- The principle that a function and its Fourier transform cannot both be too small at infinity is illustrated by the following theorem of Hardy.
If \(f\) is a function on R that satisfies
\[ f (x) = O (e ^ {- \pi x ^ {2}}) \quad \mathrm{and} \quad \hat {f} (\xi) = O (e ^ {- \pi \xi^ {2}}), \]
then f is a constant multiple of \(e ^ { - \pi x ^ { 2 } }\) . As a result, if \(f ( x ) = O ( e ^ { - \pi A x ^ { 2 } } )\) , and \(\hat { f } ( \xi ) = O ( e ^ { - \pi B \xi ^ { 2 } } )\) , with \(A B > 1\) and \(A , B > 0\) , then f is identically zero.
- If f is even, show that \(\hat { f }\) extends to an even entire function. Moreover, if \(g ( z ) = \hat { f } ( z ^ { 1 / 2 } )\) , then \(g\) satisfies
\[ | g (x) | \leq c e ^ {- \pi x} \quad \text { and } \quad | g (z) | \leq c e ^ {\pi R \sin^ {2} (\theta / 2)} \leq c e ^ {\pi | z |} \]
when \(x \in \mathbb { R }\) and \(z = R e ^ { i \theta }\) with \(R \geq 0\) and \(\theta \in \mathbb { R }\)
- Apply the Phragmén–Lindelöf principle to the function
\[ F (z) = g (z) e ^ {\gamma z} \quad \mathrm{where} \gamma = i \pi \frac {e ^ {- i \pi / (2 \beta)}}{\sin \pi / (2 \beta)} \]
and the sector \(0 \leq \theta \leq \pi / \beta < \pi\) , and let \(\beta \downarrow 1\) to deduce that \(e ^ { \pi z } g ( z )\) is bounded in the closed upper half-plane. The same result holds in the lower half-plane, so by Liouville’s theorem \(e ^ { \pi z } g ( z )\) is constant, as desired.
- If \(f\) is odd, then \({ \hat { f } } ( 0 ) = 0\) , and apply the above argument to \(\hat { f } ( z ) / z\) to deduce that \(\boldsymbol { f } = \boldsymbol { \hat { f } } = 0\) . Finally, write an arbitrary \(f\) as an appropriate sum of an even function and an odd function.
Solution
We use the Fourier-transform convention \[ \widehat f(\zeta)=\int_{\mathbb R}f(x)e^{-2\pi i x\zeta}\,dx. \] The hypothesis \(|f(x)|\le Ce^{-\pi x^2}\) implies that this integral converges for every \(\zeta\in\mathbb C\), locally uniformly in \(\zeta\), because for \(\zeta=\xi+i\eta\), \[ |f(x)e^{-2\pi i x\zeta}| \le C e^{-\pi x^2+2\pi |\eta||x|}. \] Hence \(\widehat f\) extends to an entire function. Completing the square gives the global estimate \[ |\widehat f(\xi+i\eta)| \le C\int_{\mathbb R}e^{-\pi x^2+2\pi\eta x}\,dx =C' e^{\pi\eta^2}. \tag{1} \]
Assume first that \(f\) is even. Then \(\widehat f\) is even. Its Taylor series therefore contains only even powers, so there is an entire function \(g\) such that \[ g(z)=\widehat f(\sqrt z); \] this notation is independent of the choice of square root. If \(x\ge0\), the assumed real-axis decay of \(\widehat f\) gives \[ |g(x)|\le C e^{-\pi x}. \tag{2} \] If \(x<0\), choose \(\sqrt x=i\sqrt{|x|}\) and use (1): \[ |g(x)|\le C'e^{\pi|x|}=C'e^{-\pi x}. \tag{3} \] More generally, for \(z=Re^{i\theta}\) choose \(\sqrt z=\sqrt R e^{i\theta/2}\). Then (1) yields \[ |g(z)|\le C'e^{\pi R\sin^2(\theta/2)}\le C'e^{\pi|z|}. \tag{4} \]
Fix \(\beta>1\) and consider the sector \[ S_\beta=\{z:0\le\arg z\le\pi/\beta\}. \] Put \[ \alpha=\frac{\pi}{2\beta}, \qquad \gamma=i\pi\frac{e^{-i\alpha}}{\sin\alpha} =\pi(1+i\cot\alpha), \] and \[ F_\beta(z)=g(z)e^{\gamma z}. \] By (4), \(F_\beta\) has exponential growth \(O(e^{C_\beta|z|})\). Since the sector has opening \(\pi/\beta\) and \(1<\beta\), the Phragmén–Lindelöf principle applies.
On the ray \(z=r\ge0\), \(\Re\gamma=\pi\), so by (2) \[ |F_\beta(r)|\le C. \tag{5} \] On the other boundary ray \(z=re^{2i\alpha}\), a direct calculation gives \[ \Re(\gamma e^{2i\alpha})=-\pi. \] Together with (4), whose exponent there is \(\pi r\sin^2\alpha\), this gives \[ |F_\beta(re^{2i\alpha})| \le C' e^{-\pi r+\pi r\sin^2\alpha} =C'e^{-\pi r\cos^2\alpha}\le C'. \tag{6} \] Thus Phragmén–Lindelöf and (5)–(6) imply that \(F_\beta\) is bounded on \(S_\beta\).
Now let \(\beta\downarrow1\). Then \(S_\beta\) exhausts the open upper half-plane and \[ \gamma\longrightarrow\pi. \] For each fixed \(z\) in the upper half-plane, the preceding bound may be taken with the same boundary constant \(\max(C,C')\); hence \[ |e^{\pi z}g(z)|\le C'' \] in the closed upper half-plane. Applying the reflected argument in the lower half-plane gives the same bound there. Consequently the entire function \(e^{\pi z}g(z)\) is bounded on \(\mathbb C\), so Liouville’s theorem gives \[ g(z)=c e^{-\pi z}. \] Substituting \(z=w^2\) yields \[ \widehat f(w)=c e^{-\pi w^2}. \] The Gaussian is its own Fourier transform in this normalization, so Fourier inversion gives \[ f(x)=c e^{-\pi x^2}. \tag{7} \]
Now suppose \(f\) is odd. Then \(\widehat f\) is odd and \(\widehat f(0)=0\). Hence \[ h(z)=\frac{\widehat f(z)}{z} \] extends to an even entire function. From (1) and the real-axis Gaussian bound, after increasing the constant on a neighborhood of \(0\), \(h\) satisfies the same type of estimates as the even function treated above: \[ |h(x)|\le C e^{-\pi x^2}, \qquad |h(\xi+i\eta)|\le C'e^{\pi\eta^2}. \] Therefore \[ h(z)=c e^{-\pi z^2}, \qquad \widehat f(z)=cz e^{-\pi z^2}. \] But the assumed bound \(|\widehat f(x)|\le C e^{-\pi x^2}\) on the real axis forces \(c=0\), since otherwise division by \(e^{-\pi x^2}\) would give the unbounded function \(|c||x|\). Thus \(\widehat f=0\) and \(f=0\).
For a general \(f\), write \[ f=f_e+f_o, \qquad f_e(x)=\frac{f(x)+f(-x)}2, \quad f_o(x)=\frac{f(x)-f(-x)}2. \] Both parts satisfy the same Gaussian bound, and \[ \widehat{f_e}(\xi)=\frac{\widehat f(\xi)+\widehat f(-\xi)}2, \qquad \widehat{f_o}(\xi)=\frac{\widehat f(\xi)-\widehat f(-\xi)}2, \] so their Fourier transforms do also. The even case gives \(f_e=ce^{-\pi x^2}\) and the odd case gives \(f_o=0\). This proves Hardy’s theorem.
Finally suppose \[ |f(x)|\le Ce^{-\pi A x^2}, \qquad |\widehat f(\xi)|\le Ce^{-\pi B\xi^2}, \qquad AB>1. \] Define \(q(x)=f(x/\sqrt A)\). Then \[ |q(x)|\le Ce^{-\pi x^2}, \] and the scaling rule gives \[ \widehat q(\xi)=\sqrt A\,\widehat f(\sqrt A\,\xi), \] so \[ |\widehat q(\xi)|\le C\sqrt A\,e^{-\pi AB\xi^2} \le C\sqrt A\,e^{-\pi\xi^2}. \] Hardy’s theorem gives \(q(x)=ce^{-\pi x^2}\), hence \[ \widehat q(\xi)=ce^{-\pi\xi^2}. \] The stronger estimate with \(AB>1\) is incompatible with \(c\ne0\) as \(|\xi|\to\infty\). Hence \(c=0\), so \(q=0\) and therefore \(f=0\).