Locally bounded and separately holomorphic implies holomorphic

Fix $a\in U$. Local boundedness and [L6] give $\epsilon >0$ and $M\ge 0$ with $B(a,\epsilon )\subseteq U$ and $\mathrm{\mid }f\mathrm{\mid }\le M$ on $B(a,\epsilon )$, after intersecting the bounding neighbourhood with a ball inside $U$. Put ${\rho }_k=\epsilon \mathrm{/}(2\sqrt{m})$; then ${\mathrm{\Delta }}_{\rho }(a)\subseteq B(a,\epsilon )\subseteq U$, since $\parallel z-a{\parallel }^2={\sum }_{k<m}\mathrm{\mid }{z}_k-a_k{\mathrm{\mid }}^2<m{\rho }_0^2={\epsilon }^2\mathrm{/}4$ for $z\in {\mathrm{\Delta }}_{\rho }(a)$ by [L5] and the dictionary.

The restriction of $f$ to ${\mathrm{\Delta }}_{\rho }(a)$ is separately holomorphic by [L3], since a slice domain inside ${\mathrm{\Delta }}_{\rho }(a)$ is an open subset of the corresponding slice domain inside $U$ and a restriction of a one-variable holomorphic function to an open subset is holomorphic; and $\mathrm{\mid }f\mathrm{\mid }\le M$ there by step 1.1.

Applying [L1] with $\theta =\frac{1}{2}$, the function $f$ is Lipschitz, hence continuous, on ${\bar{\mathrm{\Delta }}}_{\rho \mathrm{/}2}(a)$, and in particular on the open set ${\mathrm{\Delta }}_{\rho \mathrm{/}2}(a)$, which contains $a$ and is open by [L5] and [L6].

The sets ${\mathrm{\Delta }}_{\rho \mathrm{/}2}(a)$ obtained in step 3.1 as $a$ ranges over $U$ form an open cover of $U$ on each member of which $f$ is continuous, so $f$ is continuous on $U$ by [L4].

By step 4.1 the function $f$ is continuous on $U$ and separately holomorphic by hypothesis, so [L2] makes it holomorphic on $U$. The bound $M$ entered only through step 1.1 and [L1]; nothing above removes it, and this page proves no statement that would.