Prove that an operator on $L^2(\mathbb{R^2})$ is an orthogonal projection

Question

Let $f \in L^2(\mathbb{R^2})$ and let $E_{m}(f(x))= \int_{-\infty}^{m} d{\lambda}\frac{1}{\sqrt2\pi}e^{i\lambda x}\hat f(\lambda)$ , where $\hat f(\lambda)$ is the Fourier transform of $f(x)$. Prove that $E_m$ is an orthogonal projection.

Definition (Orthogonal projection): Let $H$ be an Hilbert space and let $U$ be a closed subspace of $H$ such that $H=U \oplus U^{\perp}$. Then for all $v\in H$ we have that $v=u+w$ where $u\in U$ and $w \in U^{\perp}$. Then the Orthogonal Projection Operator of $H$ onto $U$ is the linear operator $P$, defined such that $P(v)=u$ for all $v\in H$.

I really have no idea on how to prove this. Any help?

Answer 1

Say $F$ is the Fourier transform. The Plancherel Theorem says that $F$ is a Hilbert-space isomorphism of $L^2$ onto itself, so $P$ is an orthogonal projection if and only if $F^{-1}PF$ is an orthogonal projection. If you say $Pf=\chi_{(-\infty,m]}f$ then it's easy to show that $P$ is an orthogonal projection, and $E_m=F^{-1}PF$.