This might be a really silly question, but I'm currently studying for my upcoming analysis exam, and I can't quite find an answer to this.
The reason I'm asking is actually, that I've been looking over sample questions for the exam, some of which are about determining whether a claim is true or false – the one that gave rise to this question is the following:
"Let $\mathcal{H}$ be a Hilbert space, and let $M$ be a closed subspace of $\mathcal{H}$. For each $x \in \mathcal{H}$ there is a $y\in M$ such that $x−y\in M^{\perp}$."
What I'm thinking is:
Assume that for all $x \in M$ $y \in M$ exists such that $x-y \in M^{\perp}$. That means, that $\langle x-y,m \rangle = 0$ which, by linearity in the first variable, means that:
$\langle x , m \rangle – \langle y , m \rangle = 0 \Leftrightarrow \langle x , m \rangle = \langle y , m \rangle \Rightarrow x=y \Rightarrow x-y=0 \in M^{\perp}$
Thus such a $y$ exists and is in fact $y=x$ – if the zero vector is indeed always contained in the orthogonal complement.
Is that even remotely true? I appologise if this is a silly question, but I'm having a hard time grasping this particular subject.
Any help would be much appreciated, thanks! 🙂
Best Answer
Yes, $0$ is always in the complement, just by definition.
However, you should show the existence of $y \in M$ for every $x \in \mathcal{H}$ and not just for $x \in M$.