Hartogs triangle

Counterexample

The Hartogs triangle is the domain $$H = \set{(z, w) \in \C^2 : \abs{z} < \abs{w} < 1}.$$ It is a bounded pseudoconvex domain whose closure $\overline{H}$ has no basis of pseudoconvex neighbourhoods.

Check that $H$ is pseudoconvex. On $H$ the function $g(z, w) = z/w$ is holomorphic with $\abs{g} < 1$. Let $$\varphi(z, w) = -\log\big(1 - \abs{z/w}^2\big) - \log\big(1 - \abs{w}^2\big) - \log \abs{w}.$$ The first two terms are plurisubharmonic, since $-\log(1 - \abs{h}^2) = \sum_{k \ge 1} \abs{h^k}^2 / k$ for a holomorphic $h$ with $\abs{h} < 1$, and the last term is pluriharmonic. The boundary of $H$ consists of the points with $\abs{z} = \abs{w} \le 1$ and the points with $\abs{w} = 1$, $\abs{z} \le 1$. All three terms are nonnegative on $H$. Near the first part the first term tends to $+\infty$, except at the origin, where the last term does. Near the second part the second term tends to $+\infty$. So $\varphi$ is a continuous plurisubharmonic exhaustion function, and $H$ is pseudoconvex, hence a domain of holomorphy by Oka's theorem.

Check the neighbourhoods. Let $U$ be an open set containing $\overline{H} = \set{\abs{z} \le \abs{w} \le 1}$. Since $\overline{H}$ is compact, $U$ contains, for some $\varepsilon > 0$, the set $\set{\abs{z} < \varepsilon, \abs{w} < 1 + \varepsilon}$ and the set $\set{\abs{z} < 1 + \varepsilon, 1 - \varepsilon < \abs{w} < 1 + \varepsilon}$. The union of these two sets is a Hartogs figure. Let $f$ be holomorphic on $U$. The Laurent series of $w \mapsto f(z, w)$ on the annulus has no negative powers when $\abs{z} < \varepsilon$. Its coefficients are holomorphic in $z$, so by the identity theorem it has none when $\abs{z} < 1$. So $f$ extends holomorphically to the bidisc $\set{\abs{z} < 1, \abs{w} < 1 + \varepsilon}$. Hence any domain of holomorphy that contains $\overline{H}$ contains the whole unit bidisc, and the pseudoconvex neighbourhoods of $\overline{H}$ cannot shrink to $\overline{H}$.