Osgood's lemma: continuous and separately holomorphic implies holomorphic

Fix $a\in U$. By [L6] there is $\epsilon >0$ with $B(a,\epsilon )\subseteq U$; put ${\rho }_k=\epsilon \mathrm{/}(2\sqrt{m})$ for every $k<m$. If $z\in {\mathrm{\Delta }}_{\rho }(a)$ then $\parallel z-a{\parallel }^2={\sum }_{k<m}\mathrm{\mid }{z}_k-a_k{\mathrm{\mid }}^2<m{\rho }_0^2={\epsilon }^2\mathrm{/}4$ by [L5] and the dictionary, so ${\mathrm{\Delta }}_{\rho }(a)\subseteq B(a,\epsilon )\subseteq U$.

The restriction of $f$ to ${\mathrm{\Delta }}_{\rho }(a)$ is continuous, and it is separately holomorphic there: for $b\in {\mathrm{\Delta }}_{\rho }(a)$ and $k<m$ the slice domain inside ${\mathrm{\Delta }}_{\rho }(a)$ is an open subset of the slice domain inside $U$, on which the slice is holomorphic by hypothesis, and a restriction of a holomorphic function of one variable to an open subset is holomorphic.

Put $r_k={\rho }_k\mathrm{/}2$, so $r_k<{\rho }_k$. By [L1] applied on ${\mathrm{\Delta }}_{\rho }(a)$ there are coefficients $c_{\alpha }$ with $\mathrm{\mid }{c}_{\alpha }\mathrm{\mid }\le M{\prod }_{k<m}{r}_k^{-{\alpha }_k}$ and $f(z)={\sum }_{\alpha }{c}_{\alpha }(z-a{)}^{\alpha }$ for every $z\in {\mathrm{\Delta }}_r(a)$.

By [L2] the sum of that series is holomorphic on ${\mathrm{\Delta }}_r(a)$; by step 2.1 it is $f$ there, so $f$ is complex differentiable at every point of ${\mathrm{\Delta }}_r(a)$, in particular at $a$. Since $a\in U$ was arbitrary, [L4] makes $f$ holomorphic on $U$.

For the equivalence, let $f$ be continuous on the open $U$. If $f$ is holomorphic then it is separately holomorphic by [L3]; if it is separately holomorphic then step 2.1 gives the local power-series representation and step 3.1 gives holomorphy; and if it is locally such a sum then [L2] makes it holomorphic on a polydisc about each point, hence on $U$ by [L4]. So the three conditions are equivalent.