Hyperfunction

Remarks

Up to canonical isomorphism, $\mathcal{B}(U)$ does not depend on the choice of $V$. Write $F_+$ and $F_-$ for the restrictions of $F$ to the parts of $V$ above and below the real axis. One thinks of $[F]$ as $F_+(x + i0) - F_-(x - i0)$. When both boundary values exist as distributions, this difference is a distribution, and every distribution on $U$ arises this way.

Example: on $U = \R$ with $V = \C$, let $F(z) = -1/(2\pi i z)$ on $\C \setminus \R$. The boundary values give $F_+(x + i0) - F_-(x - i0) = -\frac{1}{2\pi i}\left(\frac{1}{x + i0} - \frac{1}{x - i0}\right)$, and this is the Dirac delta. So $\delta = [F]$. A function holomorphic on all of $V$ gives the zero hyperfunction.

Non-example: for $U = (0,1)$, the set $V = \C$ does not qualify, because $V \cap \R = \R \neq U$. With that choice, $\mathcal{O}(\C \setminus (0,1)) / \mathcal{O}(\C)$ would also see singularities at the end points $0$ and $1$, which lie outside $U$.

In several variables the definition uses relative sheaf cohomology of the sheaf of holomorphic functions instead of a quotient.