Cfrgtkky

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.

Being closed is easy. For each $ain A$, let $f_a(x)=langle a,xrangle$. Then $f_a$ is continuous and there fore $A^perp$ is continuous, since it is equal to$$bigcap_{ain A}f_a^{-1}bigl({0}bigr).$$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.

$A$ 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.

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

Let $(x_n)$ be a sequence in $A^{perp}$ such that $x_nto x$. Thus for all $ain A, langle a,x_nrangle=0$ for all $nin mathbb N$. Since inner product is a continuous function, therefore $langle a,xrangle=limlimits_{nto infty}langle a,x_nrangle=0$ for all $ain A$. Hence $xin A^{perp}$ and so $A^{perp }$ is closed.

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