[proofplan]
We prove the contrapositive at an arbitrary point of $U \cap H$. If $u$ does not vanish at such a point, continuity gives a full polydisc around that point on which $|u|$ is bounded below by a positive constant. On that polydisc, the weighted integral dominates the integral of $|z_n|^{-2}$ over a punctured complex disc, which diverges logarithmically. This contradicts the assumed finiteness of the singular weighted $L^2$ norm.
[/proofplan]
[step:Fix a point of the hypersurface and assume the value is nonzero]
Let $p \in U \cap H$ be arbitrary. Write
\begin{align*}
p = (p_1,\dots,p_{n-1},0),
\end{align*}
with the convention that the first block is absent when $n=1$. We prove $u(p)=0$.
Assume, for contradiction, that $u(p) \neq 0$. Since $u: U \to \mathbb{C}$ is holomorphic, it is continuous. Hence there exists $\rho>0$ such that the closed polydisc
\begin{align*}
P_\rho := \{z \in \mathbb{C}^n : |z_j-p_j|<\rho \text{ for } 1 \leq j \leq n-1,\ |z_n|<\rho\}
\end{align*}
satisfies $P_\rho \subset U$ and
\begin{align*}
|u(z)| \geq \frac{|u(p)|}{2}
\end{align*}
for every $z \in P_\rho$.
[guided]
We want to prove that $u$ vanishes at the chosen point $p \in U \cap H$. Because $p$ was arbitrary, this will prove vanishing on all of $U \cap H$.
Suppose instead that $u(p)\neq 0$. Holomorphic functions are continuous as maps $u: U \to \mathbb{C}$. Since $U$ is open and $p \in U$, there is a radius $\rho>0$ small enough that the polydisc
\begin{align*}
P_\rho := \{z \in \mathbb{C}^n : |z_j-p_j|<\rho \text{ for } 1 \leq j \leq n-1,\ |z_n|<\rho\}
\end{align*}
is contained in $U$. By continuity at $p$, after decreasing $\rho$ if necessary, we also have
\begin{align*}
|u(z)-u(p)| < \frac{|u(p)|}{2}
\end{align*}
for every $z \in P_\rho$. The [reverse triangle inequality](/theorems/2300) gives
\begin{align*}
|u(z)| \geq |u(p)|-|u(z)-u(p)| > \frac{|u(p)|}{2}
\end{align*}
for every $z \in P_\rho$. Thus a nonzero value on the hypersurface forces $u$ to stay uniformly away from zero in a full neighbourhood of that point.
[/guided]
[/step]
[step:Lower bound the singular integral on the polydisc]
Since $P_\rho \subset U$ and the integrand is nonnegative,
\begin{align*}
\int_U |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
\geq
\int_{P_\rho} |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z).
\end{align*}
Using the lower bound for $|u|$ on $P_\rho$, we obtain
\begin{align*}
\int_{P_\rho} |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
\geq
\frac{|u(p)|^2}{4}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z).
\end{align*}
[guided]
The weighted norm is finite by hypothesis, so we now show that the assumption $u(p)\neq 0$ forces it to be infinite. Since $P_\rho \subset U$ and the function
\begin{align*}
z \mapsto |u(z)|^2 |z_n|^{-2}
\end{align*}
is nonnegative and measurable on $U$, restricting the domain of integration can only decrease the integral:
\begin{align*}
\int_U |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
\geq
\int_{P_\rho} |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z).
\end{align*}
On $P_\rho$ we proved $|u(z)| \geq |u(p)|/2$, so squaring gives
\begin{align*}
|u(z)|^2 \geq \frac{|u(p)|^2}{4}.
\end{align*}
Multiplying by the nonnegative weight $|z_n|^{-2}$ and integrating over $P_\rho$ yields
\begin{align*}
\int_{P_\rho} |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
\geq
\frac{|u(p)|^2}{4}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z).
\end{align*}
Thus it remains only to prove that the last integral diverges.
[/guided]
[/step]
[step:Compute the logarithmic divergence in the normal variable]
Let
\begin{align*}
Q_\rho := \{z' \in \mathbb{C}^{n-1} : |z_j-p_j|<\rho \text{ for } 1 \leq j \leq n-1\}
\end{align*}
when $n\geq 2$, and let $Q_\rho$ be a one-point space with measure $1$ when $n=1$. Then $P_\rho = Q_\rho \times \{w \in \mathbb{C}: |w|<\rho\}$, where $w=z_n$.
By Tonelli's theorem for nonnegative [measurable functions](/page/Measurable%20Functions) (citing a result not yet in the wiki: Tonelli's Theorem),
\begin{align*}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
=
\mathcal{L}^{2n-2}(Q_\rho)
\int_{\{w \in \mathbb{C}: |w|<\rho\}} |w|^{-2}\, d\mathcal{L}^2(w),
\end{align*}
with the convention $\mathcal{L}^{0}(Q_\rho)=1$ when $n=1$.
Using polar coordinates in $\mathbb{C}\cong\mathbb{R}^2$, with $w=re^{i\theta}$ and
\begin{align*}
d\mathcal{L}^2(w)=r\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta),
\end{align*}
we get
\begin{align*}
\int_{\{w \in \mathbb{C}: |w|<\rho\}} |w|^{-2}\, d\mathcal{L}^2(w)
&=
\int_0^{2\pi}\int_0^\rho r^{-2} r\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta) \\
&=
\int_0^{2\pi}\int_0^\rho r^{-1}\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta) \\
&=
+\infty.
\end{align*}
Since $\mathcal{L}^{2n-2}(Q_\rho)>0$, it follows that
\begin{align*}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z)=+\infty.
\end{align*}
[guided]
The singularity is entirely in the normal coordinate $z_n$. We separate the tangential variables from this normal variable. Define
\begin{align*}
Q_\rho := \{z' \in \mathbb{C}^{n-1} : |z_j-p_j|<\rho \text{ for } 1 \leq j \leq n-1\}
\end{align*}
when $n\geq 2$. When $n=1$, there are no tangential variables, and we take $Q_\rho$ to be a one-point space with measure $1$. Then
\begin{align*}
P_\rho = Q_\rho \times \{w \in \mathbb{C}: |w|<\rho\},
\end{align*}
where $w=z_n$.
The function $(z',w)\mapsto |w|^{-2}$ is nonnegative and measurable, so Tonelli's theorem applies to the product measure decomposition of Lebesgue measure on the product polydisc. Thus
\begin{align*}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
=
\mathcal{L}^{2n-2}(Q_\rho)
\int_{\{w \in \mathbb{C}: |w|<\rho\}} |w|^{-2}\, d\mathcal{L}^2(w),
\end{align*}
where $\mathcal{L}^{0}(Q_\rho)=1$ in the case $n=1$. The tangential factor has positive finite measure because $Q_\rho$ is a nonempty product of open discs when $n\geq 2$, and by convention it has measure $1$ when $n=1$.
It remains to compute the one-dimensional complex integral. We use polar coordinates in $\mathbb{C}\cong\mathbb{R}^2$: write $w=re^{i\theta}$, where $0<r<\rho$ and $0\leq \theta<2\pi$. Under this substitution, two-dimensional Lebesgue measure transforms as
\begin{align*}
d\mathcal{L}^2(w)=r\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta).
\end{align*}
Therefore
\begin{align*}
\int_{\{w \in \mathbb{C}: |w|<\rho\}} |w|^{-2}\, d\mathcal{L}^2(w)
&=
\int_0^{2\pi}\int_0^\rho r^{-2} r\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta) \\
&=
\int_0^{2\pi}\int_0^\rho r^{-1}\, d\mathcal{L}^1(r)\, d\mathcal{L}^1(\theta).
\end{align*}
For every $\varepsilon \in (0,\rho)$,
\begin{align*}
\int_\varepsilon^\rho r^{-1}\, d\mathcal{L}^1(r)
=
\log \rho-\log \varepsilon,
\end{align*}
and this tends to $+\infty$ as $\varepsilon \downarrow 0$. Hence
\begin{align*}
\int_{\{w \in \mathbb{C}: |w|<\rho\}} |w|^{-2}\, d\mathcal{L}^2(w)=+\infty.
\end{align*}
Multiplying by the positive tangential measure gives
\begin{align*}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z)=+\infty.
\end{align*}
[/guided]
[/step]
[step:Derive the contradiction and conclude vanishing on the hypersurface]
Combining the preceding estimates gives
\begin{align*}
\int_U |u(z)|^2 |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
\geq
\frac{|u(p)|^2}{4}
\int_{P_\rho} |z_n|^{-2}\, d\mathcal{L}^{2n}(z)
=
+\infty,
\end{align*}
contradicting the hypothesis that the weighted integral is finite. Therefore the assumption $u(p)\neq 0$ is false, so $u(p)=0$. Since $p \in U \cap H$ was arbitrary, $u(z)=0$ for every $z \in U \cap H$.
[/step]
