Orthogonal projection on the Hilbert space .

Solution 1:

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