Hilbert Spaces – Orthogonal Projection on the Hilbert Space

functional-analysishilbert-spaceslinear algebra

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.

Best Answer

The first part is often called the Orthogonal Decomposition Theorem and is found in just about any textbook on Hilbert spaces. Here (look at 3.6 and right below 3.9) is a readily available proof from the web.

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 $\langle Px_1,x_2\rangle=\langle x_1,Px_2\rangle$.
Related Question