Orthogonal projection on the Hilbert space .

I want to prove the following:

1. If $X$ is a Hilbert space and $Y$ is a closed subspace of $X$, then every $x\in X$ can be written as $x=y+z$ where $y\in Y$, $z \in Y^\perp$.
2. The projection (into $Y$) map $P:X\to Y$, given by $P(x)=y$ is linear, bounded, $P^2=P$, and $\langle x_1 , Px_2\rangle =\langle Px_1 , x_2\rangle$.

Here I have avoided subscripts (but the projection is always onto the $Y$ space):

Consider $x$ in $X$ , then there is a closest point to $x$ in $Y$ . Let us say that point as $Px$ , now we prove that $x-Px$ is orthogonal to $Y$ . Choose $y \in Y$ and $|y|=1$

Now $|x-(Px+ y)|^2 =|x-Px|^2-2Re \alpha(x-Px, y) + |\alpha|^2 y^2$ Let us choose $\alpha = (y, x-Px)$ then it becomes , $:= |x-Px|^2 - 2|(x-Px, y) |^2 + |(y, x-Px)|^2 =|x-Px|^2 - |(x-Px, y) |^2$

Which implies that distance of $x$ from $Px+y \in Y$ is less than the $|x-Px|$ , unless $|(x-Px, y) |^2 =0$ which gives us that $x$ and $x-Px$ are orthogonal .

Now let us see if $P : x \to Px$ is linear , Define $Qx =x-Px$ , We have already shown that $Qx$ is orthogonal to $Y$

Then $P(ax+by)+Q(ax+by) =ax+by =a(Px+Qx)+b(Py+Qy$, moving $P$ and $Q$ on two sides we get $P(ax+by)-(aPx +bPy) = Q(ax+by) -(aQx+bQy)$, since the right side is in $Y$ and the left side is not in $Y$, both the side should be equal to $0$ ,

$P(ax+by)-(aPx +bPy)=0$ , hence we show that $P$ is linear .

And the boundedness follows because $|x|^2=|Px|^2+|Qx|^2$ , is that right ?

Am i right so far ? I am having a bit of difficulty in proving rest of the stuff. Thanks for your help.

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It is not clear in your problem statement exactly which part(s) you want to show. Is it the claims about the properties of the projection map? –  JohnD Jan 10 '13 at 20:19
@JohnD : I want to show that if $Y$ is a closed subspace of hilbert space $X$ , then we know for every $x$ we can write it as $x=y+z$ , $y\in Y$ and $z \in Y\perp$ . Now i want to prove that if i define a map $P$ such that $P$ takes $x$ to $y$ , then it is a orthogonal projection, and rest of the thing as i have stated . –  Theorem Jan 10 '13 at 20:22

For the second part, we can establish the following properties about $P$ rather quickly:
• linear: Let $x_i=y_i+z_i$, where $x_i\in X$, $y_i\in Y$, $z_i\in Y^\perp$, and $\alpha,\beta$ be scalars. Then \begin{align}P(\alpha x_1+\beta x_2)&=P(\alpha(y_1+z_1)+\beta(y_2+z_2))\\&=P(\alpha y_1+\beta y_2+\alpha z_1+\beta z_2)=\alpha y_1+\beta y_2=\alpha P(x_1)+\beta P(x_2).\end{align}
• bounded: Since $x=0$ is trivial, suppose $x\not=0$. Because the projection is orthogonal, the (generalized) Pythagorean Theorem says $\|x\|^2=\|y\|^2+\|z\|^2$, so $$\|Px\|^2=\|y\|^2=\|x\|^2-\|z\|^2\le \|x\|^2.$$ Therefore, $${\|Px\|^2\over \|x\|^2}\le 1 \implies \|P\|=\max_{x\not =0}{\|Px\|\over \|x\|}\le 1,$$ and hence $P$ is bounded.
• idempotent: $P^2x=P(Px)=Py=y=Px$, so $P^2=P$.
• self-adjoint: $$\langle Px_1,x_2\rangle=\langle y_1,y_2+z_2\rangle=\langle y_1,y_2\rangle+\langle y_1,z_2\rangle=\langle y_1,y_2\rangle+0=\langle y_1,y_2\rangle$$ and $$\langle x_1,Px_2\rangle=\langle y_1+z_1,y_2\rangle=\langle y_1,y_2\rangle+\langle z_1,y_2\rangle=\langle y_1,y_2\rangle+0=\langle y_1,y_2\rangle,$$ so $\langle Px_1,x_2\rangle=\langle x_1,Px_2\rangle$.