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:

 $\ds \map \phi x$ $=$ $\ds \map \phi 0 + \int_0^x \dfrac {\d \map \phi x} {\d x} \rd x$ $\ds$ $=$ $\ds \map \phi 0 + x \int_0^1 \dfrac {\d \map \phi {t x} } {\d \paren{t x} } \rd t$

Furthermore:

 $\ds \int_{a \mathop \ge \size x \mathop > \epsilon} \frac 1 x \rd x$ $=$ $\ds \int_\epsilon^a \frac 1 x \rd x + \int_{-a}^{-\epsilon} \frac 1 x \rd x$ $\ds$ $=$ $\ds \int_\epsilon^a \frac 1 x \rd x + \int_a^\epsilon \frac 1 x \rd x$ $\ds$ $=$ $\ds 0$

Existence of the limit

 $\ds \int_{\size x \mathop > \epsilon} \frac {\map \phi x} x \rd x$ $=$ $\ds \int_{a \mathop \ge \size x \mathop > \epsilon} \frac {\map \phi x} x \rd x$ $\ds$ $=$ $\ds \map \phi 0 \int_{a \mathop \ge \size x \mathop > \epsilon} \frac 1 x \rd x + \int_{a \mathop \ge \size x \mathop > \epsilon} \rd x \int_0^1 \dfrac {\d \map \phi {t x} }{\d \paren {t x} } \rd t$ $\ds$ $=$ $\ds 0 + \int_{a \mathop \ge \size x \mathop > \epsilon} \rd x \int_0^1 \dfrac {\d \map \phi {t x} }{\d \paren {t x} } \rd t$ $\ds$ $=$ $\ds \int_{a \mathop \ge \size x \mathop > \epsilon} \rd x \int_0^1 \dfrac {\d \map \phi {t x} }{\d \paren {t x} } \rd t$

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

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:

 $\ds \size {\map T {\phi_n} }$ $=$ $\ds \size {\int_{-a}^a \int_0^1 \dfrac {\d \map {\phi_n} {t x} } {\d \paren {t x} } \rd t \rd x}$ $\ds$ $\le$ $\ds 2a \cdot 1 \cdot \sup_{x \mathop \in \R} \size {\map {\phi_n'} x}$

Take the limit $n \to \infty$.

Then:

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

$\blacksquare$