# Relationship between volume and area

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 $$\mathbb {C}^n$$?

• 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. Commented Jun 16, 2020 at 21:43
• 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).
– YCor
Commented Jun 17, 2020 at 11:35
• 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, $B_\mu(r) = ???$ Commented Jun 17, 2020 at 18:17
• $Vol(B_\mu(r))$ is the volume of the ball with respect to the measure $\mu dV$ Commented Jun 18, 2020 at 8:43
• $B(r) ={z\in\mathbb {C^n} ;|z|<r}$ Commented Jun 18, 2020 at 8:45

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 $$2\pi t$$.
• Ok, the situation is clearer now. My answer above solves both your particular (case $n=1$) and general problems (cases $n > 1$). Commented Jun 18, 2020 at 10:30