Linear Subspace Dense iff Zero Orthocomplement

Theorem

Let $H$ be a Hilbert space.

Let $K$ be a linear subspace of $H$.

Then $K$ is everywhere dense if and only if $K^\perp = \paren 0$, where $K^\perp$ is the orthocomplement of $K$, and $\paren 0$ denotes the zero subspace.

Proof

Sufficient Condition

Assume that $K$ is everywhere dense.

Let $x \in K^\perp$.

Then:

\(\ds x\)

\(\in\)

\(\ds K^\perp\)

\(\ds \leadsto \ \ \)

\(\ds \forall y \in K: \, \)

\(\ds x\)

\(\perp\)

\(\ds y\)

\(\ds \leadsto \ \ \)

\(\ds \forall y \in H: \, \)

\(\ds x\)

\(\perp\)

\(\ds y\)

as $K$ is everywhere dense, and Inner Product is Continuous

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds 0\)

$\Box$

Necessary Condition

Assume $K^\perp = 0$.

Then by Double Orthocomplement is Closed Linear Span:

$\vee K = \paren 0^\perp = H$

Hence $K$ is everywhere dense.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Corollary $2.11$