Set of finite linear combination is dense in $L^2(0,\infty)$

Question

This exercise is from Rudin's Real and Complex analysis at chapter 15, problem 22.

$A=\{ f_{n} = t^n exp(-t)\}$ where $(0 \leq t \leq \infty, n \in \mathbb{N}).$ Prove that $A$ is dense in $L^{2}(0,\infty)$.

Hint : If $g \in L^{2}(0,\infty)$ is orthogonal to each $f_n$ and if $F(z) = \int_{0}^{\infty}exp(-tz)\overline{g(t)}dt,$ (Re $z>0$), then all derivatives of $F$ are $0$ at $z=1$. Consider $F(1+yi)$.

I understand the hint want me to use Hanh-Banach theorem, but I don't know how the $F(z)$ is related with the theorem. Even if I accept $F'(1)=0$, what can I do that? Could you give me more hints to approach this problem?

Edit: Thank you for all of you. I'll give a try for solving problem.

Let $g \in L^{2}(0,\infty)$ is orthogonal to each $f_n,$ e.g. $\int_{0}^{\infty}t^{n}e^{-t}\overline{g(t)}dt = 0$ for all $n=0,1,2, \cdots$. Define $F(z) = \int_{0}^{\infty}\exp(-tz)\overline{g(t)}dt$ where Re $z >0$.

1) F is well-defined since $(\int_{0}^{\infty}|exp(-tz)|^{2}dt)^{1/2} = \frac{1}{\sqrt{2 Re z}} <\infty$, so $e^{-tz}$, $g(t)$ are in $L^{2}$.

2) $F^{(n)}(w) = \int_{0}^{\infty}(-1)^{n}t^n e^{-tz} \overline{g(t)}dt$

since $\lim_{z \to w}\frac{F(z) - F(w)}{z-w} = \lim_{z \to w} \int_{0}^{\infty}\frac{e^{-tz}-e^{-tw}}{z-w}\overline{g(t)}dt $

and this part is what I do not know how to deal with, but I think I can apply dominated convergence theorem or any other theorem, to derive that

$\lim_{z \to w} \int_{0}^{\infty}\frac{e^{-tz}-e^{-tw}}{z-w}\overline{g(t)}dt = \int_{0}^{\infty}\lim_{z \to w} \frac{e^{-tz}-e^{-tw}}{z-w}\overline{g(t)}dt= \int_{0}^{\infty}-te^{-tz}\overline{g(t)}dt$, and by repeating such differentiating, we get desired result.

3) $F^{(n)}(1) = 0$ for any $n \in \mathbb{N}\cup \{0\}$ since $F^{(n)}(1)$ is just $<f_{n},g>$, which is 0 by assumption.

4) So $F$ is holomorphic on half plane where real part is greater than 0, and $F$'s power series representation at $z=1$ implies $F \cong 0$.

So, if we think $0= F(1+iy)$ for $y\in \mathbb{R}$, $F(i+y) = \int_{0}^{\infty}e^{-t}\overline{g(t)}e^{-iyt}dt$, a Fourier transform of $e^{-t}\overline{g(t)}$, call $\mathbb{F}(e^{-t}\overline{g(t)}$. Hence by fourier inversion formula, $e^{-t}\overline{g(t)} = \int_{0}^{\infty}\mathbb{F}(e^{-t}\overline{g(t)}e^{iyt} = \int 0 dt =0$. So $e^{-t}\overline{g(t)} \cong 0$. But $e^{-t}\neq 0$ for any $t \in (0,\infty)$, $g(t) = 0$ almost everywhere.

5) Conclusion : so any function which is not spanned by closure of span$\{f_n \}$ is zero almost everywhere. Hence the span is dense.

Re-question : You know that my proof was not completed, especially on interchanging limit and integral when I calculate $F^{(n)}$. Could you give some more idea?

Answer 1

Not only the first derivative, but also the evaluation and all higher derivatives are zero: $F(1)=F'(1)=F''(1)=\ldots=0.$ This implies that $F(z)$ is identically zero on its domain.

Now the last part of the hint invites you to write out $F(1+yi)$. For $y$ running through $\mathbb R$ this is actually a Fourier transform and by Fourier inversion $g$ must be zero almost everywhere on the positive reals.

Answer 2

An alternative approach. If we define Laguerre polynomials through the Rodrigues formula: $$ L_n(x) = \frac{e^x}{n!}\frac{d^n}{dx^n}\left(x^n e^{-x}\right) $$ we have: $$ \int_{0}^{+\infty} L_n(x)L_m(x)e^{-x}\,dx = \delta(n,m) $$ by integration by parts, hence Laguerre polynomials give an orthogonal base of $L^2(0,+\infty)$ with the inner product $\langle f,g\rangle = \int_{0}^{+\infty}f(x)g(x)e^{-x}\,dx$. The leading term of $L_n(x)$ is just $\frac{(-1)^n x^n}{n!}$ and the previous base is a complete base. Rudin's claim easily follows.

Answer 3

By convergence dominated theorem, the limit and integral commutes. Hint: you may use the following inequality to dominate the function sequence: \begin{equation*} |\exp(z)-z-1|\leq\frac{|z|^2\exp(|z|)}{2},\quad z\in \mathbb{C}. \end{equation*}