- This exercise generalizes some of the properties of \(e ^ { - \pi x ^ { 2 } }\) related to the fact that it is its own Fourier transform.
Suppose \(f ( z )\) is an entire function that satisfies
\[ | f (x + i y) | \leq c e ^ {- a x ^ {2} + b y ^ {2}} \]
for some \(a , b , c > 0\) . Let
\[ \hat {f} (\zeta) = \int_ {- \infty} ^ {\infty} f (x) e ^ {- 2 \pi i x \zeta} d x. \]
Then, \(\hat { f }\) is an entire function of ζ that satisfies
\[ | \hat {f} (\xi + i \eta) | \leq c ^ {\prime} e ^ {- a ^ {\prime} \xi^ {2} + b ^ {\prime} \eta^ {2}} \]
for some \(a ^ { \prime } , b ^ { \prime } , c ^ { \prime } > 0\)
[Hint: To prove \(\hat { f } ( \xi ) = O ( e ^ { - a ^ { \prime } \xi ^ { 2 } } )\) , assume \(\xi > 0\) and change the contour of integration to \(x - i y\) for some \(y > 0\) fixed, and \(- \infty < x < \infty\) . Then
\[ \hat {f} (\xi) = O (e ^ {- 2 \pi y \xi} e ^ {b y ^ {2}}). \]
Finally, choose \(y = d \xi\) where d is a small constant.]
Solution
For real \(x\), the hypothesis gives \(|f(x)|\le c e^{-ax^2}\), so \(f|_{\mathbb R}\in L^1(\mathbb R)\). For \(\zeta=\xi+i\eta\), \[ |f(x)e^{-2\pi i x\zeta}|\le c e^{-ax^2+2\pi |\eta||x|}, \] and the right side is integrable. Uniform domination on compact subsets of the \(\zeta\)-plane therefore permits differentiation under the integral sign to all orders; hence \(\widehat f\) is entire.
We first obtain Gaussian decay on the real axis. Fix real \(\xi\). For any real \(y\), Cauchy’s theorem on the rectangle with vertices \(\pm R\) and \(\pm R-iy\), followed by \(R\to\infty\), gives \[ \widehat f(\xi)=\int_{-\infty}^{\infty}f(x-iy)e^{-2\pi i(x-iy)\xi}\,dx. \tag{1} \] Indeed, on the vertical sides the factor \(f(z)\) is bounded by \(c e^{-aR^2+by^2}\), while \(|e^{-2\pi i z\xi}|\) is bounded there by a constant depending only on \(y,\xi\), so those side integrals tend to \(0\).
For \(\xi>0\), choose \(y=d\xi\) with \(d>0\). From (1), \[ |\widehat f(\xi)| \le c e^{-2\pi d\xi^2+bd^2\xi^2} \int_{-\infty}^{\infty}e^{-ax^2}\,dx. \] Choose \(0<d<2\pi/b\). Then \[ a':=2\pi d-bd^2>0, \] and therefore \(|\widehat f(\xi)|\le C e^{-a'\xi^2}\). For \(\xi<0\), shift upward by writing \(y=d\xi\) in the same formula; the identical estimate results. Thus \[ |\widehat f(\xi)|\le C e^{-a'\xi^2}\qquad(\xi\in\mathbb R). \tag{2} \]
Finally let \(\zeta=\xi+i\eta\). Directly from the defining integral, \[ |\widehat f(\xi+i\eta)| \le c\int_{-\infty}^{\infty}e^{-ax^2+2\pi\eta x}\,dx =C_0 e^{\pi^2\eta^2/a}. \tag{3} \] To retain decay in \(\xi\) simultaneously, repeat the contour shift above with \(y=d\xi\). Then \[ |e^{-2\pi i(x-iy)(\xi+i\eta)}| =e^{2\pi\eta x-2\pi y\xi}, \] so completing the square in \(x\) gives \[ |\widehat f(\xi+i\eta)| \le C_0 \exp\!\left((-2\pi d+bd^2)\xi^2+\frac{\pi^2}{a}\eta^2\right). \] With the same choice of \(d\) this is \[ |\widehat f(\xi+i\eta)|\le c' e^{-a'\xi^2+b'\eta^2}, \qquad b'=\frac{\pi^2}{a}>0. \] This is the required estimate.