1. ## Annihilator of subspace

Dear friends,
can u help me to solve the following problem

Show that Y={x|x=(\lambda_j \in l^2, \lambda_2n=0, n\in N} is closed subspace of l^2 and find annihilator Y*. What is the annihilator Y* if Y=span{\epsilon_1,......\epsilon_n}\subset \subset l^2, where \epsilon_j = (\delta_jk))?

2. Originally Posted by kinkong
Dear friends,
can u help me to solve the following problem

Show that Y={x|x=(\lambda_j \in l^2, \lambda_2n=0, n\in N} is closed subspace of l^2 and find annihilator Y*. What is the annihilator Y* if Y=span{\epsilon_1,......\epsilon_n}\subset \subset l^2, where \epsilon_j = (\delta_jk))?
You have to try friend? The first becomes clear if one notes that $f_n:\ell^2\to\mathbb{R}$ given by $f((\lambda_j))=\lambda_n$ is continuous because then $\displaystyle Y=\bigcap_{n\in\mathbb{N}}\ker f_{2n}$. To find $\text{Ann}(Y)$ note that $\left\{e_{2n}\right\}_{n\in\mathbb{N}}$ is a basis for $Y$ and so some $\varphi\in\text{Hom}\left(\ell^2,\mathbb{R}\right)$ will be identically zero if and only if it annihilates that basis. Thus $\text{Ann}\left(Y\right)=\left\{\varphi\in\text{Ho m}\left(\ell^2,\mathbb{R}\right):\varphi(e_{2n})=0 \text{ for all }n\in\mathbb{N}\right\}$.

3. thanks a lot...i tried it but couldnt get the right way....