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Ok, you said you knew how to prove it. Others might not. And you may like this better than what you have (it's the second thing I always think of when this comes up, and I like it a lot better than the first thing I think of...) Say $u$ is a distribution and $u'=0$. By definition $u(\phi')=0$ for any test function $\phi$. Hence $u(\phi)=0$ for any test ...

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HINT: Split the integral as $$\int_{-\infty}^{\infty} \text{sgn} (x-y)e^{-|x-y|}f(y) \, dy = \int_{-\infty}^x e^{-(x-y)}f(y)\,dy-\int_x^{\infty} e^{(x-y)}f(y) \, dy$$ and use Leibnitz's Rule for differentiating under an integral. SPOILER ALERT: SCROLL OVER SHADED AREA TO SEE ANSWER

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No, for instance if $S(x) = x$ then $\hat S = i\delta'$ (up to a multiplicative constant depending on how you define the fourier transform). More generaly if $S$ is a polynomial $S(x) = \sum a_k x^k$ then $\hat S = \sum a_k i^k\delta^{(k)}$

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By the (multi-dimensional) Leibniz formula, we have \begin{align*} \partial^{\alpha} \phi_k&= \sum_{\beta+\gamma = \alpha} \underbrace{c_{\beta,\gamma} (\partial^{\beta} \phi) \cdot (\partial^{\gamma} (1-\psi(k \bullet)))}_{=:S_{\beta,\gamma}} \end{align*} for some constants $c_{\beta,\gamma}$ (which can be calculated expliticly). If $\gamma=0$, ...

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Notation: For a fixed function $f$ defined on $\mathbb{R}^{n}$ and $y\in\mathbb{R}^{n}$, we write $\tilde{f}(x)=f(-x)$ and $(\tau^{y}f)(x)=f(x-y)$. I am assuming that you are familiar with the following result. Theorem. For $\phi\in\mathcal{S}(\mathbb{R}^{n})$, the tempered distribution $u\ast\phi$ coincides with a $C^{\infty}(\mathbb{R}^{n})$ ...

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APPROACH 1: We can show that $J_{\nu}(kr)$ and $J_{\nu}(sr)$ are orthogonal by appealing to the governing ODE $$\frac{d}{dr}\left(r\frac{dJ_{\nu}(kr)}{dr}\right)+\left(k^2r-\frac{\nu}{r}\right)J_{\nu}(kr)=0 \tag 1$$ $$\frac{d}{dr}\left(r\frac{dJ_{\nu}(sr)}{dr}\right)+\left(s^2r-\frac{\nu}{r}\right)J_{\nu}(sr)=0 \tag 2$$ Multiplying $(1)$ by ...

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Tempered distributions have Fourier transforms. If $P(D)\,F=0$, take the Fourier transform to get $P(\xi)\,\hat F(\xi)=0$ for all $\xi\in\mathbb{R}$. Since $P(\xi)\ne0$ if $\xi\ne0$ the support of $\hat F$ is $\{0\}$. Can you take it fromhere?

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