Answer to Question #163049 in Functional Analysis for Ebenezer

Question

Let H be a Hilbert space.

(i) Let S ⊆ H be any non-empty subset of S. Show that S ^⊥ is a subspace of H.

(ii) Let L ⊆ H be a linear manifold. Show that L^ ⊥⊥ = L

Answer 1

1. Let S ⊆ H be any non-empty subset of S. $S^\perp=\{x\in H: \forall s\in S \langle x,s\rangle=0\}$

Let $x,y\in S^\perp$. Then for any $s\in S$ we have $\langle x+y,s\rangle=\langle x,s\rangle+\langle y,s\rangle=0+0=0$ and for any scalar $\lambda$ we have $\langle\lambda x,s\rangle=\lambda\langle x,s\rangle=0$. Therefore, $S^\perp$ is a linear.

If $\{x_n\}_{n=1}^{+\infty}\subset S^\perp$ converges to $x\in H$, then for any $s\in S$ we have $\langle x,s\rangle=\lim_{n\to+\infty}\langle x_n,s\rangle=0$.

This implies that $S^\perp$ is a closed subspace of H.

2. Let L ⊆ H be a linear manifold (i.e. closed linear subspace) of H. For all $x\in L$ and $y\in L^\perp$ we have $\langle x,y\rangle=0$, therefore, by the definition, $L\subset L^{\perp\perp}$.

If $x\notin L$ then let $x_L$ be an orthogonal projection of to L. It can be obtained as follows.

L is a closed linear subspace of H, therefore L is a Hilbert space. Let $\{v_1,v_2,...\}$ - the orthonormal basis of L, $c_n=\langle x, v_n\rangle$, $x_n=\sum_{k=0}^{n}c_kv_k$. Then if k<n then $\langle x_n, v_k\rangle=\langle x, v_k\rangle$ and $\langle x-x_n, v_k\rangle=0$. From the Pythagorean theorem $|x|^2=|(x-x_n)+x_n|^2=|x-x_n|^2+|x_n|^2$, from where $|x_n|^2\leq |x|^2$ . From the other side, $|x_n|^2=\sum_{k=0}^{n}|c_k|^2$ , therefore the partial sums of the series $\sum_{k=0}^{+\infty}|c_k|^2$ are bounded and the sequence $x_n$ is fundamental in L, hence, it converges to some vector $x_L\in L$. Since $\langle x-x_n, v_k\rangle=0$ , then $\langle x-x_L, v_k\rangle=\lim_{n\to+\infty}\langle x-x_n, v_k\rangle=0$ for any k. Then $x-x_L\in L^\perp$ and $\langle x-x_L,x\rangle=\langle x-x_L,x\rangle-\langle x-x_L,x_L\rangle=\langle x-x_L,x-x_L\rangle=|x-x_L|^2>0$

This shows that $x\notin L^{\perp\perp}$ and therefore, $L=L^{\perp\perp}$.