\in\mathbb{C}\mid-\frac{1}{2}<\text{Re}\left(s\right)<\frac{1}{2}\right\} \to\mathbb{C},s\mapsto\int_{\mathbb{R}}e^{sx}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx. $$ Note that $\Gamma$ is well-defined because of \begin{eqnarray*} \int_{\mathbb{R}}\left|e^{sx}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\right|\, dx & \leq & \sqrt{\int_{\mathbb{R}}\frac{\left|f\left(x\right)\right|^{2}}{e^{x}+e^{-x}}\, dx}\cdot\sqrt{\int_{\mathbb{R}}\frac{\left|e^{sx}\right|^{2}}{e^{x}+e^{-x}}\, dx}\\ & = & \sqrt{\int_{\mathbb{R}}\frac{\left|f\left(x\right)\right|^{2}}{e^{x}+e^{-x}}\, dx}\cdot\sqrt{\int_{\mathbb{R}}\frac{e^{2\text{Re}\left(sx\right)}}{e^{x}+e^{-x}}\, dx}\\ & \leq & \sqrt{\int_{\mathbb{R}}\frac{\left|f\left(x\right)\right|^{2}}{e^{x}+e^{-x}}\, dx}\cdot\sqrt{\int_{\mathbb{R}}e^{\left(2\left|\text{Re}\left(s\right)\right|-1\right)\left|x\right|}\, dx}<\infty. \end{eqnarray*} Also observe that we can bound the integrand on each strip $\left\{ s\in\mathbb{C}\mid-c<{\rm Re}\left(s\right)<c\right\} $, with $c\in\left(0,\frac{1}{2}\right)$, by $\frac{e^{2c\left|x\right|}\cdot\left|f\left(x\right)\right|}{e^{x}+e^{-x}}$ independently(!) of $s$ in that strip, so that standard arguments yield continuity of $\Gamma$.

Now, let $\gamma:\left[a,b\right]\to U$ be a closed (piecewise) $C^{1}$-curve in $U:=\left\{ s\in\mathbb{C}\mid-\frac{1}{2}<{\rm Re}\left(s\right)<\frac{1}{2}\right\} $. Then \begin{eqnarray*} \int_{\gamma}\Gamma\left(z\right)\, dz & = & \int_{a}^{b}\Gamma\left(\gamma\left(t\right)\right)\cdot\gamma'\left(t\right)\, dt\\ & = & \int_{a}^{b}\int_{\mathbb{R}}e^{\gamma\left(t\right)\cdot x}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\cdot\gamma'\left(t\right)\, dx\, dt\\ & \overset{\left(\ast\right)}{=} & \int_{\mathbb{R}}\int_{a}^{b}e^{\gamma\left(t\right)\cdot x}\cdot\gamma'\left(t\right)\, dt\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx\\ & = & \int_{\mathbb{R}}\int_{\gamma}e^{xz}\, dz\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx\\ & = & 0. \end{eqnarray*} In the last step, we used that $z\mapsto e^{xz}$ is an entire function so that $\int_{\gamma}e^{xz}\, dz$ vanishes by Cauchy's integral theorem. The application of Fubini's theorem in $\left(\ast\right)$ is justified because of \begin{eqnarray*} \int_{\mathbb{R}}\int_{a}^{b}\left|e^{\gamma\left(t\right)\cdot x}\cdot\gamma'\left(t\right)\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\right|\, dt\, dx & = & \int_{a}^{b}\int_{\mathbb{R}}\left|e^{\gamma\left(t\right)\cdot x}\cdot\gamma'\left(t\right)\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\right|\, dx\, dt\\ & \leq & \left\Vert \gamma'\right\Vert _{\sup}\cdot\left(b-a\right)\cdot\int_{\mathbb{R}}\frac{e^{2c\left|x\right|}\left|f\left(x\right)\right|}{e^{x}+e^{-x}}\, dx, \end{eqnarray*} where $c>0$ is chosen with $\left|{\rm Re}\left(\gamma\left(t\right)\right)\right|\leq c<\frac{1}{2}$ for all $t\in\left[a,b\right]$ (such a $c$ exists, because $\gamma\left(\left[a,b\right]\right)\subset U$ is compact). Note that Tonelli's theorem is always applicable for non-negative (measurable) functions.

By Morera's theorem, we conclude that $\Gamma$ is holomorphic on $U$. But for $\left|s\right|<\frac{1}{4}$, we have $$ \sum_{n=0}^{\infty}\frac{\left|\left(sx\right)\right|^{n}}{n!}=e^{\left|sx\right|}\leq e^{\frac{1}{4}\left|x\right|} $$ with $\int_{\mathbb{R}}\frac{e^{\frac{1}{4}\left|x\right|}\left|f\left(x\right)\right|}{e^{x}+e^{-x}}\, dx<\infty$ (as above), so that dominated convergence yields \begin{eqnarray*} \Gamma\left(s\right) & = & \int_{\mathbb{R}}e^{sx}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx\\ & = & \int_{\mathbb{R}}\lim_{n\to\infty}\sum_{\ell=0}^{n}\frac{\left(sx\right)^{n}}{n!}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx\\ & = & \lim_{\ell\to\infty}\int_{\mathbb{R}}\left[\sum_{\ell=0}^{n}\frac{\left(sx\right)^{n}}{n!}\right]\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx=0. \end{eqnarray*} By the identity theorem for holomorphic functions, we conclude $\Gamma\equiv0$ and hence $$ 0=\Gamma\left(i\xi\right)=\int_{\mathbb{R}}e^{i\xi x}\cdot\frac{f\left(x\right)}{e^{x}+e^{-x}}\, dx, $$ so that the Fourier transform of the $L^{1}$-function(!) $x\mapsto\frac{f\left(x\right)}{e^{x}+e^{-x}}$ vanishes, which implies $\frac{f\left(x\right)}{e^{x}+e^{-x}}\equiv0$ (a.e.) and hence $f\equiv0$ (a.e.), a contradiction.

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  • $\begingroup$ I just noted that one can probably replace the application of Morera's theorem by simply differentiating under the integral sign. $\endgroup$ – PhoemueX Sep 30 '14 at 10:36
  • $\begingroup$ Thanks, @PhoemueX. I had noticed that it is straightforward to show the Fourier transform is holomorphic in a neighborhood of the origin. This approach follows the Wikipedia proof for the Hermite case, except the Hermite Fourier transform is entire. $\endgroup$ – Steven Heston Sep 30 '14 at 18:23

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