If $M$ is a non-empty subset of a Hilbert space $H$, the span of $M$ is dense in $H$ iff $M^{\perp} = \{0\}$.

Question

I need help only with the converse of the proof. Suppose, $M^{\perp} = \{0\}$, and $V=span(M)$, then if $x \perp V$, this implies $x \perp M$ (why?) so that $x \in M^{\perp}$ and $x=0$. Hence $V^{\perp} = \{0\}$ (again why?). Noting that $V$ is a subspace of $H$, we obtain $\bar{V}=H.$

The last equality is obtained from $H= \bar{V}\oplus \bar{V}^{\perp}$, I think, but I don't understand this very well either. Please explain the logic behind the steps indicated with a (why?) that I'm having trouble with. Do we have that $V$ is closed? Also, Is this true that $V^{\perp \perp} = \bar{V}$? How do I you prove this? Thanks for all your help.

Edit: I've proven this in the following way and need to know if this is correct:

$V^{\perp}$ is closed, $\bar{V}$ is a closed subspace of $H$. Then

$H = \bar{V} \oplus \bar{V^{\perp}} = \bar{V} \oplus V^{\perp}=\bar{V} \oplus \{0\}$ implies $H = \bar{V}$.

Answer 1

Your edit is correct modulo the proof that $(\overline{V})^\perp=V^\perp$ (i.e. be careful on the order of applying bars and perps).

To prove this, note first that since $V\subset \overline{V}$, we trivially have $\overline{V}^\perp \subset V^\perp$. To prove the converse, let $f \in V^\perp$, and let $g \in \overline{V}$. We must show that $\langle f, g \rangle = 0$. But $g=\lim_{n \to \infty}g_n$ for some sequence $\{g_n\}$ in $V$, hence $0 = \lim \langle f, g_n \rangle = \langle f, \lim g_n \rangle = \langle f,g \rangle$, and we are done.

Also, you could have used that $V^\perp$ is closed and $(V^\perp)^\perp=\overline{V}$, so $\mathcal{H}=V^\perp \oplus (V^\perp)^\perp = \{0\} \oplus \overline{V}=\overline{V}$ (Of course, this is really the same proof).

Answer 2

These sketches may help you...

(Taken from my notes.)

Answer 3

As I also encounter this proof in Kreyszig's book, let me just add some remarks to clarify things a bit.

We now have a non-empty set $M \subset H$ where $H$ is Hilbert. And we wish to prove $V=span(M)$ dense in $H$, i.e. $\bar{V} = H$.

Step(i) Use the direct sum decomposition theorem. If $Y$ is a $\textbf{closed} $ subspace of Hilbert $H$, one has $H = Y \oplus Y^{\perp}$. In our case, one has $H = \bar{V} \oplus (\bar{V})^{\perp}$.

Step(ii) You only need to prove $(\bar{V})^{\perp} \subset M^{\perp}$, but that is basically free. As $M \subset V \subset \bar{V}$, $x \in \bar{V}^{\perp} \Rightarrow x \in M^{\perp}$. The assumption $M^{\perp} = \{0\}$ then forces $(\bar{V})^{\perp} =\{0\}$.

Step(iii) Back to (i), immediately you have $H = \bar{V}$.