Does this sequence span $L^2$?

Question

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).

Answer 1

Not an answer, but maybe helpful.

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 $n$.

Answer 2

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?

Answer 3

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.

Answer 4

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.

Answer 5

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...