If $f$ is a integrable function, then, $f$ is a tempered distribution by
$$T_f(\varphi)=\int f\varphi,\quad \varphi \text{ Schwartz function }$$
However, I read that the function $f(x)=1/x,\, x\in\mathbb{R}$, despite not being integrable, is still a tempered distribution. Why? Where I cand found a reference in which this result appears? Thank.
Best Answer
It can define a tempered distribution, in the sense of the Cauchy principle value. Define the distribution $f$, sometimes called $\mathrm{P.\!\!V.} \frac 1 x$, by the rule $$\begin{align*} f[\varphi] &:= \lim_{\varepsilon \searrow 0} \int_{|x|>\varepsilon} \frac{\varphi(x)}{x} \, \mathrm{d} x \\ &\equiv \lim_{\varepsilon \searrow 0} \int_{-\infty}^{-\varepsilon} \frac{\varphi(x)}{x} \, \mathrm{d} x + \int_{\varepsilon}^{+\infty} \frac{\varphi(x)}{x} \, \mathrm{d} x \\ &\equiv \lim_{\varepsilon \searrow 0} \int_\varepsilon^{+\infty} \frac{\varphi(x) - \varphi(-x)}{x} \, \mathrm{d} x \\ &=2 \varphi'(0) \end{align*}$$ for $\varphi \in \mathcal{C}_c^\infty(\mathbb{R})$. For $\varphi$ a Schwarz function instead, we end up with instead $$ |f[\varphi]| \le 2 \sup_{x \in \mathbb{R}} |\varphi'(x)| \qquad |f[\varphi]| \le 2 \sup_{x \in \mathbb{R}} |x \varphi (x)| $$ (I don't know if we can do better than a bound in the Schwarz case.)
This particular example is discussed in more detail on the Wikipedia page linked above. (I got it from Vretblad's Fourier Analysis and Its Applications as I'm working through it recently.)