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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.
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    It's a common abuse of notation. If $f$ is a nice enough function, we define the distribution $f\cdot T$ by $(f\cdot T)[\varphi] = T[f\cdot \varphi]$. Then $f\cdot T$ is often written as $f(x)T(x)$. – 2017-01-01
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    So what the statement is saying is that $S(\phi(x)) = e^{i \alpha x}T(\phi(x))$ for all $\phi \in \mathcal{S}(\mathbb{R})$? – 2017-01-01
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    No, it's $S[\phi(x)] = T[e^{i\alpha x}\phi(x)]$. If for example $T = \delta'$, i.e. $T[\phi] = - \phi'(0)$ for all $\phi\in \mathcal{S}(\mathbb{R})$, then $S[\phi] = -i\alpha \phi(0) - \phi'(0)$, so $S = -i\alpha\delta + \delta'$. – 2017-01-01
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    The function $f_{\alpha}$ induces a tempered distribution $T \colon \varphi \mapsto \int \sin(\alpha x)\varphi(x)\,dx$. It's the Fourier transform of that distribution that is asked for. By a common abuse of notation, they identify the function $f_{\alpha}$ with the distribution it induces. – 2017-01-02

1 Answers 1

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

  1. 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 $\hat f_\alpha = \frac{\sqrt{2\pi}}{2i} (\delta_\alpha- \delta_{-\alpha})$.

  1. 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 $\alpha \to \pm \infty$.