Density of the Span of Closed Subsets in Hilbert Spaces

# Density of the Span of Closed Subsets in Hilbert Spaces

 Lemma 1: Let $H$ be a Hilbert space and let $H \subseteq S$. Then $S^{\perp} = (\mathrm{span}(S))^{\perp}$.
• Proof: Let $x \in S^{\perp}$. Then $x \perp S$, that is, for every $s \in S$ we have that $\langle x, s \rangle = 0$. Let $s' \in \mathrm{span} (S)$. Then $s' = a_1s_1 + a_2s_2 + ... + a_ns_n$ for $a_1, a_2, ..., a_n \in \mathbb{C}$ and for $s_1, s_2, ..., s_n \in S$. So:
(1)
\begin{align} \quad \langle x, s' \rangle &= \langle x, a_1s_2 + a_2s_2 + ... + a_ns_n \rangle \\ &= \overline{a_1} \underbrace{\langle x, s_1 \rangle}_{= 0} + \overline{a_2} \underbrace{\langle x, s_2 \rangle}_{=0} + ... + \overline{a_n} \underbrace{\langle x, s_n \rangle}_{= 0} \\ &= 0 \end{align}
• Therefore $x \in (\mathrm{span} S)^{\perp}$ and so:
(2)
• Now observe that $S \subseteq \mathrm{span} (S)$. Let $x \in (\mathrm{span}(S))^{\perp}$. Then $\langle x, s \rangle = 0$ for every $s \in \mathrm{span} (S)$, and so $\langle x, s \rangle = 0$ for every $s \in S$. Hence:
(3)
• From $(*)$ and $(**)$ we conclude that:
 Theorem 2: Let $H$ be a Hilbert space and let $S \subseteq H$ be a closed subset of $H$. Then $\mathrm{span} (S)$ is dense in $H$ if and only if $S^{\perp} = \{ 0 \}$.
• Proof: $\Rightarrow$ Suppose that $\mathrm{span} (S)$ is dense in $H$. Then $\overline{\mathrm{span} (S)} = H$. Observe that $\overline{\mathrm{span} (S)}$ is a closed subspace of $H$, and so:
• Therefore $(\overline{\mathrm{span}(S)})^{\perp} = \{ 0 \}$ and so $(\mathrm{span} (S))^{\perp} = \{ 0 \}$ and lastly, $S^{\perp} = \{ 0 \}$.
• $\Leftarrow$ Suppose that $S^{\perp} = \{ 0 \}$. Then $\mathrm{span} (S)^{\perp} = \{ 0 \}$ and so $\overline{\mathrm{span} (S)}^{\perp} = \{ 0 \}$. Since $H = \overline{\mathrm{span} (S)} \oplus (\overline{\mathrm{span} (S)})^{\perp}$ we have that:
• So $\mathrm{span} (S)$ is dense in $H$. $\blacksquare$