# Thread: Question from Hilbert spaces

1. ## Question from Hilbert spaces

$e_m(t)=e^{imt}, f_n=e_{-n} + n e_n (m \in Z, n \in N)$ and $A = \overline{Lin \{ e_n| n \in N\}},
B = \overline{Lin \{ f_n| n \in N\}}$
. (Clousures in $L^2(-\pi, \pi))$. Are A+B closed? What is $\overline{A+B}$ ?

2. Originally Posted by veljko
$e_m(t)=e^{imt}, f_n=e_{-n} + n e_n (m \in Z, n \in N)$ and $A = \overline{Lin \{ e_n| n \in N\}},
B = \overline{Lin \{ f_n| n \in N\}}$
. (Clousures in $L^2(-\pi, \pi))$. Are A+B closed? What is $\overline{A+B}$ ?
If you include 0 in the natural numbers then A+B will contain all the trigonometric polynomials and therefore $\overline{A+B}= L^2(-\pi, \pi)$. If you do not allow 0 as a natural number then $\overline{A+B}= \{x\in L^2(-\pi, \pi): \int_{-\pi}^\pi\!\!\!x(t)\,dt=0\}$ (the orthogonal complement of the constant functions in $L^2(-\pi, \pi)$).

To see that A+B is not closed, you only need to find some element $x\in L^2(-\pi, \pi)$ such that $x\notin A+B$. For that, notice that every x in A+B must be of the form $\sum_{m\in\mathbb{Z}}x_me_m$ with $\sum_{n=1}^\infty n^2|x_{-n}|^2 < \infty$ (and of course $\sum_{m\in\mathbb{Z}}|x_m|^2<\infty$ to ensure that $x\in L^2(-\pi, \pi)$). So for example $x = \sum_{n=1}^\infty \frac{e_{-n}}n \notin A+B$.