If $H$ is a Hilbert space, $U≤H$ is closed and $E≤U^⊥$ such that $x∈H$ with $x⊥_H E$ implies $x∈U$, then $U^⊥=\overline E^{\langle\;⋅\;,\;⋅\;\rangle}$

Question

Let

• $\left(H,\langle\;\cdot\;,\;\cdot\;\rangle\right)$ be a separable Hilbert space
• $U$ be a closed subspace of $H$
• $E$ be a subspace of $$U^\perp:=\left\{x\in H:\langle x,u\rangle=0\text{ for all }u\in U\right\}$$

Suppose that we can show the following Lemma:

Let $x\in H$ with $x\perp E$, i.e. $$\langle x,e\rangle=0\;\;\;\text{for all }e\in E\;.$$ Then, $x\in U$.

Why can we conclude that $U^\perp=\overline E^{\langle\;\cdot\;,\;\cdot\;\rangle}$?

Answer 1

The short answer is: $E^\perp\subset U$ implies $U^\perp\supset \overline{E}=E^{\perp\perp}\supset U^\perp$.

Let's see what's going on here. First, if $X\subset Y\subset H$, then $X^\perp\supset Y^\perp$ just by definition. Applied to the subspaces in question we get $E^{\perp\perp}\supset U^\perp$.

Further, $E^{\perp\perp}$ is a closed subspace containing $E$ (once again easy to check using only the definition), so $E^{\perp\perp}\supset \overline E$. On the other hand, every $x\in E^{\perp\perp}$ can be decomposed as $x=y+z$ with $y\in \overline{E}$ and $z\in \overline {E}^\perp\cap E^{\perp\perp}$. As $\overline{E}^{\perp}\cap E^{\perp\perp}=\{0\}$ we conclude that $\overline{E}=E^{\perp\perp}$.

Finally, $U^\perp$ is a closed set containing $E$, hence $U^\perp\supset \overline{E}$.