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 $)
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 $$ T(\varphi_n) = \int \frac{\varphi_n(x)}{|x|}\mathrm{d}x. $$