- Suppose f is continuous and of moderate decrease, and \({ \hat { f } } ( \xi ) = 0\) for all \(\xi \in \mathbb { R }\) Show that \(f = 0\) by completing the following outline:
- For each fixed real number t consider the two functions
\[ A (z) = \int_ {- \infty} ^ {t} f (x) e ^ {- 2 \pi i z (x - t)} d x \quad \text { and } \quad B (z) = - \int_ {t} ^ {\infty} f (x) e ^ {- 2 \pi i z (x - t)} d x. \]
Show that \(A ( \xi ) = B ( \xi )\) for all \(\xi \in \mathbb { R } .\)
Prove that the function F equal to A in the closed upper half-plane, and B in the lower half-plane, is entire and bounded, thus constant. In fact, show that \(F = 0\)
Deduce that
\[ \int_ {- \infty} ^ {t} f (x) d x = 0, \]
for all t, and conclude that \(f = 0\)
Solution
Goal. Show \(\hat f = 0\) forces \(f = 0\) for continuous \(f\) of moderate decrease.
1. (a) \(A(\xi) = B(\xi)\) for all \(\xi \in \RR\).
1.1. \(A(\xi) = \int_{-\infty}^t f(x) e^{-2\pi i \xi(x-t)}\,dx\) and \(B(\xi) = -\int_t^\infty f(x) e^{-2\pi i \xi(x-t)}\,dx\).
definitions.
1.2. \(A(\xi) - B(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi(x-t)}\,dx = e^{2\pi i \xi t}\hat f(\xi) = 0\).
\(A - B\) is the full Fourier transform of \(f\) (with a phase), and \(\hat f(\xi) = 0\) by hypothesis.
1.3. Hence \(A(\xi) = B(\xi)\).
1.2.
2. (b) \(F\) (equal to \(A\) on the closed upper half-plane and \(B\) on the lower half-plane) is entire, bounded, hence constant, and in fact \(F = 0\).
2.1. \(A\) and \(B\) agree on the real axis (by (a)), so \(F\) is well-defined and continuous.
1.3.
2.2. \(A\) is holomorphic on the upper half-plane and \(B\) on the lower half-plane.
each is an integral of a holomorphic function in \(z\) (the integrand \(e^{-2\pi i z(x-t)}\) is entire in \(z\)), and the integrals converge uniformly (moderate decrease).
2.3. \(F\) is entire.
\(A\) and \(B\) agree on \(\RR\) and are holomorphic on their respective half-planes, so \(F\) is holomorphic across the real axis (Morera’s theorem).
2.4. \(F\) is bounded.
\(|A(z)| \le \int_{-\infty}^t |f(x)| e^{2\pi \Im(z)(x-t)}\,dx\), which is bounded for \(\Im z \ge 0\) (moderate decrease of \(f\)); similarly for \(B\) on the lower half-plane.
2.5. Hence \(F\) is constant (Liouville’s theorem).
a bounded entire function is constant.
2.6. \(F = 0\).
\(F(\xi) = A(\xi) = 0\) for all real \(\xi\) (since \(A(\xi) = B(\xi)\) and \(A(\xi) - B(\xi) = 0\)… more directly, \(A(\xi) = \int_{-\infty}^t f(x)e^{-2\pi i\xi(x-t)}dx\), and as \(\xi \to \infty\) this tends to \(0\) by Riemann–Lebesgue, so the constant \(F\) is \(0\)).
3. (c) \(f = 0\).
3.1. \(F(0) = A(0) = \int_{-\infty}^t f(x)\,dx = 0\).
\(F = 0\), and \(A(0) = \int_{-\infty}^t f(x) e^0\,dx = \int_{-\infty}^t f(x)\,dx\).
3.2. Hence \(\int_{-\infty}^t f(x)\,dx = 0\) for all \(t\).
3.1 holds for every \(t\).
3.3. Differentiating in \(t\) gives \(f(t) = 0\) for all \(t\).
the fundamental theorem of calculus: \(\frac{d}{dt}\int_{-\infty}^t f(x)\,dx = f(t) = 0\).
4. Q.E.D.
3.3 shows \(f = 0\).