Question on Hilbert Space

Question

$S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $H\Leftrightarrow S^{\bot}=\{0\}$

I have been able to show $\Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $v\in S^{\bot}, \exists$ a sequence $v_n$ in $S_1$ such that $v_n\rightarrow v$ as $n\rightarrow\infty$, $v_n\bot v $ $\forall n\in\mathbf{N}$ $\therefore \lim_{n\rightarrow\infty}\|v_n-v\|^2=\lim_{n\rightarrow\infty} \|v_n\|^2+\|v\|^2=2\|v\|^2=0 \Rightarrow v=0$

I am struggling with the other implication, We have $S^{\bot\bot}=H$, need to show closure of $S_1$ is $S^{\bot\bot}=H$

Answer 1

Notice that $\left(\overline{\operatorname{span}S}\right)^\perp = S^\perp$. The Riesz projection theorem implies $$\left(\overline{\operatorname{span}S}\right)\oplus \left(\overline{\operatorname{span}S}\right)^\perp = H$$ so $$\text{linear span of } S \text{ is dense in } H \iff\overline{\operatorname{span}S} = H \iff \left(\overline{\operatorname{span}S}\right)^\perp = \{0\} \iff S^\perp = \{0\}$$

Answer 2

The adherence $\bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $\bar{S_1}$. The linear form defined on $\bar{S_1}\oplus\mathbb{C}x$ by $f(\bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=\langle x_0,.\rangle$. This implies that $x_0$ is in the orthogonal of $\bar{S_1}$.