# Prove that, for a closed subspace $W$ of a vector space $V$, $(W^{\perp})^{\perp} = W$

Prove that, for a closed subspace $W$ of a (infinite dimensional) vector space $V$, (with inner product) $(W^{\perp})^{\perp} = W$.

The notion 'closed' is in relation to the natural metric $d (x,y) = \|x-y\|$ with $\|\cdot\|$ defined as $\sqrt{\langle\cdot,\cdot\rangle}$

This is a basic fact that I have found referenced but not proven. I know that, for all subspaces $U$, $U^{\perp}$ is closed.

Thank you.

• What do you mean by closed subspace? Commented Mar 28, 2016 at 2:24
• @aboat Presumably this is a topological vector space, in which case it means topologically closed. The OP should be more precise about the nature of the ambient space, e.g. whether it's a Banach space or more general. Commented Mar 28, 2016 at 2:47
• Thanks-edited to be more precise (has inner product) Commented Mar 28, 2016 at 3:35
• Commented Aug 11, 2017 at 12:07

I will show that for any subspace $$W$$, $$W^{\perp\perp} = W^-$$. (Closure)
Observe that $$W^{\perp\perp} = \{v \in V\mid \forall w \in W^\perp, \langle v, w \rangle = 0\} = \{v \in V\mid \forall w \in V, (\forall u \in W, \langle u,w \rangle = 0 \implies \langle w,v \rangle = 0)\}$$ Then $$u \parallel v$$, so that $$W \subseteq W^{\perp\perp}$$. Since any orthogonal complement is closed, $$W^{\perp\perp}$$ is closed. Since $$W^-$$ is the smallest closed set containing $$W$$, we have $$W^- \subseteq W^{\perp\perp}$$.
Conversely, let $$w \in W^{\perp\perp}$$. Assume that $$w \not\in W^-$$. By the Hilbert Projection Theorem, ($$W^-$$ is convex) there exists $$w_0 \in W^-$$ and some $$w_1\mid w_1\neq 0, w_1 \perp W^-$$ such that $$w = w_0 + w_1$$. Thus, $$\langle w, w_1 \rangle = \langle w_0 + w_1, w_1 \rangle = \langle w_1, w_1 \rangle > 0$$ hence $$w \not\in W^{\perp\perp}$$. Therefore $$W^- \supseteq W^{\perp\perp}$$.