Question #163049

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


1
Expert's answer
2021-02-17T02:54:46-0500

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

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

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

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


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

If xLx\notin L then let xLx_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 {v1,v2,...}\{v_1,v_2,...\} - the orthonormal basis of L, cn=x,vnc_n=\langle x, v_n\rangle, xn=k=0nckvkx_n=\sum_{k=0}^{n}c_kv_k. Then if k<n then xn,vk=x,vk\langle x_n, v_k\rangle=\langle x, v_k\rangle and xxn,vk=0\langle x-x_n, v_k\rangle=0. From the Pythagorean theorem x2=(xxn)+xn2=xxn2+xn2|x|^2=|(x-x_n)+x_n|^2=|x-x_n|^2+|x_n|^2, from where xn2x2|x_n|^2\leq |x|^2 . From the other side, xn2=k=0nck2|x_n|^2=\sum_{k=0}^{n}|c_k|^2 , therefore the partial sums of the series k=0+ck2\sum_{k=0}^{+\infty}|c_k|^2 are bounded and the sequence xnx_n is fundamental in L, hence, it converges to some vector xLLx_L\in L. Since xxn,vk=0\langle x-x_n, v_k\rangle=0 , then xxL,vk=limn+xxn,vk=0\langle x-x_L, v_k\rangle=\lim_{n\to+\infty}\langle x-x_n, v_k\rangle=0 for any k. Then xxLLx-x_L\in L^\perp and xxL,x=xxL,xxxL,xL=xxL,xxL=xxL2>0\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 xLx\notin L^{\perp\perp} and therefore, L=LL=L^{\perp\perp}.


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