# Does this sequence span $L^2$?

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).

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Harald Hanche-Olsen and coudy, thank you for your comments. I've been too hasty. – Andrey Rekalo Jun 10 '10 at 20:41
A better term than "span" might be to ask if the sequence is total. – Nate Eldredge Jun 10 '10 at 20:42
@Guy Katriel: could you possibly comment on why you are interested in the sequence? Where does it come from? – Andrey Rekalo Jun 10 '10 at 21:21
@Andrey: In conversation with a physicist, I was told that someone had used such a sequence for numerically approximating eigenfunctions of a radial Schroedinger operator, and obtained bad results. The explanation offered was that it does not form a basis. I therefore wonder how to show this (if its true). – Guy Katriel Jun 10 '10 at 21:31
@Guy: Thanks for the comment. – Andrey Rekalo Jun 10 '10 at 21:34

I cannot answer your question fully but for numerical computations this 'basis' is not usable since it is not stable. A sequence $(f_n)$ of functions is stable if $$\|g\|_2^2 \lesssim \sum_n |\langle f_n, g\rangle |^2\quad \mbox{for all }g\in L_2.$$

Consider your system and set $g=\chi_{[0,\lambda]}$, the indicator function of the interval $[0,\lambda]$. Then $$\sum_n |\langle f_n, g\rangle |^2 \le \|g\|_2^2\sum_n\int_0^\lambda x^{2n}=\|g\|_2^2\sum_n\frac{\lambda^{2n+1}}{2n+1}\to 0 \mbox{ for }\lambda \to 0.$$ This implies instability. It is interesting that in this argument the problems occur near zero. I would like to know if problems occur also near infinity?

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But if $\lambda \to 0$, does the norm of $g$ not also approach 0? Can you clarify what you mean by "this implies stability"? – Willie Wong Jun 11 '10 at 13:57
... I meant, of course, what you mean by "this implies instability"? (Sorry about the typo.) – Willie Wong Jun 11 '10 at 14:00
Well, it is was stable'' then by definition, there should be a constant $c>0$ with $\sum_n |\langle f_n,g\rangle|^2 \geq c \|g\|^2_2$, which Philipp has shown can't happen for this sequence $(f_n)$. – Matthew Daws Jun 11 '10 at 15:37
I am not familiar with the notion of stability, but don't you need to assume that $\|f_n\|=1$ in the inequality that defines it? (Presumably, it's invariant under rescalings.) Also, what happened to the exponential factor in the second displayed formula? – Victor Protsak Jun 13 '10 at 12:18

These questions can be much harder than they look. There is a simple-looking, explicit set of functions $\{f_\alpha \} \subset L^2(0,1; dx/x)$ (see Operators, Functions, and Systems: an Easy Reading: Volume 1 by Nikolai K. Nikolski) with the property: the Riemann Hypothesis is equivalent to the density of the span of $\{f_\alpha \}$.

Of course it doesn't mean your problem is so difficult; but it's certainly interesting and non-trivial (I think). The necessary results should be known and exist somewhere [UPDATE: I've changed my mind, maybe not!!] I haven't found anything yet; but the general problem I describe below is certainly very "natural" and I'm sure I'm not the first to think of it, so if it isn't known then it's an interesting open problem; I would be very interested to know the solution!

The Laplace transform gives (a multiple of) a unitary operator from $L^2(0, \infty)$ onto the Hardy space $H^2( \{ \mathrm{Re}(z)>0 \})$ (depending on your normalisations).

The Laplace transform of $x^{n} e^{-x/n}$ is, up to a constant, $(z+1/n)^{-(n+1)}$.

So you're asking whether the orthogonal complement of these functions is zero in $H^2$. The scalar product of $F(z)$ with $(z+\overline{\lambda})^{-k-1}$ is, up to a constant, the derivative $F^{(k)}(\lambda)$.

Thus (assuming my algebra is correct), your question is equivalent to asking whether there is any non--trivial $F \in H^2$ satisfying

$$F^{(n)}(1/n) = 0, \qquad n=1,2,3,\ldots$$

If you just wanted some non-trivial analytic function $F$ on the half-plane $\{ x+iy : x>0 \}$ to satisfy this, it's possible (I think, if I remember correctly!) - at each point of a countable set without limit point in the domain, we can prescribe values of finitely many derivatives. The extra condition $F \in H^2$ is the difficult part.

More generally we have:

Problem classify all sequences $(z_n)$, $(k_n)$ such that we have a uniqueness result: $$F \in H^2, \quad F^{(k_n)}(z_n) =0 \quad (n=1,2,3,\ldots) \qquad \Rightarrow \qquad F \equiv 0.$$

Of course there might be a special trick for your particular case $z_n = 1/n$, $k_n = n$.

Special cases are well-known. For example: there is no non-trivial $G \in H^2$ satisfying $G(z_n)=0$ if and only if $\sum_n \frac{\mathrm{Re}(z_n)}{|1+z_n|^2} = +\infty$ (the Blaschke condition). e.g. consider $z_n$ converging to zero, or out to infinity; if it does this so quickly that the sum is finite, then the set $\{ z_n \}$ is sparse enough to allow non-trivial $G$. Thus the classification is known if $k_n = 0$ for all $n$.

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Thank you. The connection with complex analysis is very interesting, and I am indeed interested in a general approach to such problems, not only in the specific question. – Guy Katriel Jun 11 '10 at 9:54

My two cents. I would guess that $f_0(x)=exp(-x)$ is not in the vector space generated by the f_n because of its behavior at zero. There is a way to get some confidence at least numerically.

The distance from f to $F_n=vect(f_1,...,f_n)$ can be computed with a Gram determinant. If I remember well,

$d(f_0, F_n)^2 = det((\langle f_i,f_j\rangle )_{0..n})/det((\langle f_i,f_j\rangle)_{1..n})$

Everything is explicit, but there are factorials everywhere, so it is tedious to make it by hand. So I suggest to plug the formula into a computer algebra software (which I can't access at the moment) to get the result. This may hint at the solution.

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@Wadim: what weak limit do you mean? $e^{-x/m}$ is unbounded in $L^2$ and so not weakly convergent. (And if you renormalize it, the weak limit is 0.) – Nate Eldredge Jun 10 '10 at 23:23
Thanks coudy. I tried implementing your suggestion (on MAPLE). The distance from $e^{-x}$ to $F_n$ do seem to decrease very slowly with $n$: for $n=10$ it is $0.209$, for $n=40$ it is $0.201$ (I can't do much higher $n$ because the determinant computation takes a long time). So this at least suggests that $e^{-x}$ is not in the span. – Guy Katriel Jun 11 '10 at 8:49

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.

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Thank you Jeff, that's very interesting. So one could conjecture that $\sum_{n=1}^\infty n \frac{a_n}{(1+a_n)^2} =\infty$ is a necessary and sufficient condition for the sequence $f_n$ as you defined to span $L^2[0,\infty)$. The that $a_n$ is constant is the most famous one, with the Laguerre functions used to solve the Hydrogen atom. So you gave a very nice proof of the completeness of the Lagueree functions. – Guy Katriel Jun 13 '10 at 11:28
I hesitate to make the conjecture because when $\sum n \frac{a_n}{(1+a_n)^2} <\infty$, we get more than that the span of your sequence has dense range. In fact, the span of the larger collection $f_{n,k}=x^k e^{-a_n x}$ where $1\le k\le n<\infty$ fails to be dense. However that could focus your problem a bit: if you can show that $x^k e^{-a_n x}$ for $k<n$ is in the closure of the span of your functions then you should be able to conclude that this condition is necessary and sufficient. – Jeff Schenker Jun 13 '10 at 16:14

Let me try again (I am surprised to see no moves around the question since my morning): assume that the above functions $f_n(x)$ form a basis of $L_2[0,+\infty)$. Pick an arbitrary function $f(x)$ from $L_2[0,+\infty)$ which is continuous on $[0,+\infty)$. Writing it as $f(x)=\sum_{n=1}^\infty c_nf_n(x)$ and using the continuity of all the functions involved we conclude that $\sum_{n=1}^\infty c_nf_n(x)$ converges to $f(x)$ pointwise on $[0,+\infty)$. But then, restricted to a compact set --- say $[0,1]$, the convergence of $\sum_{n=1}^\infty c_nf_n(x)$ to $f(x)$ is uniform. Since all the functions $f_n(x)$ vanish at $x=0$, so does their uniform limit $f(x)$. On the other hand, there are plenty of continuous functions $f(x)$ in $L_2[0,+\infty)$ which do not vanish at $x=0$, a contradiction.

Summarising, any continuous function from $L_2[0,+\infty)$ that does not vanish at the origin cannot belong to the span of $f_n$'s. In particular, $e^{-x}$ does the job.

By the way, the functions $g_n(x)=e^{-x/n}x^{n-1}$, $n=1,2,\dots$, seem to span $L_2[0,+\infty)$. But I have no idea on how to show this, except the standard orthogonolisation...

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How does the $L^2$-convergence of the sum imply the pointwise convergence? – Philipp Jun 11 '10 at 9:06
Well, assuming that the derivative of a function $f(x)$ exists and is square integrable implies the absolute convergence of the Fourier series... This is specific for standard bases however. But if we pose the condition on $f(x)$ to be analytic on the whole $\mathbb R$, then everything will be absolutely and uniformly convergent (example $e^{-x^2}$). I don't know why do I answer the question; after all these I find it boring. The other answers give hints and nobody complains, while I am immediately asked to give more details. It's not a solution but a hint! – Wadim Zudilin Jun 11 '10 at 9:46
The above argument would seem to imply, for example, that the sequence $x,x^2,x^3,...$ does not span $L^2[0,1]$ (since all these functions vanish at $0$) - but in fact it does, by simple arguments. The problem is, as pointed out by Philipp, the implication from $L^2$ to pointwise convergence. – Guy Katriel Jun 11 '10 at 9:48
An obvious point... but the original question certainly does not ask that (f_n) forms an orthonormal basis: only if the linear span is dense. – Matthew Daws Jun 11 '10 at 9:55
Wadim, I did not want to criticize your argument but understand it better. This was not meant as an attack or anything, so no reason to be angry. You can also see the asking of questions positively: I asked because I found your answer interesting! – Philipp Jun 11 '10 at 10:04