Principal value distribution one over x

Changing $x$ to $-t$ in the negative interval, which follows directly by reflecting its tagged partitions, gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }\phi (x)\mathrm{/}xdx={\int }_{\epsilon }^R(\phi (t)-\phi (-t))\mathrm{/}tdt$ for $0<\epsilon <R$. The quotient extends continuously to $t=0$ with value $2{\phi }^{\mathrm{\prime }}(0)$ by the definition of derivative. Its modulus is at most $2\sup \mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$ by F2. F3 therefore gives an integral on $[0,R]$, and the omitted interval has integral tending to zero, bounded by a constant times $\epsilon \sup \mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$. This proves existence of the principal value.

The expression is linear in $\phi$, since each truncation is linear and limits preserve finite sums. For any fixed compact $K\subseteq \mathbb{R}$, choose $R>0$ with $K\subseteq [-R,R]$. The integral representation from step 1.1 gives $\mathrm{\mid }\mathrm{pv}(1\mathrm{/}x)(\phi )\mathrm{\mid }\le 4R{p}_1(\phi )$ for $\phi \in {\mathcal{D}}_K$, by separately bounding real and imaginary integrals; the sharper $2R$ bound also follows from the complex integral triangle inequality but is unnecessary. F1 proves continuity. An even test gives zero because the quotient vanishes. Tests supported away from zero give their ordinary integral against $1\mathrm{/}x$, and the zero test gives zero. The limit requires symmetric removal at zero, with no assertion about independently varying two cutoffs. $\mathrm{\square }$