[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 $z \in P_\rho$.[/step]

[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: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*}[/step]

[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: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*}[/step]

[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: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]