I want to prove the following:
- 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 $y\in Y$, $z \in Y^\perp$.
- 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.