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    $E^\perp$ is a closed subspace of a Hilbert space and $(E^\perp)^\perp=\overline{E}$

    P-THQJ3

    Area
    Real Analysis
    Topics
    Closure, Hilbert Spaces
    Status
    draft

    Let \(E \subseteq H\) a Hilbert space.

    1. Show that \(E\perp \subseteq H\) is a closed subspace.

    2. Show that \((E^\perp)^\perp = \cl_H(E)\).

    Solution

    1. \(E^\perp\) is a subspace of \(H\).

    \(E^\perp = \{x \in H : \langle x, e\rangle = 0 \ \forall e \in E\}\).

    2. \(E^\perp\) is closed.

    if \(x_n \in E^\perp\) and \(x_n \to x\), then for each \(e \in E\), \(\langle x, e\rangle = \lim_n \langle x_n, e\rangle = 0\) (continuity of the inner product), so \(x \in E^\perp\).

    3. \(E \subseteq (E^\perp)^\perp\), hence \(\cl_H(E) \subseteq (E^\perp)^\perp\).

    for \(e \in E\) and \(x \in E^\perp\), \(\langle e, x\rangle = \overline{\langle x, e\rangle} = 0\), so \(e \in (E^\perp)^\perp\).

    4. \((E^\perp)^\perp \subseteq \cl_H(E)\).

    let \(M = \cl_H(E)\), a closed subspace.

    5. Q.E.D.

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