# “Contradiction” to Bochner's theorem for distributions

I recently asked a question "For what values of $\lambda$ the distribution $(x-i\epsilon)^{\lambda}$ is positive?". User Marcel was kind enough to point out in his answer that one uses Bochner's theorem for that.

According to the theorem, $(x-i\epsilon)^{\lambda}$ is positive if and only if $\lambda\le 0$.

I did not know about this theorem and was not sure what test functions the authors meant, so I calculated it for simple cases $\exp(-x^2)$ and a bump function $$f(x)=\begin{cases}\exp\left(\frac{1}{x^2-1}\right),\quad \text{if}\; |x|<1\\ 0,\quad\quad\quad\quad\quad\quad \text{otherwise.} \end{cases}$$

In Gelfand, Shilov "Generalized functions, Vol I" the authors define $$(x-i0)^{-n}=x^{-n} + \frac{i\pi(-1)^{n-1}}{(n-1)!}\delta^{(n-1)}(x),$$ for positive integers $n$. If $n=2m$, i.e. $n$ is even, then $$(x^{-2m},\phi)=\int_0^\infty x^{-2m}\left(\phi(x)+\phi(-x)-2\left[\phi(0)+\frac{x^2}{2!}\phi''(0)+\dots+\frac{x^{2m-2}}{(2m-2)!}\phi^{(2m-2)}(0)\right]\right)dx.$$

Taking $n=2$ and $\phi(x)=\exp(-x^2)$ I get $$((x-i0)^{-2},\phi)=(x^{-2},\phi)=\\ \int_0^\infty \frac{\phi(x)+\phi(-x)-2\phi(0)}{x^2}dx=\int_0^\infty \frac{2\exp(-x^2)-2}{x^2}dx=-2\sqrt\pi<0$$

and for $\phi(x)=f(x)$ (bump function) I used Mathematica to get

$$((x-i0)^{-2},\phi)=(x^{-2},\phi)=\\ \int_0^\infty \frac{\phi(x)+\phi(-x)-2\phi(0)}{x^2}dx=\int_0^{\infty}\frac{2\exp\left(\frac{1}{x^2-1}\right)-2}{x^2}dx\approx-1.56537083433<0$$ which I guess is $-\frac{\pi}{2}$. Since I don't believe that Bochner or Schwartz have not considered the most trivial examples, I would like to ask

$$\textbf{what am I doing wrong?}$$

Both $\exp(-x^2)$ and $f(x)$ (bump function) are symmetric and their convolution is again the function of the same type, so in these examples it does not matter if we are considering a positive distribution or a distribution of positive type.

Background mathematics:

From Reed and Simon "I: Functional Analysis", 1980:

Theorem (Bochner-Schwartz). A distribution $T\in \mathscr{D}'(\mathbb{R}^n)$ is a distribution of positive type if and only if $T\in\mathscr{S}'(\mathbb{R}^n)$ and $T$ is the Fourier transform of a positive measure of at most polynomial growth.

Definition. A distribution $T\in\mathscr{D}'(\mathbb{R}^n)$ is of positive type if $T(\bar{\tilde{\varphi}}*\varphi)\ge 0$ for all $\varphi\in\mathscr{D}(\mathbb{R}^n)$. Here $\tilde{\varphi}(x)=\varphi(-x)$ and $*$ denotes convolution.

Definition. Let $K_n$ be an increasing family of compact sets with $\cup K_n^o = \mathbb{R}^n$. Put Frechet topology on $C_0^{\infty}(K_n)$ generated by the $||D^{\alpha}f||_{\infty}$ norms. The set $C_0^{\infty}$ with the inductive limit topology obtained by $C_0^{\infty}(\mathbb{R}^n)=\cup C_0^{\infty}(K_n)$ is denoted $\mathscr{D}(\mathbb{R}^n)$.

Definition of positive distribution (Lieb and Loss)
A distribution $T\in \mathcal{D}'(\mathbb{R}^n)$ is positive if $T(\phi) \ge 0\; \forall \phi \ge 0$, where $\phi\in \mathcal{D}(\mathbb{R}^n)$. Here $\phi \ge 0$ means that $\phi(x) \ge 0 \;\, \forall x\in \mathbb{R^n}$.