I have this problem:
Let $S$ be a subset of a Hilbert $H$ and let $M$ be the closed subspace generated by $S$. Show that
- $M^{\perp} = S^{\perp}$
- $M = (S^{\perp})^{\perp}$
- if $V$ is a subspace of $H$, then $H = \overline{V}\oplus V^{\perp}$.
I have some doubts, because $H$ don't have finite dimension. For example, for 1. its clear that $S \subseteq M$ and then $M^{\perp} \subseteq S^{\perp}$. Later, if $x\in S^{\perp}$ then $\langle x, a\rangle = 0$, for all $a\in S$. Now in finite dimension I know how justify that $\langle x, b\rangle = 0$, for all $b\in M$, but in a Hilbert I really don't know. Thanks in advance for your help
Best Answer
Hint: Use the completeness assumption; that is, use a Cauchy sequence of finite linear combinations in $S$ approximating a given element in $M$.
Some more elaboration: As you said, $M^{\perp} \subseteq S^{\perp}$. To show the reverse inclusion, we do the following: using your notation, $\langle x, a \rangle = 0$ for every $x \in S^{\perp}, a \in S$. Now we have to use the crucial assumption: $M$ is the closed subspace generated by $S$. This means that for every element $b \in M$, there exists a sequence $\{b_n\}$ of elements in the span of $S$ (e.g. linear combinations of elements of $S$) such that $\|b_n - b\| \rightarrow 0$. This is precisely the power of Hilbert spaces; they are complete, so you are allowed to take limits of Cauchy sequences and they will converge.
Notice that $\langle x, b_n \rangle = 0$ for every $n$, since $b_n$ is just a finite sum of elements of $S$, each of which are orthogonal to $x$. Then $$ |\langle x, b \rangle - \langle x, b_n \rangle | = | \langle x, b - b_n \rangle | \le \|x\| \|b - b_n \| \rightarrow 0$$ so it follows that $\langle x, b \rangle = 0$, and hence $x \in M^{\perp}$, so $S^{\perp} \subseteq M^{\perp}$.