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Let $\mu(z) dV_n$ be a measure in $\mathbb{C} ^n$. Let $B_n(r) := \{z \mid \|z\| < r\}$ be the ball of radius $r$ in $\mathbb C^n$, and $\partial B_n(r) $ be the corresponding sphere. In $\mathbb{C} $ how can we find the following inequality? $$ \operatorname{Vol}_{\mu}(B_1(r))=\int_{B_1(r)} \mu(z) dV_1(z)= \int_0^r\left(\int_{\partial B_1(t)} \mu dz\right)dt\geq \int_0^r \left[\int_{\partial B_1(t)}(\mu)^{ \frac{1}{2}} \right]^2\frac{1}{2\pi t} dt $$ And can we generalize this inequality in $\mathbb {C}^n$?

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    $\begingroup$ I tried to proofread this (for example, don't mix text and math in an equation; write $\operatorname{Vol}(B)$ $\operatorname{Vol}(B)$, not Vol$(B)$ Vol$(B)$), but I'm not sure I got everything right. For example, you asked whether you could generate the inequality in $\mathbb C^n$, and I changed that to 'generalize'. Please feel free to revert or re-edit if I got it wrong. $\endgroup$
    – LSpice
    Commented Jun 16, 2020 at 21:43
  • $\begingroup$ Please don't self-vandalize your own post. You can delete it if you like, as long as it has no answer (and also undelete it later). $\endgroup$
    – YCor
    Commented Jun 17, 2020 at 11:35
  • $\begingroup$ You say "Let $\mu(z)dV$ is a measure on $\mathbb C^n$". So I guess $\mu$ is a function (namely, the density of your measure w.r.t Lebesgue). Q: What does it then mean to say "Let $B_\mu(r)$ be a ball in $\mathbb C^n$ ?". More precisely, $B_\mu(r) = ???$ $\endgroup$
    – dohmatob
    Commented Jun 17, 2020 at 18:17
  • $\begingroup$ $Vol(B_\mu(r)) $ is the volume of the ball with respect to the measure $\mu dV$ $\endgroup$
    – Marouani
    Commented Jun 18, 2020 at 8:43
  • $\begingroup$ $B(r) ={z\in\mathbb {C^n} ;|z|<r} $ $\endgroup$
    – Marouani
    Commented Jun 18, 2020 at 8:45

1 Answer 1

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The problem can be solved via co-area formula and Jensen's inequality. We will do it Bourbaki style, i.e from $n$-dimensional case to particular case $n=1$.


Instead of $\mathbb C^n$, we can equivalently see the problem as a problem in $\mathbb R^m$, where $m=2n$ (i.e we isomorphically map real dimensions for each complex dimension). So, let $dV_m$ denote volume measure in $\mathbb R^m$ and $dS_{m-1}$ denote the corresponding surface area measure, i.e $(m-1)$-dimensional Hausdorff measure. The mapping $F: z \mapsto \|z\|$ on $\mathbb R^m$ has jacobian determinant $1$ (except at $z=0$, where it is undefined). Also note that for all $t \ge 0$, we have $$ F^{-1}(\{t\})=\{z \in \mathbb R^m \mid F(z) = t\} = \{z \in \mathbb R^m \mid \|z\| = t\} = \partial B_m(t). $$ By the coarea-formula (see Corollary 1.4, for example), we have $$ \begin{split} \int_{B_m(r)}\mu(z)dV_m(z) &= \int_{0}^r\left(\int_{F^{-1}(\{t\})}\frac{\mu(z)}{|Jac_F(z)|}dS_{m-1}(z)\right)dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)dS_{n-1}(z)\right)dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)S_{n-1}(\partial B_m(t))dt\\ &\ge\int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)^{1/2}}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)^2S_{m-1}(\partial B_m(t))dt\\ &= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)^{1/2}dS_{m-1}(z)\right)^2\frac{1}{S_{m-1}(\partial B_m(t))}dt, \end{split} $$ where the inequality is an applicaiton of Jensen's inequality on the convex function $x \mapsto x^2$ and the probability measure $A \mapsto S_{m-1}(A \cap \partial B(t))/S_{m-1}(\partial B(t))$.

In particular, if $n=2$, we have $m=2\cdot 1 = 2$, $dS_{m-1} = dS_1$ which is the arc-length measure, and so $S_1(\partial B(t)) =$ length or circle of radius $t$, which equals $2\pi t$.

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  • $\begingroup$ So there is no square! $\endgroup$
    – Marouani
    Commented Jun 18, 2020 at 9:23
  • $\begingroup$ The data in your question are still not clear. See my comments in the question above. $\endgroup$
    – dohmatob
    Commented Jun 18, 2020 at 9:30
  • $\begingroup$ Ok, the situation is clearer now. My answer above solves both your particular (case $n=1$) and general problems (cases $n > 1$). $\endgroup$
    – dohmatob
    Commented Jun 18, 2020 at 10:30
  • $\begingroup$ Thank you for the answer $\endgroup$
    – Marouani
    Commented Jun 18, 2020 at 11:30
  • $\begingroup$ OK, cool. If it anwsers your question, don't forgot to upvote and accept it. The is a button to the left for that :) $\endgroup$
    – dohmatob
    Commented Jun 18, 2020 at 13:44

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