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I am working with Strichartz's "A Guide to Distribution Theory and Fourier Transforms" (self-study -> not a homework question). He says none of the distributions that correspond to $1/|x|$ are non-negative. I would appreciate some help in order to understand why that is.

( we say the distribution f is non-negative if $\langle f,\psi \rangle \ge 0 $ for every test function $\psi$ that is non-negative with smooth test function that are compactly supported. Further we say a distribution $T$ corresponds to $1/|x|$ iff $T(\varphi)=∫\varphi(x)|x|dx$ for every test function $\varphi$ with $\varphi(0)=0 $)

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What are the distributions corresponding to $1/|x|$? Does he correspond different distributions to the same function? –  Hui Yu Sep 19 '12 at 14:42
    
@HuiYu yes, he definitely means for different distribution to correspond to the same function ! unfortunately understanding the whole set of distributions that can be associated with $1/|x|$ is part of what I struggle with. So far he only introduced the correspondence via integration which works because of the compact support requirement. (i.e. $\langle f, \psi \rangle = \int_\Omega f(x) \psi(x) \;dx)$ –  Beltrame Sep 19 '12 at 15:14
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@Beltrame: I have checked the book online: The author says a distribution $T$ corresponds to $1/|x|$ iff $T(\varphi)=\int\frac{\varphi(x)}{|x|} dx$ for every test function $\varphi$ with $\varphi(0)=0$. You should mention this in your question. –  Vobo Sep 19 '12 at 19:50
    
@Vobo thanks, ammended ! –  Beltrame Sep 21 '12 at 12:13

1 Answer 1

Let $\psi\geq0$ be a test function supported in $[-2,2]$ and is identically $1$ on $[-1,1]$. Let $\theta\geq0$ be another test function supported in $[-1,1]$, with $\theta(0)=1$. Define $$ \theta_n(x) = \theta(nx), \qquad\textrm{and}\qquad \varphi_n = \psi - \theta_n. $$ Clearly $\theta_n$ and $\varphi_n$ are nonnegative test functions, with $\varphi(0)=0$. The idea is to show that $T(\varphi_n)\to\infty$ as $n\to\infty$, and use $$ T(\psi) = T(\theta_n) + T(\varphi_n), $$ to force $T(\theta_n)\to-\infty$. But the claim is almost obvious since region where $\varphi_n=1$ grows closer and closer to $0$, and we have $$ T(\varphi_n) = \int \frac{\varphi_n(x)}{|x|}\mathrm{d}x. $$

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Nice, I was on the same track but didn't get the argument that $T(\theta_n)$ is really forced to go to $-\infty$. –  Vobo Sep 21 '12 at 20:36

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