Double orthogonal complement of a finite dimensional subspace w.r.t. a nondegenerate symmetric bilinear form on an infinite dimensional space

Let $$U$$ be a finite-dimensional subspace of an infinite-dimensional space $$V$$ equipped with a nondegenerate symmetric bilinear form $$\phi$$. Is it necessarily true that $${(U^{\perp})}^{\perp}=U$$?

If the bilinear form happens to be an inner product, then the statement is true (see here). If $$V$$ is finite dimensional, the statement follows from dimension count. I believe the statement as formulated is false. Anyone knows a counterexample? What if I impose the extra condition that the restriction of $$\phi$$ to $$U$$ is nondegenerate?

Yes, it is necessarily true

First, note that $$U \subseteq U^{\perp\perp}$$ will always hold, regardless of degeneracy. Indeed: by the definition of $$U^{\perp}$$, $$(x,y) = 0$$ holds for all $$x \in U$$ and $$y \in U^\perp$$. Thus, if $$x \in U$$, then $$(x,y) = 0$$ holds for all $$y \in U^\perp$$. Thus, $$x$$ is an element of $$(U^\perp)^\perp$$.

The more complicated part is showing that $$U^{\perp\perp} \subseteq U$$.

Let $$d = \dim(U)$$. Let $$\{x_1,\dots,x_d\}$$ be a basis of $$U$$. Consider the map $$\phi:V \to \Bbb F^d$$ given by $$\phi(v) = [(x_1,v),\dots,(x_d,v)].$$ Notably, $$\ker(\phi) = U^\perp$$. $$\phi$$ is a surjective (onto) map. Indeed, suppose for the purpose of contradiction that $$\phi$$ fails to be surjective. It would follow that there exists there exists a non-zero vector $$c = (c_1,\dots,c_d)$$ in $$\operatorname{im}(\phi)^\perp$$, where the orthogonal complement is taken relative to the "dot-product" over $$\Bbb F^d$$ (i.e. $$c,d \mapsto c^Td$$). For this $$c$$, it holds that for all $$v \in V$$, $$c^T\phi(v) = 0 \implies\\ c_1(x_1,v) + \cdots + c_d(x_d,v) = 0 \implies\\ (c_1 x_1 + \cdots + c_d x_d,v) = 0.$$ By the non-degeneracy of the bilinear form, this implies that $$c_1 x_1 + \cdots + c_d x_d = 0$$. Because $$\{x_1,\dots,x_d\}$$ is a basis, this implies that $$c = 0$$, contradicting our statement that $$c$$ is non-zero.

This allows us to deduce that the quotient space $$V/\ker(\phi) = V/U^{\perp}$$ has dimension $$d$$, and that $$\phi$$ induces an isomorphism $$\bar \phi:(V/U^\perp) \to \Bbb F^d$$ via the quotient map.

Now, suppose that $$x \in U^{\perp\perp}$$. Consider the linear map $$\alpha:V \to \Bbb F$$ given by $$\alpha(v) = (x,v)$$. Because $$U^\perp \subseteq \ker(\alpha)$$, $$\alpha$$ induces a linear map $$\bar \alpha:(V/U^\perp) \to \Bbb F$$. Because $$\bar \phi$$ is an isomorphism, there exists a vector $$c = (c_1,\dots,c_d) \in \Bbb F^d$$ such that $$\bar \alpha(v + U^\perp) = c^T\bar\phi(v + U^\perp)$$. Correspondingly, we may conclude that for all $$v \in V$$, we have $$\alpha(v) = c^T\phi(v) \implies (x,v) = (c_1x_1 + \cdots + c_dx_d,v) \implies\\ (c_1 x_1 + \cdots + c_d x_d - x,v) = 0 \quad \text{for all } v \in V.$$ Because the bilinear form is non-degenerate, we may conclude that $$c_1 x_1 + \cdots + c_d x_d - x = 0$$, which is to say that $$x = c_1x_1 + \cdots + c_dx_d$$, which means that $$x \in U$$.

Thus, we have $$U^{\perp \perp} \subseteq U$$. Thus, we have shown that $$U = U^{\perp\perp}$$.

• If you are not comfortable with quotient spaces, then this proof rewritten without them if we extend the basis $(x_1,\dots,x_d)$ to an (infinite) basis of $V$. Apr 25 at 13:25