Here, $\mathscr{S}(\mathbb{R})=\{f\in C^{\infty}(\mathbb{R}):\lVert f\rVert_{(N,\alpha)}<\infty\text{ for all non-negative integers }N\text{ and }\alpha\}$ is the Schwartz space ; where $$ \lVert f\rVert_{(N,\alpha)}=\sup_{x\in\mathbb{R}}\left\lbrack (1+|x|)^N \cdot\left\lvert f^{(\alpha)}(x)\right\rvert \right\rbrack. $$
Let $W_0:\mathscr{S}(\mathbb{R})\to\mathbb{C}$ be a map defined as :
$$ \langle W_0,\phi\rangle=\frac{1}{\pi}\lim_{\epsilon\to 0}\int_{\epsilon\leq |x|\leq 1}\frac{\phi(x)}{x}dx+\frac{1}{\pi}\int_{|x|>1}\frac{\phi(x)}{x}dx, $$for $\phi\in\mathscr{S}(\mathbb{R})$.
Our goal is to prove that $W_0$ is a tempered distribution.
For any $0<\epsilon<1$, we notice that $$ \int_{\epsilon\leq |x|\leq 1}\frac{1}{x}dx=\int_{\epsilon\leq x\leq 1}\frac{1}{x}dx+\int_{-1\leq x\leq -\epsilon}\frac{1}{x}dx=\int_{\epsilon\leq x\leq 1}\frac{1}{x}dx+\int_{1\leq x\leq \epsilon}\frac{1}{x}dx=0 .$$
Thus, we can write, for $\phi\in\mathscr{S}(\mathbb{R})$, $$ \langle W_0,\phi\rangle=\frac{1}{\pi}\lim_{\epsilon\to 0}\int_{\epsilon\leq |x|\leq 1}\frac{\phi(x)-\phi(0)}{x}dx+\frac{1}{\pi}\int_{|x|>1}\frac{\phi(x)}{x}dx. \hspace{2cm}(\ast)$$
We note that $$ \left\lvert\int_{|x|>1}\frac{\phi(x)}{x}dx\right\rvert\leq\int_{|x|>1}|\phi(x)|dx=\int_{|x|>1}\frac{|\phi(x)|\cdot(1+|x|)^2}{(1+|x|)^2}dx\leq\lVert \phi\rVert_{(2,0)}\cdot\left(\int_{\mathbb{R}}\frac{dx}{(1+|x|)^2} \right)=2\lVert \phi\rVert_{(2,0)}. $$
So, the second term in the right side of the equality in $(\ast)$ is bounded by $\frac{2}{\pi}\cdot\lVert\phi\rVert_{(2,0)}$.
Prof. Grafakos has directly written in his book $\textbf{Classical Fourier Analysis}$ that the fraction $\frac{\phi(x)-\phi(0)}{x}$ is controlled by $\lVert\phi'\rVert_{\infty}$ so we have the estimate $$ |\langle W_0,\phi\rangle|\leq\frac{2}{\pi}\cdot\lVert\phi'\rVert_{\infty}+\frac{2}{\pi}\cdot\lVert\phi\rVert_{(2,0)}. $$
I am not able to prove that the fraction $\frac{\phi(x)-\phi(0)}{x}$ is controlled by $\lVert\phi'\rVert_{\infty}$. It seems very easy and trivial but I am unable to prove this rigorously. Please give me some hints or solution. Thanks.
Since $\phi\in \mathbb{S}$, we can write $\phi(x)-\phi(0)=\phi'(\xi(x))x$ for $\xi(x)$ between $0$ and $x$. Then, we have
$$\begin{align} \left|\int_{\varepsilon\le |x|\le 1}\frac{\phi(x)-\phi(0)}{x}\,dx\right |&=\left|\int_{\varepsilon\le|x|\le1} \phi'(\xi(x))\,dx\right|\\\\ &\le\int_{\varepsilon \le |x|\le 1}|\phi'(\xi(x))|\,dx\\\\ &\le ||\phi'||_{\infty}\int_{\varepsilon \le |x|\le 1}1\,dx\\\\ &\le 2 ||\phi'||_{\infty} \end{align}$$
as was to be shown!