Functional Analysis: Projection linear operator

Question

$\newcommand\dag\dagger$ We have in $L^2 [-\pi , \pi ]$ the linear operator $$(Pf)(x)= \frac{1}{2\pi} \int_{-\pi}^{\pi} [4\cos^2 (x-y) -1] f(y) dy.$$

Show that $P$ is an orthogonal projector.

I have no idea about how to show that $P=P^2 =P^{\dag}$ .

So I think it's better to suppose I have an orthonormal system $\{e_n\}$, it will be $Pf= \sum\limits_n e_n \ \langle e_n , f \rangle$.

How can I use this fact in this case?

Answer 1

Just apply the integral twice and show the double integral reduces back down to the single integral via Fubini's theorem like so:

$$(P^2f)(x) = \frac{1}{4\pi^2}\int_{-\pi}^\pi \int_{-\pi}^\pi [4\cos^2(x-y)-1][4\cos^2(y-z)-1]f(z)\:dz\:dy$$

$$\frac{1}{4\pi^2}\int_{-\pi}^\pi f(z)\:dz \int_{-\pi}^\pi [4\cos^2(x-y)-1][4\cos^2(y-z)-1]\:dy$$

$$\frac{1}{4\pi^2}\int_{-\pi}^\pi f(z)\:dz \int_{-\pi}^\pi [2\cos(2y-2x)+1][2\cos(2y-2z)+1]\:dy$$

(for formatting purposes only pay attention to the $y$ integral for now)

$$\int_{-\pi}^\pi 2\cos(2z-2x)+2\cos(4y-2x-2z)+2\cos(2y-2x)+2\cos(2y-2z)+1\:dy$$

$$\left[y(2\cos(2z-2x)+1)+\frac{1}{2}\sin(4y-2x-2z)+\sin(2y-2x)+\sin(2y-2z)\right]_{-\pi}^\pi$$

(now bring the $z$ integral back into play)

$$\implies \frac{1}{2\pi}\int_{-\pi}^\pi [2\cos(2x-2z)+1]f(z)\:dz=\frac{1}{2\pi}\int_{-\pi}^\pi [4\cos^2(x-z)-1]f(z)\:dz \equiv (Pf)(x)$$

where we freely used the evenness of cosine and the $2\pi$-periodicity of sine to make the other terms vanish.

Answer 2

Before you show that P is a projector, you may notice that

$2\cos^2(2x-2y)-1=2\cos(2x-2y)+1$

$=2\cos(2x)\cos(2y)+2\sin(2x)\sin(2y)+1$

Hence

$P(f)(x)=\frac{1}{\pi}\cos(2x)\int_{-\pi}^{\pi}\cos(2y)f(y)dy+\frac{1}{\pi}\sin(2x)\int_{-\pi}^{\pi}\sin(2y)f(y)dy+\frac{1}{\pi}\int_{-\pi}^{\pi}\frac12f(y)dy$

We can consider the basis: $\{x\mapsto\cos(nx),x\mapsto\sin(nx)|n\in\mathbb{N}^*\}\cup\left\{x\mapsto\frac{1}{\sqrt{2}}\right\}$

and the scalar product $<f,g>=\frac{1}{\pi}\int_{-\pi}^{\pi} f(y)g(y)dy$

The linerity of P is clear.

Let's denote $e_0:x\mapsto 1/\sqrt{2}$ and for $n\geq 1$ $e_n:x\mapsto\cos(nx)$, $e'_n:x\mapsto\sin(nx)$.

We have for $n\geq 0$, $<e_n,e_n>=1$ and for $n\geq 1$, $<e'_n,e'_n>=1$

Moreover for $p\neq q$, $<e_p,e_q>=0$, $<e'_p,e'_q>=0$

and for all $p,q$, $<e_p, e'_q>=0$

We remark that

$P(f)=<f,e_0>.e_0+<f,e_2>e_2+<f,e'_2>e'_2$

To prove that $P^2=P$

$P^2(f)=<f,e_0><e_0,e_0>e_0+<f,e_2><e_2,e_2>e_2+<f,e'_2><e'_2,e'_2>e'_2$

$=<f,e_0>e_0+<f,e_2>e_2+<f,e'_2>e'_2$

$=P(f)$

$P$ is the orthogonal projection on $\text{Span}\{e_0, e_2, e'_2\}$

Answer 3

May I offer some comments designed to complement the answer of Jean-Claude Collette. Let $(H,\langle \cdot,\cdot\rangle)$ be a Hilbert space and $V\subset H$ a subspace. An orthogonal projector $\Pi:H\to V$ will satisfy for all $v\in H$ $$\langle \Pi v-v,w \rangle = 0 \quad\forall\,w\in V.$$ As Jean-Claude Collette showed, your projector $P$ satisfies $$ Pf = \frac{1}{\pi}\cos(2x)\int_{-\pi}^{\pi}\cos(2y)f(y)dy+\frac{1}{\pi}\sin(2x)\int_{-\pi}^{\pi}\sin(2y)f(y)dy+\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)dy.$$ From here it is fairly straight forward that \begin{align} \int_{-\pi}^\pi Pf(x)\,dx = & \, \int_{-\pi}^{\pi}f(y)\,dy,\\ \int_{-\pi}^\pi \sin (2x) Pf(x)\,dx = & \, \int_{-\pi}^{\pi}\sin (2y)f(y)\,dy,\\ \int_{-\pi}^\pi \cos (2x)Pf(x)\,dx = & \, \int_{-\pi}^{\pi}\cos (2y)f(y)\,dy.\end{align} So if we endow the space $L^2[-\pi,\pi]$ with the usual inner-product $\langle u,v \rangle:=\int_{-\pi}^{\pi}uv\,dx$ and define the space $V:=\textrm{span}\{1,\sin 2x,\cos 2x \}$ then your orthogonal projector $P:L^2[-\pi,\pi]\to V$ satisfies $$\langle P f-f,w \rangle = 0 \quad\forall\,w\in V.\tag{1}\label{1}$$ I would note that if $P$ was defined as the solution to \eqref{1}, then existence and uniqueness would follow from the Riesz Representation Theorem. From here your problem is easy. We have showed that $Pf-f$ is orthogonal to $V$, and linearity and idempotency follow trivially. (For the latter show that $Pf=f$ for all $f\in V$).