Let \(f, g\in L^1(\RR^n)\) and give a definition of \(f\ast g\).
Prove that if \(f, g\) are integrable and bounded, then \[ (f\ast g)(x) \converges{\abs x\to\infty}\to 0 .\]
In parts:
- Define the Fourier transform of an integrable function \(f\) on \(\RR^n\).
- Give an outline of the proof of the Fourier inversion formula.
- Give an example of a function \(f\in L^1(\RR^n)\) such that \(\hat{f}\) is not in \(L^1(\RR^n)\).
Solution
For \(f, g \in L^1(\RR^n)\), \((f \ast g)(x) = \int_{\RR^n} f(x - y)\,g(y)\,dy\). By Tonelli’s theorem \(\iint |f(x-y)||g(y)|\,dy\,dx = \|f\|_1\|g\|_1 < \infty\), so the integral converges absolutely for a.e. \(x\), and \(\|f \ast g\|_1 \le \|f\|_1\|g\|_1\).
1. Fix \(\eps > 0\) and choose \(R\) with \(\int_{|y| \ge R}|f| < \eps\) and \(\int_{|y| \ge R}|g| < \eps\). Then \(\int_{|y| \ge R}|f(x-y)||g(y)|\,dy \le \|f\|_\infty\eps\) for every \(x\).
\(R\) exists by dominated convergence, since \(|f|\chi_{\theset{|y| \ge R}} \to 0\) pointwise as \(R \to \infty\) with dominating function \(|f|\), and likewise for \(g\). The bound uses \(|f| \le \|f\|_\infty\).
2. For \(|x| \ge 2R\), \(\int_{|y| < R}|f(x-y)||g(y)|\,dy \le \|g\|_\infty\eps\).
If \(|y| < R\) and \(|x| \ge 2R\), then \(|x - y| \ge R\). Substituting \(z = x - y\) gives \(\int_{|y| < R}|f(x-y)||g(y)|\,dy \le \|g\|_\infty\int_{|z| \ge R}|f(z)|\,dz\).
3. Q.E.D.
By steps 1 and 2, \(|(f\ast g)(x)| \le (\|f\|_\infty + \|g\|_\infty)\eps\) for \(|x| \ge 2R\).
(c)1. For \(f \in L^1(\RR^n)\), \(\hat f(\xi) = \int_{\RR^n} f(x)\,e^{-2\pi i x \cdot \xi}\,dx\). The integral converges absolutely because \(|f(x)e^{-2\pi i x\cdot\xi}| = |f(x)|\).
(c)2. Let \(f \in L^1\) with \(\hat f \in L^1\). Then \(f(x) = \int \hat f(\xi) e^{2\pi i x\cdot\xi}\,d\xi\) for a.e. \(x\).
1. For \(\phi(x) = e^{-\pi|x|^2}\) and \(\phi_t(x) = t^{-n}\phi(x/t)\), \(\widehat{\phi_t}(\xi) = e^{-\pi t^2|\xi|^2}\) and \(\int \widehat{\phi_t}(\xi) e^{2\pi i x\cdot\xi}\,d\xi = \phi_t(x)\).
In one variable, \(\hat\phi\) and \(\phi\) both solve \(u' = -2\pi\xi u\) with \(u(0) = 1\), the first by differentiating under the integral and integrating by parts; so \(\hat\phi = \phi\). The \(n\)-variable case factors, and the formulas for \(\phi_t\) follow by scaling.
2. \(\int \hat f(\xi)\,\widehat{\phi_t}(\xi)\,e^{2\pi i x\cdot\xi}\,d\xi = (f \ast \phi_t)(x)\) for every \(x\).
Insert \(\hat f(\xi) = \int f(y)e^{-2\pi i y\cdot\xi}\,dy\). The double integral converges absolutely, so Fubini’s theorem and step 1 give \(\int f(y)\,\phi_t(x - y)\,dy\).
3. Q.E.D.
As \(t \to 0\), the left side of step 2 tends to \(\int \hat f(\xi) e^{2\pi i x\cdot\xi}\,d\xi\) for every \(x\), by dominated convergence with dominating function \(|\hat f|\), since \(\widehat{\phi_t} \to 1\) pointwise and \(|\widehat{\phi_t}| \leq 1\). The right side tends to \(f\) in \(L^1\), because \(\phi_t\) is an approximate identity; so a subsequence converges to \(f\) a.e. The two limits agree a.e.
(c)3. For \(n = 1\), \(f = \chi_{[-1,1]}\) is in \(L^1\) and \(\hat f(\xi) = \frac{\sin 2\pi\xi}{\pi\xi} \notin L^1(\RR)\).
\(\hat f(\xi) = \int_{-1}^1 e^{-2\pi i x \xi}\,dx = \frac{e^{2\pi i\xi} - e^{-2\pi i\xi}}{2\pi i \xi} = \frac{\sin(2\pi\xi)}{\pi\xi}\). For \(k \geq 1\) and \(\xi \in [k/2 + 1/8, k/2 + 3/8]\), \(|\sin 2\pi\xi| \ge 1/\sqrt2\) and \(|\xi| \le k/2 + 1\), so \(\int|\hat f| \ge \sum_{k \ge 1}\frac{1}{4\sqrt2\,\pi(k/2+1)} = \infty\).