Prove that in Hilbert spaces $(A^{\perp})^{\perp}=\overline{\text{span}(A)}$.

Question

zzr0ck3r · Answer

Im not quite sure what I need to show here @eliassaab

zzr0ck3r · Answer

So if we are in $\mathbb{R}^2$ and we take some  line through the origin, and we take some vector that is orthogonal to that line and call that vector b and set A={b}, then the orthogonal complement of A is all the vectors that make up the entire line and the orthogonal complement of that is all the linear combinations of b, or in this case, ab where a is in R. Am I thinking of this right?

anonymous · Answer

It is clear
$$
x\in(B^{\perp})^{\perp}\cap B^{\perp}=\{  0 \}\
<x,x>=0=||x||^2\implies x=0
$$

anonymous · Answer

Let G be a closed subspace of H then$$

(G^{\perp})^{\perp}= G
$$
All we need to prove that $$

(G^{\perp})^{\perp}\subset  G
$$
Now
$$
H= G \oplus  G^{\perp} \\text{ Let }  x\in (G^{\perp})^{\perp}\ \
x= s+ t; \quad s\in G, t\in G^{\perp}\
t= x-s \in (G^{\perp})^{\perp} \cap G^{\perp}=\{0\}\
x=s \in G
$$
We are done for that part

anonymous · Answer

Let finish your problem Let$$
 G= \overline{\text{span}(A)} 
$$
It is easy to see that $ G \subset  (A^{\perp})^{\perp}$
$$
A \subset G \
G^{\perp} \subset A^{\perp}\
(A^{\perp})^{\perp}\subset (G^{\perp})^{\perp}=G
$$
We are done.

anonymous · Answer

Let G be a closed subspace of a Hilbert space H, let h be an element in H not in G, then there exist a unique element t in G so that d(h,G)=|| h-t|| By definition one can write that $$ d(h,G)\le ||h -t_n|| \le d(h,G) + \frac 1 n $$ It is clear that |( t_n |) is Cauchy and its limit t satisfies d(h,G)=|| h-t|| Uniqueness. Suppose that there are s and t so that k=d(h,G)=|| h-t||=||h-s|| Write $$ 0\le || s-t||^2 =|| s- h + h-t||^2= 2||s-h||^2 + 2||h-t||^2 - ||s-h -(h-t)||^2=\ 2||s-h||^2 + 2||h-t||^2 - ||s+t -2 h||^2=\ 2||s-h||^2 + 2||h-t||^2 - ||2 \frac{s+t}2 -2 h||^2=\ 2||s-h||^2 + 2||h-t||^2 - 4|| \frac{s+t}2 - h||^2\le\ 2 k^2+2 k^2 -4 k^2=0\Hence\ s=t $$ This element h-t is perpendiculat to every element in G. To see why, write $$ z=h-t\\ g \in G\ \ ||z||^2 = || h-t||^2 \le || h - ( t+ \frac{ g}{||g||^2} ||^2=\ || h - t - \frac{ g}{||g||^2} ||^2=|| z - \frac{ g}{||g||^2} ||^2=\ ||z||^2 -2 \frac{||^2}{||g||^2} +\frac{||^2}{||g||^2}=\ ||z||^2 - \frac{||^2}{||g||^2} \implies =0 $$

anonymous · Answer

It is easy to show now that if G is closed subspace of H then
$$
H=G \oplus G^{\perp}
$$
Try it

zzr0ck3r · Answer

great, ill go play with this and get back to you if I have any questions. thanks again

anonymous · Answer

YW