Relationship between volume and area

Question

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 $\mathbb {C}^n$?

Answer 1

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 $n=1$.

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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 $2\pi t$.