Lately I was talking to a friend of mine and we came up with the following question

Denote by $\mathcal{P}$ the set of all real valued polynomial functions. Is the set $$ \{ p(x) e^{- \alpha \vert x \vert} \ : \ p\in \mathcal{P}, \ \alpha \in \mathbb{R}_{>0} \} $$ dense in $L^2(\mathbb{R} ,\mathbb{R})?$

My feeling is that it should be true. I was thinking about using the Stone-Weierstrass, however, I cannot control the $L^2$-norm of the polynomial function outside the compact set. Clearly it is finite (the exponential decay beats the polynomial growth), but it is not clear to me whether one can choose it to be small.

  • $\begingroup$ I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $\alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-\alpha |x|}$ is close enough to $f$, but then you need to make $\alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts $\endgroup$ – mathworker21 Nov 4 '18 at 22:13
  • $\begingroup$ @mathworker21. Given $\epsilon>0$ let $f(x) = 1_{[0,1]}(x) \, e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $\|f-p\|_{\infty} < \epsilon.$ Then $\|1_{[0,1]} - e^{-x}p\|_\infty = \|e^{-x}(f-p)\|_\infty < \|f-p\|_\infty < \epsilon$ since $\|e^{-x}\|_\infty = 1$ on $[0,1]$. $\endgroup$ – md2perpe Nov 4 '18 at 22:36
  • $\begingroup$ @md2perpe but you're doing $L^\infty$ over $[0,1]$ not over $\mathbb{R}$. My comment is saying that if you want to get a good $L^\infty$ bound over $[0,1]$, you won't have the right decay in $\mathbb{R}$. And conversely if you do have the right decay in $\mathbb{R}$, you won't have an $L^\infty$ bound on $[0,1]$. $\endgroup$ – mathworker21 Nov 4 '18 at 23:11
  • $\begingroup$ @mathworker21. Ah, you're right. $\endgroup$ – md2perpe Nov 5 '18 at 6:44

Yes, the answer is affirmative. It is equivalent to

$\{p(x)e^{-|x|} : p\in\mathcal{P}\}$ is dense in $L_2(\Bbb{R},\Bbb{R})$.

In our context of Hilbert spaces, we need to show that $$L_0 = \left\{f\in L_2(\Bbb{R}) : (\forall n\in\Bbb{Z}_{\geq 0})\ \int_{\Bbb{R}}x^n e^{-|x|}f(x)\,dx=0\right\}$$ consists of $f\equiv 0$ only. Here and below, all functions are complex-valued.

Let $\Lambda=\{\lambda\in\Bbb{C} : |\Re\lambda|<1\}$; for $f\in L_2(\Bbb{R})$, the function $$B_f(\lambda)=\int_{\Bbb{R}}e^{-|x|+\lambda x}f(x)\,dx$$ is analytic in $\Lambda$ (differentiation is admissible under the integral sign). Further, for any $n\in\Bbb{Z}_{\geq 0}$ we have $B_f^{(n)}(0)=\displaystyle\int_{\Bbb{R}}x^n e^{-|x|}f(x)\,dx$ and, therefore, $f\in L_0$ if and only if $B_f\equiv 0$.

Now let $f\in L_0$ and $g\in L_1(\Bbb{R})$ (fixme... much less is enough). Then $$0=\int_{\Bbb{R}}g(\lambda)B_f(i\lambda)\,d\lambda=\int_{\Bbb{R}}e^{-|x|}\hat{g}(x)f(x)\,dx,\quad\hat{g}(x)=\int_{\Bbb{R}}e^{i\lambda x}g(\lambda)\,d\lambda.$$ Thus, $e^{-|x|}f(x)$ is orthogonal to $\{\hat{g} : g\in L_1(\Bbb{R})\}$. This space is dense in $L_2(\Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from $$\hat{g}_0(x) = \max\{0,1-|x|\} \impliedby g_0(\lambda)=\frac{2}{\pi}\left(\frac{\sin\lambda/2}{\lambda}\right)^2$$ using linear combinations and shifts; $\hat{g}_1(x)=\hat{g}(x+a)\impliedby g_1(\lambda)=e^{i\lambda a}g(\lambda)$.

(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)

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    $\begingroup$ Great answer. Thanks! $\endgroup$ – Severin Schraven Nov 5 '18 at 13:53
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    $\begingroup$ I don't think it's stronger. The set of polynomials is invariant under $x\mapsto \frac{x}{\alpha}$. $\endgroup$ – mathworker21 Nov 5 '18 at 18:58
  • $\begingroup$ @mathworker21: +1, edited. $\endgroup$ – metamorphy Nov 5 '18 at 21:56
  • $\begingroup$ @metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions. $\endgroup$ – mathworker21 Nov 15 '18 at 15:10
  • $\begingroup$ @metamorphy also, one more question. it seems you used fourier inversion to conclude that $\widehat{B_f(i\cdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(i\cdot) \in L^1$? How would you show this? $\endgroup$ – mathworker21 Nov 15 '18 at 16:18

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