Principal Value of One over x is Distribution

Theorem
Let $\phi \in \map \DD \R$ be a test function.

Let $T : \map \DD \R \to \C$ be a mapping such that:


 * $\forall \phi \in \map \DD \R : \map T \phi = \PV \frac {\map \phi x} x \rd x : = \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \frac {\map \phi x} x \rd x$

where $PV$ denotes the Cauchy principal value.

Then $T$ is a distribution.

Proof
Let $\phi \in \map \DD \R$ be a test function with a support on $\closedint {-a} a$.

Then:

Furthermore:

Existence of the limit
Since $\phi \in \map \DD \R$, the integral exists for any $\epsilon$.

Hence, the limit exists.

Thus, we can rewrite $T$ as:


 * $\ds \map T \phi = \int_{-a}^a \int_0^1 \dfrac {\d \map \phi {t x} } {\d \paren {t x} } \rd t \rd x$

Linearity
Follows from Riemann Integral Operator is Linear Mapping.

Continuity
By Convergent Sequence Minus Limit, we can shift the sequence to set its limit to zero.

Let $\mathbf 0 : \R \to 0$ be a zero mapping.

Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence with support on $\closedint {-a} a$ such that it converges to $\mathbf 0$:


 * $\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$

Then:

Take the limit $n \to \infty$.

Then:


 * $\map T {\mathbf 0} = 0$

By definition, $\PV \frac 1 x$ is a distribution.