anihilator closure and continuity

Question

Let H be a Hilbert spcae and $f:H \rightarrow \mathbb{C}^n$ lineal such that $\overline{N(f)}=N(f)$, where N(f) is the anihilator of f. Is f continuous? n>1

Answer 1

Since $N(f)$ is closed, $H=U\oplus N(f)$ where $U$ is the finite dimensional orthogonal complement of $N(f)$. Write $U=Vect(e_1,..,e_n)$ with $\|e_i\|=1$, and $f(x)=<x,e_i>$, $f_i$ is continuous. Let $a_i=f(e_i)$, consider $g=a_1f_1+...+a_nf_n$, for every $x\in H, x=u+v, u\in U, v\in N(f)$, write $u=u_1e_1+..u_ne_n$, $f(x)=f(u+v)=f(u)=f(u_1e_1+..+u_ne_n)=$

$=u_1a_1+..u_na_n=a_1<x,e_1>+..+a_n<x,e_n>=a_1f_1(x)+..+a_nf_n(x)=g(x)$ implies that $f=g$ and $f$ is continuous since $g$ is continuous.