Completeness of the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ in $L^2[0,1]$ In the book R. M. Young, An introduction to non-harmonic Fourier series, I came across the following problem (page 18):
Problem. Show that the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ is complete in $L^2[0,1]$.
I tried to apply the Müntz-Szasz theorem but it didn't work out. Any ideas or approaches how to show completeness here? Thanks!
 A: This is a more complicate proof which works also on $(0,\infty)$. Write
$$
F(\lambda)=\int_0^\infty \frac{f(x)}{x+\lambda}\, dx=\int_0^\infty f(x)\, dx \int_0^\infty e^{-(x+\lambda)y}\, dy=\int_0^\infty e^{-\lambda y} dy \int_0^\infty f(x)e^{-xy}dx
$$
so that $F$ is the Laplace transform of $\int_0^\infty f(x)e^{-xy}dx$. By assumption $F(n)=0$ and then $F=0$ since the integers are a uniqueness set for the Laplace transform. But then, by uniqueness again, $\int_0^\infty f(x)e^{-xy}dx=0$ for $y>0$  and, by the same argument, $f=0$.
A: This is just a comment on the present responses which I am adding in the hope that it will be of interest--it will be too long for that format.  Firstly, it treats  completeness for more general functions of the form $$\frac1{x+\lambda_n}$$ where $(\lambda_n)$ is a sequence of positive numbers (even complex ones with positive real parts) and displays the connection with the Müntz-Szasz theorem.
It also uses the Hahn Banach theorem to connect  with uniqueness theorems for spaces of analytic functions as is done  above.
We begin by replacing the $L^2$-space with $C([0,1)$--anything complete in the latter remains so in the former.  We now consider for any measure $\mu$  the function $$f(z)=\int_0^1 \frac{d\mu(x)}{x+z}.$$  This is bounded and analytic in the right half plane.  We now use the fact that a sequence $(\lambda_n)$ there is the zero set of such a function if and only if $$\sum\left(1-\left|\frac{\lambda_n-1}{\lambda_n+1}\right|\right)<\infty.$$
This suffices to prove your result, together with a wide-ranging set of generalisations.
The same proof, with the definition
$$f(z)=\int_0^1 x^zd\mu(x),$$
can be used to prove the classic Müntz-Szasz theorem (which assumes that the sequence increases to infinity along the real line), but also variants which allow it to converge in more exotic manners to the boundary of the half-plane.
A: Let $S$ be the Hilbert subspace of $L^2[0,1]$ spanned by the sequence. For every positive integer $n$, the function $\frac{n}{x+n}=\frac{1}{1+x/n}$ lies in $S$. For $n$ large, this function is very close to the constant $1$ function in sup-norm, hence $1\in S$. Now we prove by induction that, for every nonnegative integer $m$, the functions $x^m$ and $\frac{x^m}{x+n}$ ($n=1,2,\dotsc$) lie in $S$. We have already established the base case $m=0$. So let us assume that $m>0$ and the functions $x^{m-1}$ and $\frac{x^{m-1}}{x+n}$ ($n=1,2,\dotsc$) lie in $S$. Then $x^{m-1}-\frac{nx^{m-1}}{x+n}=\frac{x^m}{x+n}$ lies in $S$. Moreover, $\frac{nx^m}{x+n}=\frac{x^m}{1+x/n}$ lies in $S$ as well, so by the same approximation argument as before, $x^m\in S$.
It follows that every polynomial lies in $S$, therefore $S=L^2[0,1]$ by the Weierstrass approximation theorem.
