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 $\endgroup$ Nov 4, 2018 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, 2018 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 $[0,1]$. $\endgroup$ Nov 4, 2018 at 23:11
  • $\begingroup$ @mathworker21. Ah, you're right. $\endgroup$
    – md2perpe
    Nov 5, 2018 at 6:44

2 Answers 2


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, see comments). 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...)

  • 1
    $\begingroup$ Great answer. Thanks! $\endgroup$ Nov 5, 2018 at 13:53
  • 2
    $\begingroup$ I don't think it's stronger. The set of polynomials is invariant under $x\mapsto \frac{x}{\alpha}$. $\endgroup$ Nov 5, 2018 at 18:58
  • $\begingroup$ @mathworker21: +1, edited. $\endgroup$
    – metamorphy
    Nov 5, 2018 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. $\endgroup$ Nov 15, 2018 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? $\endgroup$ Nov 15, 2018 at 16:18

There is a stronger conclusion. Every function $f(x)$ continuous on $\mathbb{R}$ and tending to $0$ at $\pm \infty$ can be approximated uniformly by a sequence of the form $p_n(x)e^{-|x|},$ where $p_n(x)$ are polynomials. This can be proved using Stone-Weierstrass theorem, but requires some effort. The above can be used to solve your problem. It suffices to consider the case $\alpha=2.$ Let $f(x)$ be a continuous and vanishing at $\infty.$ Then $$\|p_n(x)e^{-2|x|}-f(x)e^{-|x|}\|_2 \le \|p_n(x)e^{-|x|}-f(x)\|_\infty \|e^{-|x|}\|_2= \|p_n(x)e^{-|x|}-f(x)\|_\infty. $$ The functions of the form $f(x)e^{-|x|}$ are dense in $L^2(\mathbb{R})$ as they contain continuous functions with bounded support.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .