Can I form a complete set of functions using $e^{-nx}$? If I start with the set of functions $e^{-nx}$ for all integers $n>1$, can I use them as basis to create a complete set of orthogonal functions on the interval $(0,+\infty)$?  By complete I mean that they can be used in an expansion that will converge in the mean to any square integrable function on that domain.
I'm not saying that they are a good basis, of course.  They are possibly the worst...
 A: Yes, this set is complete. There's a pretty easy change of variable argument.
We need to show that if $f \in L^2[0,\infty)$ is orthogonal to $e^{-nx}$ for all $n > 1$ then $f = 0$ almost everywhere.
Given that $\int_0^\infty f(x)e^{-nx}\, dx = 0$ for $n > 1$, use the change of variable $u = e^{-x}$ to get $\int_0^1 f(-\ln u) u^n\, du = 0$ for $n > 0$. Applying the same change of variable to the condition $\int_0^\infty |f(x)|^2\, dx < \infty$ yields $\int_0^1 |f(-\ln u)|^2\frac{1}{u}\, du < \infty$, i.e., the function $\frac{f(-\ln u)}{\sqrt{u}}$ belongs to $L^2[0,1]$, and hence so does $f(-\ln u)$. Thus $f(-\ln u) \in L^2[0,1]$ is orthogonal to every polynomial with no constant term. But those polynomials are uniformly dense in the continuous functions which vanish at $0$, and the latter are dense in $L^2[0,1]$, so the fact that $f(-\ln u)$ is orthogonal to them implies $f(x) = f(-\ln u) = 0$ almost everywhere, as desired.
A very similar argument answers the question you asked in the comments about the functions $x^{-n}$.
