Direct sum decomposition of $L^2(\mathbb{R})$ using Fourier Transform

Question

Let $L_+^2(\mathbb{R})=\{f\in L^2(\mathbb{R}):supp \hat{f}\subset\mathbb{R^+}\}$ and $L_-^2(\mathbb{R})=\{f\in L^2(\mathbb{R}):supp \hat{f}\subset\mathbb{R^-}\}$, where $\hat{f}$ denotes the Fourier trans form of $f$ in $\mathbb{R}$. $\mathbb{R^+}$ and $\mathbb{R^-}$ respectively denotes set of positive and negative reals. Then I need to show that $L^2(\mathbb{R})=L_+^2(\mathbb{R})\oplus L_- ^2(\mathbb{R})$. I observed that $L_+^2(\mathbb{R})$ and $ L_- ^2(\mathbb{R})$ are orthogonal. Please help me to break $f=f_1+f_2$ where $f_1\in L_+^2(\mathbb{R})$ and $f_2\in L_-^2(\mathbb{R})$

Answer 1

Recall that the fourier transform is an isometric isomorphism $\def\F{\mathscr F}\F \colon L^2(\def\R{\mathbf R}\R) \to L^2(\R)$. For a subset $A \subseteq \R$ let $\chi_A$ denote its characteristic function. Given $f \in L^2(\R)$, define $$ f_1 = \F^{-1}\bigl(\chi_{\R^+}\F f\bigr) $$ and $$ f_2 = \F^{-1}\bigl(\chi_{\R^-}\F f\bigr) $$ As $\chi_{\R^+} + \chi_{\R^-} = 1$ almost everywhere we have $f = f_1 + f_2$.