TITLE: Does this sequence span $L^2$?
QUESTION [13 upvotes]: Consider the following sequence of functions in $L^2[0,\infty)$:
$$f_n(x)=e^{-x/n}x^n,\;\;n\geq 1$$
Does this sequence span $L^2[0,\infty)$ (that is, is the set of finite linear combinations
of these functions dense)?
(My guess is that it doesn't).

REPLY [2 votes]: This is a very interesting problem.  An intermediate generalization between that stated problem and that proposed by Zen Harper is to look at the sequence
$$f_n(x) = x^n e^{-a_n x}, \quad n=1,\ldots$$
for some given sequence $a_n$ of positive numbers.  As in Zen Harper's post, the Laplace transform of $f_n$ is the function $F_n(z)= (-1)^n \frac{1}{(z+a_n)^n}$, defined for $z$ in the right half plan $\{ \mathrm{Re}z >0\}$, and a function $f$ is orthogonal to the span of $\{f_n\}$ if and only if its Laplace transform $F$ satisfies
$$ F^{(n)}(a_n)=0.$$
Thus, if $a_n$ is a constant sequence then the span of this sequence is dense, since anything orthogonal to the span would vanish to infinite order at a point.
On the other hand if $\sum_{n=1}^\infty n \frac{a_n}{(1+a_n)^2} <\infty$ then using a bit of complex analysis one can find a non-zero function $g$ orthogonal to the span of $\{f_n\}$.  Specifically, we form the Blaschke product $B(z)$ with $n$ zeros at $a_n$ -- this is an analytic function in the right half plane, everywhere bounded by one and with zero of order $n$ at $a_n$ for each $n$.  So $B(z)$ satisfies $B^{(n)}(a_n)=0$ and in fact
$$B^{(k)}(a_n)=0,\quad k=1,\ldots,n.$$
Multiplying $B$ by an non-zero $H^2$ function $F$ produces a function $BF \in H^2$ and perpendicular to the given sequence.  Taking the inverse Laplace transform gives the desired function $g$.
But the original problem has $n a_n/(1+a_n)^2\rightarrow 1$ so this does not produce a counterexample.