Fourier Transform of Various Tempered Distributions

Question

Let $T$ be a tempered distributions, and let $S(x) = e^{i\alpha x}T(x)$, where $\alpha \in \mathbb{R}$.

1. Find a formula that relates the Fourier transforms $\hat S$ and $\hat T$.
2. Find the Fourier transform $\hat f_\alpha$ of $f_\alpha (x) = \sin \alpha x$.
3. Find $\lim_{\alpha \to \infty} \alpha \hat f_\alpha$ in the sense of (tempered) distributions.

Answer 1

1. Choose $\phi \in \mathcal{S}(\mathbb{R})$. Then $$\hat S \phi (k) = \hat T (e^{i\alpha x}\phi (k)) = T(\widehat{e^{i\alpha x}\phi}(k)) = T\big(\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{i\alpha t}\phi(t)e^{-ikt} dt\big) = T(\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \phi(t)e^{-it(k-\alpha)}dt\big) = T(\hat \phi(k-\alpha) = T(\tau_\alpha \hat \phi (k)) = \tau_\alpha \hat T\pi(k)$$

So $\hat S=\tau_\alpha \hat T$.

2. Note that $f_\alpha(x) = \sin \alpha x = \frac{1}{2i}(e^{i\alpha x}-e^{-i\alpha x})$. Using part 1 ($\alpha \in \mathbb{R}$), we infer that $$\hat f_\alpha = \frac{1}{2i} \tau_\alpha\hat 1 - \frac{1}{2i} \tau_{-\alpha}\hat 1 = \frac{1}{2i} \tau_\alpha\sqrt{2\pi} \delta - \frac{1}{2i}\tau_{-\alpha}\sqrt{2\pi}\delta = \frac{1}{2i}\sqrt{2\pi} \delta_\alpha - \frac{1}{2i}\sqrt{2\pi} \delta_{-\alpha}$$

We see then that $\hat f_\alpha = \frac{\sqrt{2\pi}}{2i} (\delta_\alpha- \delta_{-\alpha})$.

3. Again choose $\phi \in \mathcal{S}(\mathbb{R})$. $$ \lim_{\alpha \to \infty} \hat f_\alpha \phi= \lim_{\alpha \to \infty} \frac{\sqrt{2\pi}}{2i}(\delta_\alpha - \delta_{-\alpha})\phi = \lim_{\alpha \to \infty} \frac{\sqrt{2\pi}}{2i} (\phi(\alpha) - \phi(-\alpha)) =0$$ where the last is because $\phi \in \mathcal{S}(\mathbb{R})$, and thus should decay to zero as $\alpha \to \pm \infty$.