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I'm new around here (so suggestions about posting are welcome!) and want to give my contribution to this question, even though a bit old. I feel I need to because using the divergence theorem in this context is not quite rigorous. Strictly speaking $1/r$ is not even differentiable at the origin. So here's a proof using limits of distributions. Let ...


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The latter limit does not exist because you think about it pointwisely. Instead, you should check the limit in the topology of tempered distributions, where it will be seen to exist and be exactly what it should be. To this end, pick $\varphi \in \mathcal S(\Bbb R)$ and check whether (I'm omitting multiplicative constant factors) $\lim \limits _{M \to ...


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There are many equivalent ways to construct the topology on $C^\infty_c(\Bbb{R^d})$ that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive. First, we set our goal to be to find a topology on $C^\infty_c(\Bbb{R^d})$ that makes it a locally ...


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The first idea always works; this is basically a linear algebraic fact, since the BVP is essentially the system of linear equations $Lu=f,Ru=g$ where $R$ is the restriction to the boundary (or similar in the non-Dirichlet case) and $g$ is the given boundary function. Provided a Green's function exists, that approach will also always work. A Green's function ...


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The assumption: $\forall \phi $ (in the test class) $\langle f,\phi\rangle=0 \Leftrightarrow \langle g,\phi \rangle = 0$. The desired conclusion: $\exists c$ such that $\forall \phi$ the equality $\langle f,\phi \rangle =c\langle g,\phi\rangle$ holds. Plan of attack: Dispose of the trivial case in which $\forall \phi $ $\langle g,\phi\rangle=0$. ...


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If you take for example the Dirac delta distribution at $x=0$ and a continuous function $f(x)=\max(0,x)$, you find very quickly that the properties of Schwarz distributions break down. The product would have to be defined as the functional $\phi\mapsto f(0)\phi(0)$ for all $\phi\in C_0^\infty(\mathbb{R})$, which would give you zero. When you try to ...


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There are two ways to do it as far as I know, but the better way to do it is probably from definition (The other way is using conjugate Poisson kernel, see for example wikipedia: Hilbert transform) I am going to do it formally, but you could easily justify the calculation below. Since $p.v(1/x)$ is a tempered distribution, by definition, \begin{align} ...


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Well, derivation under the integral sign gives you directly $$\delta(y)=\int e^{-ixy}dx$$ $$\delta'(y)=-i\int xe^{-ixy}dx$$ $$\delta''(y)=-\int x^2 e^{-ixy}dx$$ $$\delta'''(y)=i\int x^3 e^{-ixy}dx$$ Then, by integration by parts, you can verify what it does as a distribution: $$\int \delta(y)f(y)dy=f(0)$$ $$\int ...



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