Relation of integration of parabola over multiple volumes

Question

I have a multivariate function $f(x) = \sum_i^d x_i^2 c_i$ where all $c_i$ are non-negative and $x$ is a $d$-dimensional vector. I will integrate it over volume $A$ and volume $D$. Volume $A$ is determined by: $A = \{ ||x||_2 \leq C \}$. Volume $B$ is determined by: $B = \{ \max_i |x_i| \leq C\}$. Volume $D = B \setminus A$. Note that $B$ is the rectangle enclosing volume $A$.

For me it seems obvious that the mean of $f$ on $D$ must be larger than the mean of $f$ on $A$: $$\frac{\int_D f(x) dx}{\int_D dx} \geq \frac{\int_A f(x) dx}{\int_A dx}$$ This can be made clear using a drawing, since $D$ surrounds $A$. However, showing this statement formally seems difficult. Perhaps it is possible to do a coordinate transformation to some coordinate $u_i^2 = x_i^2 c_i$ and then to polar-coordinates, and compare the integrals there?

Answer 1

It is better to avoid polar coordinates. Both the hypercube $B$ and the hypersphere $A$ are left invariant by mapping $(x_1,\ldots,x_d)$ into $(x_{\sigma(1)},\ldots,x_{\sigma(d)})$ for any $\sigma\in S_d$ or just for any cyclic permutation of the coordinates. For instance, assuming $d=3$, $$ \int_{B}\left(c_1 x_1^2+c_2 x_2^2+c_3 x_3^2\right)\,d\mu =\frac{c_1+c_2+c_3}{3}\int_{B}(x_1^2+x_2^2+x_3^2)\,d\mu$$ and the same applies to $A$. The whole problem boils down to the computation of two simple integrals, $\int_{A}\|x\|^2\,d\mu$ and $\int_{B}\|x\|^2\,d\mu$. Just pick your favourite technique to find a closed form for them.

Your second inequality is just stating that the mean value of $f$ over $D$ is greater than the mean value of $f$ over $A$ - this requires no computation at all, it is a trivial consequence of the convexity of $f$. You may also notice that the Laplacian of $f$ is non-negative, hence $f$ is a subharmonic function.