Good evening, I'm trying to solve this exercise about Hilbert space:
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
- $\overline A = H \Longrightarrow A^\perp = \{0\}$
First, notice that $H^\perp = \{0\}$ for any Hilbert space $H$. The reasoning is as follows: for $x \in H^\perp$, we have $\langle x,x \rangle =0$ because $x \in H$. Hence $x=0$ by axiom of inner product space.
Let $x \in A^\perp$ and $h \in H$. Because $\overline A = H$, there is a sequence $(a_n)$ in $A$ such that $a_n \to h$ as $n \to \infty$. Consider the sequence $\langle a_n, x\rangle_{n \in \mathbb N}$. We have $\langle a_n, x\rangle \to \langle h, x\rangle$ as $n \to \infty$. On the other hand, $\langle a_n, x\rangle = 0$ for all $n \in \mathbb N$. Hence $\langle h, x\rangle = 0$. This is true for all $h \in H$, so $x \in H^\perp = \{0\}$. Hence $x=0$ and thus $A^\perp = \{0\}$.
- $\overline A = H \Longleftarrow A^\perp = \{0\}$
It follows from $A \subseteq \overline A$ that ${(\overline A)}^\perp \subseteq A^\perp$, so ${(\overline A)}^\perp = \{0\}$. Because $\overline A$ is closed, we have $H = \overline A \oplus {(\overline A)}^\perp = \overline A \oplus \{0\} = \overline A$. This completes the proof.
Best Answer
Your attempt is fine, but for the direction $\overline{A} = H \implies A^{\perp} = \{0\}$ I would suggest not using the generic $h$ at all, apply the argument directly to $h = x \in A^{\perp}$.
Since $A$ is dense, there is a sequence $(a_n)_{n \in \mathbb{N}}$ in $A$ with $a_n \to x$. Then, since $x \in A^{\perp}$, we have
$$\lVert x\rVert^2 = \langle x, x\rangle = \lim_{n \to \infty} \langle a_n,x \rangle = \lim_{n \to \infty} 0 = 0,$$
in other words $x = 0$ for all $x \in A^{\perp}$.