Suppose $f$ a function holomorphic on the unit bidisk $\mathbb{D}\times \mathbb{D}$, such that $f$ is $\mathcal{C}^{\infty}$ on $]-1,1[\times\partial\mathbb{D}$, and has holomorphic extension on $\mathcal{U}\times \mathbb{D}$, where $\mathcal{U}$ is some neighborhood of $\partial\mathbb{D}$. Does $f$ has an extension in $\mathcal{V}\times \partial\mathbb{D}$, where $\mathcal{V}$ is some small neigborhood of $]-1,1[$, which is holomorphic with respect to the first variable. If not, can we exhibit a counter-example ?



a counter-example of a holomorphic extension
Take your favorite $C^\infty$ function $F(x)$ on the line that is not analytic and write $F(x)=\sum\_{k\ge 1}P\_k(x)$ where $P\_k$ are polynomials of degree $k$ such that for all $m\ge 0, d>0$, we have $|P\_k^{(m)}(x)|\le C\_{m,d}k^{-d}$ for all $x\in[-2,2],k\ge 1$. Then $|P\_k(z)|\le C\_{0,0}(1+|z|)^{k}$ for all $z\in\mathbb C$. Now define $f(z,w)=\sum\_{k\ge 1}P\_k(z)w^{k^2}$. This is analytic in $\mathbb C\times\mathbb D$ and $C^\infty$ on $[-2,2]\times \partial\mathbb D$ because all formal series for the derivatives
$\partial\_x^m\partial\_w^n f(x,w)=\sum\_{k\ge 1}P\_k^{(m)}(x)[k^2(k^2-1)\dots(k^2-n+1)]w^{k^2-n}$ converge uniformly (just use any $d\ge 2n+2$). However $f(x,1)=F(x)$.


