 Solution

Choose $R>0$ with $\mathrm{supp}\varphi \subset [-R,R]$. Evenness of $f$ changes the symmetrically truncated integral intoThe mean value theorem gives $|\varphi (x)-\varphi (-x)|\le 2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Since $f$ is continuous, it is bounded on $[0,R]$. Thus the Cauchy principal value exists andThis expression is linear in the test function and satisfiesFor each fixed compact support, choose one $R$ once and for all. The estimate proves that the even-weight principal-value distribution has order of a distribution at most one.

If $\varphi (0)=0$, the mean value theorem also gives $|\varphi (x)|\le |x|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Hence $f(x)\varphi (x)/x$ is absolutely integrable near zero and has compact support. The truncated integrals tend to its ordinary Lebesgue integral, so