Closure of the orthogonal complement.

Question

Let H be a Hilbert space and let $A \subset H$. Let the orthogonal complement of A be:

$A^\perp$ = {$x \in H : x \perp A$}.

How do I show that $A^\perp$ is a vector space and that it is closed? I thought I could go through the axioms of a vector space but I'm just wondering if it's a consequence of the fact that it is a subspace of a Hilbert space. Much appreciated.

Answer 1

$A$ is a subset of $H$, so to verify it is a subspace you must show:

1. $x,y \in A^\perp$ implies $x+y \in A$,
2. $x \in A^\perp$ and $\alpha \in \mathbb R$ implies $\alpha x \in A$,
3. if $\{x_k\} \subset A$ and $x_k \to x \in H$, then $x \in A$.

The first two are immediate, and the third follows from the continuity of the inner product.

Answer 2

Being closed is easy. For each $a\in A$, let $f_a(x)=\langle a,x\rangle$. Then $f_a$ is continuous and therefore $A^\perp$ is continuous, since it is equal to$$\bigcap_{a\in A}f_a^{-1}\bigl(\{0\}\bigr).$$For the rest (that is, being a subspace), all it takes is to prove that it is not empty and that it is closed under sums and under products by scalars.

Answer 3

Consider $f_a(x)=<x,a>, A^\perp=\cap\{a\in A Kerf_a\}$, so it is vector space and closed because the intersection of closed vector spaces is a closed vector space.

Answer 4

Let $(x_n)$ be a sequence in $A^{\perp}$ such that $x_n\to x$. Thus for all $a\in A, \langle a,x_n\rangle=0$ for all $n\in \mathbb N$. Since inner product is a continuous function, therefore $\langle a,x\rangle=\lim\limits_{n\to \infty}\langle a,x_n\rangle=0$ for all $a\in A$. Hence $x\in A^{\perp}$ and so $A^{\perp }$ is closed.

Answer 5

Show for $x,y\in A^\perp$ and $r\in F$ then $x+ry\in A^\perp$. This means $<x,z>=0$ and $<y,z>=0$ so $<x+ry,z>=0$ for all $z\in A^\perp$. Therefore $x+ry\in A^\perp$. Also if $\{x_n\}$ be a sequence with elements in $A^\perp$, which is converges to $x$ then $x\in A^\perp$, since for all $z\in A^\perp$, $<x_n,z>=0$. then $0=\lim <x_n,z>=<\lim x_n,z>=<x,z>$.