- Show that if \(f\in L^1\) then \(\hat f\) is bounded and uniformly continuous.
Solution
Use \(\hat f(\xi) = \int f(x)e^{-ix\xi}\,dx\).
1. \(|\hat f(\xi)| \le \|f\|_1\) for every \(\xi\).
\(|\hat f(\xi)| = \left|\int f(x)e^{-ix\xi}\,dx\right| \le \int |f(x)|\,dx\).
2. \(\sup_\xi|\hat f(\xi + h) - \hat f(\xi)| \le \int |f(x)|\,|e^{-ihx} - 1|\,dx\).
\(\hat f(\xi+h) - \hat f(\xi) = \int f(x)e^{-ix\xi}(e^{-ihx} - 1)\,dx\), and \(|e^{-ix\xi}| = 1\).
3. \(\int |f(x)|\,|e^{-ihx} - 1|\,dx \to 0\) as \(h \to 0\).
The integrand tends to \(0\) pointwise and is bounded by \(2|f| \in L^1\), so the dominated convergence theorem applies along every sequence \(h_k \to 0\).
4. Q.E.D.
Step 1 is boundedness, and steps 2 and 3 give \(\sup_\xi|\hat f(\xi + h) - \hat f(\xi)| \to 0\) as \(h \to 0\), which is uniform continuity.