# Double orthogonal complement of a finite dimensional subspace

Let $\mathbb{W}$ be a finite dimensional subspace of an inner product space $\mathbb{V}$. Prove or disprove the following: the double orthogonal complement of $\mathbb{W}$ is equal to itself, or, in other words, $(\mathbb{W}^\perp)^\perp = \mathbb{W}.$

We might begin a potential proof as follows. If $\vec{v} \in \mathbb{W}$, then $\langle \vec{v}, \vec{w}\rangle = \vec{0}$ for any vector $\vec{w} \in \mathbb{W}^\perp.$ Therefore, by definition, $\vec{v} \in (\mathbb{W}^\perp)^\perp$, and so $\mathbb{W} \subseteq (\mathbb{W}^\perp)^\perp.$

Now, it's easy to prove equality if $\mathbb{V}$ is finite dimensional. For instance, we can use the fact that $\dim \mathbb{W} + \dim \mathbb{W}^\perp = \dim \mathbb{V}$ to show that $\dim \mathbb{W} = \dim (\mathbb{W}^\perp)^\perp$ which implies the desired result. However, in this case, $\mathbb{V}$ is not necessarily finite dimensional and so this step is not valid.

In the same vein, if we were to relax the condition that $\mathbb{W}$ must be finite dimensional, then the statement is false. Let $\mathbb{V}$ be the inner product space of all polynomials with real coefficients under the inner product $$\langle a_0 + a_1x + \dots + a_n x^n, b_0 + b_1x + \dots + b_k x^k\rangle = a_0b_0 + a_1b_1 + \dots + a_jb_j$$ where $j$ is the lesser of $n$ and $k$. Then, the subspace $$\mathbb{W} = \{ f(x) \in \mathbb{V} \mid f(1) = 0\}$$ of $\mathbb{V}$ has orthogonal complement $\{ \vec{0}\}$ which shows that $(\mathbb{W}^\perp)^\perp = \mathbb{V} \neq \mathbb{W}$.

The question, then, is: does there exist a counter example where $\mathbb{V}$ is infinite dimensional, but the subspace $\mathbb{W}$ is not? My intuition tells me yes, but I'm stuck.

Hint: Generally in a Hilbert space $H$ we have that for a linear subspace $W \subset H$ that $$(W^\perp)^\perp = \overline{W}$$ where the bar denotes the closure.
• @Demophilus Since $W$ itself is closed, it must be equal to its closure, correct? Jun 12 '17 at 21:55