Orthogonal complement of a Hilbert Space

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

1. $M^{\perp} = S^{\perp}$
2. $M = (S^{\perp})^{\perp}$
3. 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

Solution 1:

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 $S^{\perp} \subseteq M^{\perp}$.

Solution 2:

It should be well known that for a closed subspace $M$ of a Hilbert space $H$, $H = M \oplus M^{\perp}$. From $ S \subset M$ follows $M^\perp \subseteq S^\perp $ so $H = M^\perp \oplus M \subseteq S^\perp \oplus M $. Then $S^\perp $ cannot be a proper subset of $M^\perp $ so $S^\perp = M^\perp $.

$M^\perp $ is closed (intersection of kernels of continuous $x \mapsto (m, x),m \in M $), so $ H = (M^\perp)^\perp \oplus M^\perp = M \oplus M^\perp $. $M$ cannot be proper subspace of $(M^\perp)^\perp$ so $M=(M^\perp)^\perp$. Therefore $M= (M^\perp)^\perp = (S^\perp)^\perp$.

If $V$ is a closed subspace of $H$ then from $V \subset \overline V, V^\perp \supseteq {\overline V}^\perp $, so $ H = \overline V \oplus (\overline V)^\perp \subseteq \overline V \oplus V^\perp$ wherefrom $H= \overline V \oplus V^\perp$.

Solution 3:

M is the subspace generated by S so M is the smallest closed subspace containing S by definition. For the first question since $S\subset M$ then if $v\in M^{\perp}$ then $v$ is perpendicular to all elements in $M$. Hence is perpendicular to all elements in $S$. So $v\in S^{\perp}$. Since $M$ is the smallest closed subspace containing $S$ and the inner product in continuous in both components then if $v\in S^{\perp}$ then I claim that $v\in M^{\perp}$. For any vector $w\in S$ we have $<w,v>=0$. All other vectors in $M$ can be obtained by sequences in $S$. So if $w\in M$ but $w\notin S$ then we can find a sequence $\{w_n\}_{n\ge1}$ of elements in $S$ converging to $w$. But $<w,v>=lim_{n\rightarrow\infty}<w_n.v>=0$. So $S^{\perp}\subset M^{\perp}$. This give us the first question. I think that is what you were having trouble justifying.