- Show that if \(\alpha > 1\) , then
\[ F _ {\alpha} (z) = \int_ {- \infty} ^ {\infty} e ^ {- | t | ^ {\alpha}} e ^ {2 \pi i z t} d t \]
is an entire function of growth order \(\alpha / ( \alpha - 1 )\)
[Hint: Show that
\[ - \frac {| t | ^ {\alpha}}{2} + 2 \pi | z | | t | \leq c | z | ^ {\alpha / (\alpha - 1)} \]
by considering the two cases \(| t | ^ { \alpha - 1 } \leq A | z | { \mathrm { ~ a n d ~ } } | t | ^ { \alpha - 1 } \geq A | z | .\) , for an appropriate constant A.]
Solution
1. Holomorphicity of \(F_\alpha(z)\):
1.1. For any \(z = x + iy \in \mathbb{C}\) and \(t \in \mathbb{R}\):\[ \left| e^{-|t|^\alpha} e^{2\pi i z t} \right| = e^{-|t|^\alpha} e^{-2\pi y t} \le e^{-|t|^\alpha + 2\pi |z| |t|}. \]
\(|e^{2\pi i (x+iy)t}| = e^{-2\pi y t} \le e^{2\pi |z| |t|}\).
1.2. Since \(\alpha > 1\), as \(|t| \to \infty\) the term \(-|t|^\alpha\) dominates \(2\pi |z| |t|\), ensuring that for any compact subset \(K \subset \mathbb{C}\), the integrand and its \(z\)-derivative \(2\pi i t e^{-|t|^\alpha} e^{2\pi i z t}\) are uniformly dominated by an \(L^1(\mathbb{R})\) function.By Morera’s Theorem and Fubini’s Theorem, \(F_\alpha(z)\) is an entire function.
differentiation under the integral sign / Morera’s Theorem.
2. Upper bound on the order of growth:
2.1. Let \(\beta = \frac{\alpha}{\alpha - 1}\), so that \(\frac{1}{\alpha} + \frac{1}{\beta} = 1\) are conjugate exponents.By Young’s inequality \(ab \le \frac{a^\alpha}{\alpha} + \frac{b^\beta}{\beta}\) with \(a = 2^{1/\alpha} |t|\) and \(b = 2^{-1/\alpha} (2\pi |z|)\):\[ 2\pi |z| |t| = (2^{1/\alpha} |t|) (2^{-1/\alpha} 2\pi |z|) \le \frac{2 |t|^\alpha}{\alpha} + \frac{(2^{-1/\alpha} 2\pi |z|)^\beta}{\beta} \le \frac{|t|^\alpha}{2} + c |z|^\beta, \]where \(c = \frac{(2\pi)^\beta}{\beta 2^{\beta/\alpha}} > 0\).
Young’s inequality for conjugate exponents.
2.2. Rearranging gives \(-|t|^\alpha + 2\pi |z| |t| \le -\frac{|t|^\alpha}{2} + c |z|^\beta\).Integrating over \(\mathbb{R}\):\[ |F_\alpha(z)| \le \int_{-\infty}^\infty e^{-|t|^\alpha + 2\pi |z| |t|} \, dt \le e^{c |z|^\beta} \int_{-\infty}^\infty e^{-|t|^\alpha / 2} \, dt = C_0 e^{c |z|^\beta}, \]where \(C_0 = \int_{-\infty}^\infty e^{-|t|^\alpha / 2} \, dt < \infty\).Thus the growth order \(\rho(F_\alpha) \le \beta = \frac{\alpha}{\alpha - 1}\).
integral estimate.
3. Lower bound on the order of growth:
3.1. Evaluate \(F_\alpha\) on the negative imaginary axis \(z = -i y\) for \(y > 0\):\[ F_\alpha(-iy) = \int_{-\infty}^\infty e^{-|t|^\alpha + 2\pi y t} \, dt \ge \int_0^\infty e^{-t^\alpha + 2\pi y t} \, dt. \]
restricting domain of integration to \([0, \infty)\).
3.2. The exponent \(g(t) = -t^\alpha + 2\pi y t\) attains its maximum at \(t_0 = \left(\frac{2\pi y}{\alpha}\right)^{1/(\alpha-1)}\), with maximum value:\[ g(t_0) = -\left(\frac{2\pi y}{\alpha}\right)^{\frac{\alpha}{\alpha-1}} + 2\pi y \left(\frac{2\pi y}{\alpha}\right)^{\frac{1}{\alpha-1}} = \left(1 - \frac{1}{\alpha}\right) \left(\frac{2\pi}{\alpha}\right)^\beta y^\beta = c' y^\beta \quad (c' > 0). \]
calculus optimization of \(g(t)\).
3.3. Integrating over the interval \([t_0, t_0 + 1]\) gives \(F_\alpha(-iy) \ge C_1 e^{c' y^\beta}\) for large \(y > 0\).Thus the order of growth satisfies:\[ \rho(F_\alpha) = \limsup_{r \to \infty} \frac{\log \log M(r)}{\log r} \ge \beta = \frac{\alpha}{\alpha - 1}. \]
maximum modulus along a ray.
4. Conclusion:\(F_\alpha\) is an entire function of exact growth order \(\rho = \frac{\alpha}{\alpha - 1}\). Q.E.D.
2 and 3.