In an example in Reed & Simon's book on functional analysis they prove that the Cauchy principal part integral $$\mathscr{P}\Big(\frac1x\Big): f \mapsto \lim_{\epsilon \rightarrow 0} \int_{|x|\geq \epsilon} \frac1x f(x) dx \tag{1}$$ is a tempered distribution.
Let $f \in \mathscr{S}(\mathbb{R}^1)$ (a Schwarz function). They first rewrite the integral in (1) as $$\int_{|x|\geq \epsilon} \frac1x f(x) dx = \int_\epsilon^\infty \frac{f(x)-f(-x)}{x}dx.$$ They then say that since $[f(x)-f(-x)]/x \rightarrow 2f'(0)$ as $x \rightarrow 0$, we can write $$\mathscr{P}\Big(\frac1x\Big)(f) = \int_0^\infty \frac{f(x)- f(-x)}{x}dx$$ which proves it is finite. I understand everything until this point.
From here, they say since $$\left|\frac{1}{x}[f(x) - f(-x)]\right| \leq \frac1x\int_{-x}^x|f'(t)|dt \leq 2 \|f'\|_\infty \tag{2}$$ we have $$\left| \mathscr{P}\left(\frac1x\right)(f)\right| \leq 2 \int_0^1\|f'\|_\infty dx + \left|\int_1^\infty(xf(x))\frac{dx}{x^2}\right| \leq 2\|f\|_{1,0} + \|f\|_{0,1}. \tag{3}$$ Thus $\mathscr{P}(1/x)$ is a distribution.
Here they are using the notation $$\|\varphi\|_{\alpha, \beta} = \sup_{x \in \mathbb{R}^n} |x^\alpha D^\beta \varphi(x)| $$ to be the seminorm on $\mathscr{S}(\mathbb{R}^n)$.
I do not understand how they establish the inequalities in (3) or how they follow from (2).
Let $$g(x) = \frac{f(x)- f(-x)}{x}$$ We have $$ \begin{align} \|f\|_{1,0} &= \sup_{x\in\Bbb{R}} |xf(x)|\\ \|f\|_{0,1} &= \sup_{x\in\Bbb{R}} |f'(x)| = \|f'\|_\infty \end{align} $$ by $(2)$ $$|g(x)| \le 2 \|f\|_{0,1}$$ for $x>0$. So $$\left|\int_0^1 g(x)dx\right|\le \int_0^1 |g(x)|dx \le 1\cdot 2 \|f\|_{0,1} = 2 \|f\|_{0,1}$$ Also for $x>0$ $$|g(x)| \le \left|\frac{f(x)}{x}\right| + \left|\frac{f(-x)}{x}\right|$$ So $$\left|\int_1^\infty g(x)dx\right|\le \int_1^\infty |g(x)|dx \le \int_1^\infty \left|\frac{f(x)}{x}\right|dx + \int_1^\infty \left|\frac{f(-x)}{x}\right|dx$$ Now $$\int_1^\infty \left|\frac{f(x)}{x}\right|dx = \int_1^\infty \left|\frac{xf(x)}{x^2}\right|dx \le \|f\|_{1,0} \int_1^\infty \frac{dx}{x^2} = \|f\|_{1,0}\cdot 1 = \|f\|_{1,0}$$ and similarly $$\int_1^\infty \left|\frac{f(-x)}{x}\right|dx\le \|f\|_{1,0}$$ so $$\int_1^\infty |g(x)|dx \le 2 \|f\|_{1,0}$$ and altogether $$\left|\int_0^\infty g(x)dx\right| \le \int_0^\infty |g(x)|dx \le 2 \|f\|_{0,1} + 2 \|f\|_{1,0}$$