Completeness and orthogonal projection

Question

a. Which are the properties that define an orthogonal projection? Give a precise definition.

b. What does completeness mean? Please state both the definition and an example (without proof) of a complete vector space.

Can someone give these precise definitions please ?

Regarding a) I know it orthogonal if it

1. Is linear
2. Is idempotent.

B) I don't have clue about it.

Answer 1

Definition [Orthogonal Projection]: Let $\mathcal{H}$ be a real or complex Hilbert space, and let $M$ be a closed subspace of $\mathcal{H}$. The orthogonal projection $P_{M} : \mathcal{H}\rightarrow\mathcal{H}$ maps $x\in\mathcal{H}$ to the unique $P_{M}x \in M$ such that $$ (x-P_{M}x) \perp M. $$ (That is, $\langle x-P_{M}x,m\rangle=0$ for all $m\in M$.)

To see that $P_M$ is well-defined, suppose $(x-m)\perp M$ and $(x-m')\perp M$ for some $m,m'\in M$. Then $m-m'=(x-m')-(x-m)$ is orthogonal to itself, making $m-m'=0$.

One can show that orthogonal projection is the same as closest point projection onto $M$. Closest point projection exists because $M$ is a complete subspace. So orthogonal projection is everywhere defined, and these two projections map $x$ to the same $m \in M$.

Linearity: Orthogonal projection is automatically linear because of uniqueness, and the fact that $$ (x-P_Mx)\perp M,\;\;(y-P_{M}y)\perp M \\ \implies \{(\alpha x+\beta y)-(\alpha P_{M}x+\beta P_{M}y)\} \perp M. $$ Idempotent: $P_{M}$ is the identity on $M$ because $(m-m)\perp M$. Therefore $P_{M}^2=P_{M}$.

Symmetry: $ \langle P_{M}x,y\rangle = \langle P_{M}x,(y-P_{M}y)+P_{M}y\rangle = \langle P_{M}x,P_{M}y\rangle.$ Hence, $$ \langle P_{M}x,y\rangle = \langle P_{M}x,P_{M}y\rangle = \langle x,P_{M}y\rangle. $$ Bounded: $\|x\|^2=\|x-P_{M}x\|^2+\|P_{M}x\|^2 \ge \|P_{M}x\|^2$.

Conversely, if $P=P^2=P^{\star}$ is linear, then you can show that $P$ satisfies the properties of an orthogonal projection onto $M=P\mathcal{H}$ because \begin{align} \langle x-Px,m \rangle & = \langle x-Px,Pm \rangle \\ &= \langle P^{\star}(x-Px),m\rangle \\ &= \langle P(x-Px),m\rangle \\ &= \langle Px-P^2x,m\rangle = 0,\;\;\; m \in M. \end{align}

Definition [Complete]: Let $X$ be normed linear space over the real or complex numbers. A sequence $\{ x_n \} \subset X$ is a Cauchy sequence if, for every $\epsilon > 0$, there exists $N$ such that $\|x_n-x_m\| < \epsilon$ whenever $n,m \ge N$. $X$ complete if every Cauchy sequence in $X$ has a limit in $X$.

Example: $\mathbb{R}^{n}$ is complete.

Answer 2

A. Your properties are characterized for projection, not orthogonal. A mapping $p :E \rightarrow E $ is orthogonal projection if $p$ is linear, idempotent, and $<x-p(x),p(y)>= \, \forall x,y \in E$ ($<,>$ is denoted for scalar products in $E$)

B. Completeness property is "Every Cauchy sequences is converged". Informally, if you have a sequence look so close at the end, it eventually got a limit in your space. For example, $\mathbb{Q}$ is not complete, take the sequence 1, 1.4, 1.414, ... It converges to $\sqrt{2}$ which is well known that irrational. $\mathbb{R}$ is complete and it is completion of $\mathbb{Q}$