Let $f,g \in L^1 (\Bbb R) \cap L^2(\Bbb R)$ and define $\phi(x) = \widehat f(x) \widehat g(x)$ for all $x\in \Bbb R$.
- First of all, I'd like to show that $\phi \in L^1 (\Bbb R) \cap L^2(\Bbb R)$.
As $f,g \in L^2(\Bbb R)$, we have $\hat f,\hat g \in L^2(\Bbb R)$. That $\phi \in L^1(\Bbb R)$ follows from $\|\phi\|_1 \le \|\hat f\|_2\|\hat g\|_2 < \infty$. Next, $|\phi| \le |\hat f \hat g| \le \|f\|_1 |\hat g|$ gives $\|\phi\|_2 \le \|f\|_1 \|\hat g\|_2 < \infty$. So, $\phi \in L^1 (\Bbb R) \cap L^2(\Bbb R)$.
- Next, suppose $f$ is even and $g$ is odd. I want to show $\lim_{\xi \to 0} \widehat \phi(\xi) = 0$.
As $f$ is even and $g$ is odd, we have $\widehat f(\xi) = \widehat f(-\xi)$ and $\widehat g(\xi) + \widehat g(-\xi) = 0$ for all $\xi \in \Bbb R$. By continuity of $\widehat g$, this gives $\widehat g(0) = \int_{\Bbb R} g(t)\, dt = 0$. As $\phi = \hat f \hat g = \widehat{f\ast g}$, we have $\widehat\phi(\xi) = \widehat{\widehat{f\ast g}}(\xi) = f\ast g(-\xi)$. That is, $$\widehat\phi(\xi) =f\ast g(-\xi) = \int_{\Bbb R} f(t) g(-\xi - t)\, dt = -\int_{\Bbb R} f(t) g(\xi + t)\, dt$$ The substitution $u = -t$ gives $$-\int_{\Bbb R} f(t) g(\xi + t)\, dt = -\int_{\Bbb R} f(u)g(\xi - u)\, du = - f\ast g(\xi)$$ and so $$\widehat \phi(\xi) = \frac{f\ast g(-\xi) - f\ast g(\xi) }{2}$$ giving $$\lim_{\xi \to 0} \widehat \phi(\xi) = \lim_{\xi \to 0}\frac{f\ast g(-\xi) - f\ast g(\xi) }{2} = 0$$
- Lastly, if $\operatorname{supp} f, \operatorname{supp} g \subset [0,1]$, I want to show $$\int_{-\infty}^\infty \phi(x)\, dx = 0$$ Indeed, $$\int_{-\infty}^\infty \hat f(x) \hat g(x)\, dx = \int_{-\infty}^\infty f(-x) g(x)\, dx = 0$$ as $\widehat{\widehat{f(x)}} = f(-x)$ and so $\operatorname{supp} \widehat{\widehat{f}} \subset [-1,0]$.
I'd like to know if my work is correct. Thank you!