Let \(\alpha>1/2\) and define \[ f_\alpha(x)=\int_{\mathbb R}(1+\xi^2)^{-\alpha}e^{2\pi i x\xi}\,d\xi. \] Without explicitly evaluating the integral, determine and prove which of the following properties hold:
- \(\lim_{|x|\to\infty}|f_\alpha(x)|=0\); (ii) \(f_\alpha\in L^2(\mathbb R)\).
- Without appealing to general Fourier-transform properties, prove that (i) \(f_\alpha\in C(\mathbb R)\) and (ii) \(f_\alpha\) is bounded on \(\mathbb R\).
Solution
For every \(\alpha>1/2\), all four asserted properties hold.
Set \[ g_\alpha(\xi)=(1+\xi^2)^{-\alpha}. \] Because \(g_\alpha(\xi)\asymp |\xi|^{-2\alpha}\) as \(|\xi|\to\infty\) and \(2\alpha>1\), one has \[ g_\alpha\in L^1(\mathbb R). \] Also \(g_\alpha\in L^2(\mathbb R)\) because \[ |g_\alpha(\xi)|^2=(1+\xi^2)^{-2\alpha}\asymp |\xi|^{-4\alpha}, \] and \(4\alpha>1\).
1. \(f_\alpha(x)\to0\) as \(|x|\to\infty\).
The function \(f_\alpha\) is the Fourier transform of the \(L^1\) function \(g_\alpha\), with the convention \[ \widehat g(x)=\int_{\mathbb R}g(\xi)e^{2\pi i x\xi}\,d\xi. \] Hence the Riemann–Lebesgue lemma gives \[ f_\alpha(x)\longrightarrow0 \qquad(|x|\to\infty). \]
2. \(f_\alpha\in L^2(\mathbb R)\).
Since \(g_\alpha\in L^2(\mathbb R)\), Plancherel’s theorem gives \[ \|f_\alpha\|_2=\|g_\alpha\|_2<\infty \] for this Fourier-transform normalization. Thus \(f_\alpha\in L^2(\mathbb R)\).
3. Directly, \(f_\alpha\) is bounded.
For every \(x\in\mathbb R\), \[ |f_\alpha(x)| \le \int_{\mathbb R}(1+\xi^2)^{-\alpha}\,d\xi =\|g_\alpha\|_1<\infty. \] The right-hand side is independent of \(x\), so \(f_\alpha\) is bounded.
4. Directly, \(f_\alpha\) is continuous.
Let \(x_n\to x\). For each \(\xi\), \[ e^{2\pi i x_n\xi}g_\alpha(\xi) \longrightarrow e^{2\pi i x\xi}g_\alpha(\xi). \] Moreover, \[ \left|e^{2\pi i x_n\xi}g_\alpha(\xi)\right|=g_\alpha(\xi), \] and \(g_\alpha\in L^1(\mathbb R)\). Dominated convergence therefore yields \[ f_\alpha(x_n)\longrightarrow f_\alpha(x). \] Hence \(f_\alpha\in C(\mathbb R)\).