Sokhotski-Plemelj formulas

Prove that

Prove the Sokhotski formulas:

Let $[a,b]$ be a finite interval in $\mathbb {R}$ and define, for $z$ in $D = \mathbb {C} - [a,b]$,

Show that $f(z)$ is analytic in $D$. Given $c$, $a < c < b$, calculate the limit of $f(z)$ as $z$ tends to $c$ from the upper half plane and as $z$ tends to $c$ from the lower half plane.

For each $z \in U = \left\{ z : \im z > 0\right\}$ define

Determine which points $a \in \mathbb {R}$ have the following property: there exist $\varepsilon > 0$ and an analytic function $f$ on $D(a,\varepsilon )$ such that $f(z) = g(z)$ for all $z \in U \cap D(a,\varepsilon )$.

Prove the Sokhotskii identity $1/(x\pm i0)=P\frac1x\mp \pi i\delta$.

In the space $D^{\prime }$ there exists the limit $\lim_{y \rightarrow +0} \frac{1}{x \pm i y}$ (it is denoted $\frac{1}{x \pm i 0}$), and the formula

holds (see problem 57).

Suppose $I = [a, b]$ is an interval on the real axis, $\varphi$ is a continuous function on $I$, and

Show that

exists for every real $x$, and find it in terms of $\varphi$. How is the result affected if we assume merely that $\varphi \in L^{1}$? What happens then at points $x$ at which $\varphi$ has right- and left-hand limits?