To show principal value integral has order 1 as a distribution
I have proved that $|p.v\int \frac{\phi(x)}{x}dx|\leq C\displaystyle\sum_{|\alpha|\leq 1}|\partial^\alpha \phi|$ for some constant $C$
Now, this implies the order is at most 1, but how can I prove that the order is not 0.
I am trying to use contradiction. Suppose its order is 0. Then for any compact set $K\subseteq\mathbb{R}$, there exist constant $C>0$ such that $$|p.v.\int\frac{\phi(x)}{x}dx|\leq C \sup|\phi|$$ for all $\phi\in C^\infty_c(\mathbb{R})$ with $supp(\phi)\subseteq K$
But how to choose suitable $K$ and $\phi$ to get contradiction?
Define $f_k : \mathbb{R} \to \mathbb{R}$ by $$ f_k(x) = \begin{cases} kx, & |x| \leq \frac{1}{k} \\ \operatorname{sign}(x), & \frac{1}{k} \leq |x| \leq 1 \\ 0, & 1 < |x| \end{cases} $$
Even if $f_k \not\in C_c^\infty(\mathbb{R})$ we can apply the definition of $\text{p.v.}\frac{1}{x}$ on it: $$ \langle \text{p.v.}\frac{1}{x}, f_k \rangle := \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{1}{x} \, f_k(x) \, dx. $$
What is the result? What is the limit as $k \to \infty$? What is $\sup |f_k|$?
And finally: How can we create a $\phi_k \in C_c^\infty(\mathbb{R})$ that is close to $f_k$?