Projection Operator and Perpendicularity

Question

$M\neq \emptyset$ including $M^\bot=\{x \in X\mid x \bot M \}$ is a vector space show that? And it is closed show that?

How can i do show that?

Answer 1

I am not sure I understood your question but if you want to show that $M^{\perp}$ is closed just take $x$ in its closure, and so you will have $x_n\rightarrow x$ where $x_n \in M^{\perp}$ and so $\langle x_n,m\rangle =0$ for every $n$ and $m\in M$, and then using the fact the inner product is continuous you get that $\langle x,m\rangle =\langle\lim\limits_{n\rightarrow \infty}x_n,m\rangle=\lim\limits_{n\rightarrow\infty}\langle x_n,m\rangle =0$, and so $x\in M^{\perp}$ and the set is closed.

I assumed you are working with an Hilbert space but if you are using the definition of $\perp$ for a Banach space the idea is the same.