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$G$ itself is a function, but the derivative $\Delta_y$ is taken in the sense of distributions, so the resulting object $\Delta_y G(x,y)$ need not be a function. (A simpler example: the Heaviside function $H(x)$ is a function, and its derivative in the ordinary sense is zero for all $x\neq 0$ and undefined at the origin. But one can instead interpret the ...

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First of all note that $\ln |x|$ is integrable near $0$ so that for any $\phi \in \mathcal{D}'$ the integral $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx$$ is defined, and that $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx.$$ Now, $$\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = ... 0 I hope to have found the answer. Let A \subseteq \mathbb{R}^n be an arbitrary set and choose an arbitrary open set \Omega \subseteq \mathbb{R}^n with A \subseteq \Omega (e.g. \Omega = \mathbb{R}^n). Set \mathcal{D} := \mathcal{D}(\Omega) and \mathcal{D}' := \mathcal{D}'(\Omega). We want to find a pre-dual for the space of distributions T \in ... 1 Suppose the spikes in the smooth approximation to \delta'(x) are located at x=-h and x=h. When \bar{x} \approx x+h, the smooth approximation to \delta'(x-\bar{x}) will be large and positive, so the integral will roughly pick up "something large" times f(x+h). Similary, for \bar{x} \approx x-h, the integral will pick up the same large factor ... 4 For K \subset \mathbb{R}^n compact, consider the space$$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$endowed with the seminorms$$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$for k \in \mathbb{N}^n. It is straightforward to show that \mathcal{D}_K is then a Fréchet ... 4 Suppose that we have a function \psi(x) so that$$ \delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1} $$Let 0\le\eta(x)\le1, \eta(x)=0 for |x|\le\frac12, and \eta(x)=1 for |x|\ge1. Define \phi_\lambda(x)=\psi(x)\eta(\lambda x). Then, since \phi_\lambda is 0 in a neighborhood of 0, we have$$ \begin{align} 0 ...

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You have the equation $$e^x(e^{-x}u)'=\delta_0+1.$$ First step is to notice that the function $x\to e^{x}$ is $C^\infty$, strictly positive everywhere, hence we can safely divide by it both sides of the equation without producing and/or losing solutions. Thus, we get $$(e^{-x}u)'=e^{-x}\delta_0+e^{-x}=\delta_0+e^{-x}.$$ It is easy to take the ...

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(I would usually post this as a comment, but apparently I need an account for that.) Suppose that $f \in \mathscr{S}'(\mathbb{R})$ is a solution to the equation $$f'-f = \delta_0 + 1.$$ Since you are following Gerd Grubb's textbook, have a look at the technique described on page 108, starting around equation (5.41). We can set the differential operator ...

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