Prove that \(\hat f = \hat g\) implies that \(f=g\) a.e.
Show that if \(f, g \in L^1\) then \[\int \hat f g = \int f\hat g.\]
Solution
Use \(\hat f(\xi) = \int f(x) e^{-ix\cdot\xi}\,dx\) on \(\RR^n\).
1. If \(f, g \in L^1\) and \(\hat f = \hat g\), then \(f = g\) a.e.
1.1. Let \(h = f - g\). Then \(h \in L^1\) and \(\hat h = 0\).
The Fourier transform is linear.
1.2. Let \(\phi \geq 0\) be continuous with compact support and \(\int\phi = 1\), and put \(\phi_\eps(x) = \eps^{-n}\phi(x/\eps)\) and \(u_\eps = h \ast \phi_\eps\). Then \(u_\eps \to h\) in \(L^1\) as \(\eps \to 0\).
This is the \(L^1\) convergence of approximations to the identity.
1.3. \(\widehat{u_\eps}(\xi) = \hat h(\xi)\hat\phi(\eps\xi) = 0\) for every \(\xi\).
By Fubini’s theorem the Fourier transform of a convolution of \(L^1\) functions is the product of the transforms, and \(\hat h = 0\) by step 1.
1.4. \(u_\eps = 0\) identically.
\(u_\eps\) is continuous, as the convolution of an \(L^1\) function with a bounded uniformly continuous function. It lies in \(L^1 \cap L^\infty \subseteq L^2\). By Plancherel’s theorem and step 3, \(\|u_\eps\|_2 = (2\pi)^{-n/2}\|\widehat{u_\eps}\|_2 = 0\), so \(u_\eps = 0\) a.e., and a continuous function that vanishes a.e. vanishes everywhere.
1.5. Q.E.D.
By steps 2 and 4, \(h\) is the \(L^1\) limit of the zero functions \(u_\eps\), so \(h = 0\) a.e.
2. If \(f, g \in L^1\), then \(\int \hat f\,g = \int f\,\hat g\).
\(\iint |f(x)g(\xi)|\,dx\,d\xi = \|f\|_1\|g\|_1 < \infty\), so Fubini’s theorem gives \[ \int \hat f(\xi)g(\xi)\,d\xi = \int_\xi \int_x f(x)g(\xi)e^{-ix\cdot\xi}\,dx\,d\xi = \int_x \int_\xi f(x)g(\xi)e^{-ix\cdot\xi}\,d\xi\,dx = \int f(x)\hat g(x)\,dx. \]
3. Q.E.D.
Step 1 proves the first part and step 2 the second.