I'm doing an exercise in $H=L^2(\mathbb R^+, \mu)$, where $\mu$ is the measure with density $e^{-x}$. In the previous questions I showed that $L_n$ (the Laguerre polynomials) are orthonormal and that the functions $e^{-\alpha x}$ are in the closure of their span. The next question is this:
Let $F$ be the set of smooth and compactly supported functions on $\mathbb R^+$. Then prove that span of the functions $e^{-nx}$ is dense in $F$ under the uniform norm $\lVert\cdot\rVert_\infty$.
I think I can get this through Stone-Weierstrass, but then the final question is:
Show that $(L_n)$ is a Hilbert basis in $H$.
I don't see what that has to do with the previous question, which was about a totally different space of functions, with a totally different norm. Is there some density result I'm not aware of that I should be using?
You were given that $d\mu = e^{-x}dx$. Let $y=e^{-x}$. Then $$ \int_{0}^{\infty}e^{-nx}f(x)d\mu(x) = \int_{0}^{1}y^{n}f(-\ln y)dy. $$ The function $f \in L^2_{\mu}[0,\infty)$ gives $g(y)=f(-\ln(y))$ in $L^2[0,1]$ because $$ \int_{0}^{\infty}|f(x)|^2e^{-x}dx=\int_{0}^{1}|f(-\ln(y))|^2dy. $$ You can now invoke standard results for $L^2[0,1]$ to get what you want.