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Let $H$ be a Hilbert Space. $S\subseteq H$ be a finite set .Show that $(S^\perp)^\perp=\overline {\operatorname{span} (S)}$ .

Now $\operatorname{span}(S)$ is the smallest set which contains $S$ and $\overline{\operatorname{span}(S)}$ is the smallest closed set containing $S$. Also $S^{\perp\perp}$ is a closed set containing $S$. Thus $\overline{\operatorname{span (S)}}\subset S^{\perp\perp}$ .

How to do the reverse?

Emily
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1 Answers1

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You don't need $S$ to be a finite set. This is true for any subset $S\subseteq \mathscr{H}$.

First of all notice that $S^{\perp}$ is a closed subspace of $\mathscr{H}$ (using continuity of inner product) and $S \subseteq (S^{\perp})^{\perp}$ for any subset $S \subseteq \mathscr{H}$. Thus $Span(S) \subseteq (S^{\perp})^{\perp} \Rightarrow \overline{Span(S)} \subseteq (S^{\perp})^{\perp}$. Further notice that if $C \subseteq D$, then $D^{\perp} \subseteq C^{\perp}$. So, we have \begin{align*} S \subseteq \overline{Span(S)} & & \Longrightarrow & & \left(\overline{Span(S)}\right)^{\perp} \subseteq S^{\perp} & & \Longrightarrow & & (S^{\perp})^{\perp} \subseteq \left(\left(\overline{Span(S)}\right)^{\perp}\right)^{\perp}. \end{align*} But $\left(\left(\overline{Span(S)}\right)^{\perp}\right)^{\perp} = \overline{Span(S)}$ (because if $M$ is a closed subspace of a Hilbert space $\mathscr{H}$, then $(M^{\perp})^{\perp} = M.$) Thus, $(S^{\perp})^{\perp} = \overline{Span(S)}.$

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  • Neat and clean +1 – Learnmore Apr 30 '15 at 02:15
  • If we change $H$ from Hilbert space to any inner product space, then the statement remains true or not? – TRUSKI Jun 21 '18 at 06:14
  • @TRUSKI Then the statement is not true. Remember $M^{{\perp}{\perp}}=M$ is true for every closed subset of Hilbert space. That's why in non-Hilbert case, you will find a counter-example. – Error 404 Jun 26 '18 at 07:03
  • Can you suggest me where I can get a counter example. What you said does not effect the proof stated above as $M^{\perp}$ is always a closed subspace. – TRUSKI Jun 26 '18 at 09:09