An inequality for mean Lebesgue norms

Question

Denote by $\mathcal{Q}$ the set of all cubes in $\mathbb{R}^n$ and $\ell(Q)$ for the side-length of the cube $Q$. What is the proof of the following proposition:

For all $Q\in \mathcal{Q}$ and $0

Answer 1

First we get rid of $p$ putting $g(y)=|f(y)|^p$ for each $y\in Q$ and taking $p$-th powers if both sides of the inequality. Thus we have to show that given a cube $Q\in \mathcal{Q}$ and $0

From your formulation is not clear what is $|Q|$ and $|R|$, about which integral (Riemann’s or Lebesque’s) you are asking, and what a kind of a function is $f$. But I expect than the following simply observations hold when the function $f$ (and so the function $g$) is sufficiently good (I guess it suffices to assume that $g$ in integrable on $Q$).

Let $k$ be the smallest integer such that $s’=l(Q)/2^k\le s$. Then $s’\le s<2s’$ and the cube $Q$ is a union of a family $\mathcal F(Q)$ consisting of $2^{kn}$ mutually interior disjoint cubes of side $s’$, like Rubik’s cube. Since

$$\int_{Q}g(y) dy=\sum_{ P\in \mathcal F(Q)}\int_{P}g(y) dy,$$

there exists a cube $P\in \mathcal F(Q)$ such that $$\int_{P}g(y) dy\ge\frac 1{2^{kn}} \int_{Q}g(y) dy.$$

Let $R$ be an arbitrary cube with side $s$ such that $P\subset R\subset Q$. Since $g(y)=|f(y)|^p$ for each $y\in Q$, the function $g$ is non-negative, so

$$\int_{P}g(y) dy\le \int_{R}g(y)dy.$$

Thus

$$\int_{Q}g(y) dy\le 2^{kn} \int_{P}g(y) dy\le \left(\frac{l(Q)}{s’}\right)^n \int_{R}g(y)dy\le \left(\frac{2l(Q)}{s}\right)^n \int_{R}g(y)dy.$$

Since each smaller homothetic copy of the cube $Q$ can cover at most one vertex of $Q$, I guess that assuming that the cube $R$ is homothetic to the cube $Q$, we cannot diminish multiplier $2^n$ at the right hand side to $2^n-\varepsilon$ for any $\varepsilon>0$, as shows the example when the function $g$ maps $Q=[0,1]^n$ into $[0,1]$ and for each vertex $v$ of $Q$ exist sufficiently small neighborhoods $V_v\subset U_v$ such that $g|V_v$ is close to $1$ and $g|Q\setminus\bigcup V_v$ is close to $0$. For instance, I guess that we can take a function $g(y)\prod_{i=1}^n\cos^{2k} \pi y_i$ (where $y=(y_1,\dots,y_n)$) for sufficiently large natural $k$.

On the other hand, if we relax the condition that the cube $R$ is homothetic to the cube $Q$ then the covering arguments show that there exists a cube $R$ (not necessarily contained in $Q$) with side-length $s=s(r)$ such that

$$\int_{P}g(y) dy\le r\int_{R}g(y)dy$$

provided a cube of side-length $l(Q)$ can be covered by $r$ cubes of side-length $s$. This naturally arise a question which is the smallest such $s=s(r)$ for given $r$. For $n=1$ clearly, $s(r)= l(Q)/r$, but already for $n=2$ it becomes, as far as I know, a combinatorial geometry problem, which is unsolved due to its complexity. See, for instance known best upper bounds for $s(r)=\frac{l(Q)}{\sqrt{A(n)}}$ at Erich Friedman’s Packing Center.