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I proved that the linear span of $\{e^{-n x}\}_{n\in\mathbb N}$ is dense in $L^2((0,\infty))$ (see below).

Question: Given $f\in L^2((0,\infty))$, can I find "expansion coefficients" $f_n$ such that $$f(x) = \sum_{n\in\mathbb N} f_ne^{-n x} ?$$

Proof of density: Using the Stone-Weierstrass theorem for locally compact Hausdorff space, the linear span of $\{e^{-n x}\}_{n\in\mathbb N}$ is dense in $C_0((0,\infty))$, the space of continuous functions vanishing at infinity. Since $C_0((0,\infty))$ is dense in $L^2((0,\infty))$, we prove the result.

Note: After very nice comments, the question is simplified.

Laplacian
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    Did you try looking at https://en.wikipedia.org/wiki/Laplace_transform? – Pedro Apr 13 '23 at 13:39
  • @Pedro Thanks for comment. I know the basics of Laplace transform. Could you explain how the theory of Laplace transform helps to answer this question? – Laplacian Apr 13 '23 at 13:42
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    Wait a minute. It is the linear span of ${ e^{-\alpha x}}_{\alpha\ge 0}$ that is dense. Not the set itself. – Giuseppe Negro Apr 13 '23 at 13:46
  • @GiuseppeNegro Yes, you are right. I modified the question. – Laplacian Apr 13 '23 at 13:46
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    The collection $E$ of functions in your problem separates points; hence the algebra it generates (polynomials in $P$) is dense in $C_0(0,\infty)$ (this is a version of the Stone-Weierstrass theorem for locally compact Hausdorff spaces) $C_0$ is dense in $L_p(0,\infty)$ for all $0<p<\infty$. – Mittens Apr 13 '23 at 14:19
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    If $\lim_{T\to\infty}\int_0^T e^{-\alpha x} \tilde f(\alpha) d\mu(\alpha)$ converges for some $x_0$ then it is analytic for $x > x_0$. So no there is often no such $\tilde{f}$. – reuns Apr 13 '23 at 18:23
  • @OliverDíaz Thanks for pointing out that. Your argument is simpler to prove that the linear span of ${e^{-\alpha x}}_{\alpha>0}$ is dense in $L^2((0,\infty)$. Is there way to actually write down the expansion coefficient $\tilde f(\alpha)$? – Laplacian Apr 14 '23 at 00:52
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    @eigenvalue: I think that even the span of ${f_n(x)=e^{-nx}:n\in\mathbb{N}}$ is dense on $L_2$ then you can use the methods of Gram-Schmidt to find an orthonormal basis in $L_2$. – Mittens Apr 14 '23 at 00:57
  • @reuns Thanks for very interesting comment. However, I cannot reconcile your statement with the fact that linear span of ${e^{-\alpha x}}_{\alpha>0}$ is dense. Could you elaborate on this? – Laplacian Apr 14 '23 at 01:23
  • @OliverDíaz Yes, according to the Stone-Weierstrass for locally compact Hausdorff space. By the way, how do you think of reuns's comment that if $f$ is written as $f(x) = \sum_{n\in \mathbb N} \tilde f(n) e^{-nx}$, then it is likely to be analytic? – Laplacian Apr 14 '23 at 01:24
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    @eigenvalue $\sum_{n\ge 1} c_n e^{-nx}$ is analytic for $x > x_0$ if it converges at $x_0$ (in that case it converges in $L^2(x_0,\infty)$ norm). The $e^{-nx}$ are dense because any $L^2(0,\infty)$ function is of the form $\sum_{m \ge 1} \sum_{n=1}^{m} c_{m,n} e^{-nx}$ where the series converges in $L^2$ norm. The latter form is very different to the former. – reuns Apr 14 '23 at 10:43
  • @reuns Can't I simplify the double sum by changing the order as $$\sum_{m\geq 1} \sum_{n=1}^m c_{m,n} e^{-nx} = \sum_{n=1}^\infty \left( \sum_{m=n}^\infty c_{m,n} \right) e^{-nx}?$$ Then it really looks like $\sum_{n\geq 1} \tilde c_n e^{-nx}$, assuming $\left( \sum_{m=n}^\infty c_{m,n} \right)$ is convergent. I cannot figure out why "the latter form is very different to the former". – Laplacian Apr 15 '23 at 01:46

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