Looking for (information about) long diamonds

Question

I was given an open problem as a birthday present recently. While I can probably handle spoilers at this point, what I really want are literature and other references. Also acceptable would be suggestions for approaches. Ideally, search terms for the web are most welcome. (It is my lack of imagination for search terms for this problem that is most keeping me from doing preliminary research on it.)

Question: Where and how can I find out more about the Problem ($C_p \lt 2$?) below?

Problem: Let $n$ be a an integer parameter sufficiently large. Let $x$ and $y$ be vectors in $n$-dimensional real space. ( Do complex if it makes it easier. ) We will be using a family of $L_p$ norms $|z|$ for $p$ real and greater than $1$ ( $p$th root of the sum of the $p$th powers of the $n$ coordinates of the vector $z$), although other norms are also of interest. Fixing $p$ and its associated norm, there is a value $C_p$ ( which may be independent of $n$ and all other parameters ) such that for all $x$ and $y$ with $|x|=|y|=|x-y|=1$, one has $|x+y|\leq C_p$. The triangle inequality gives $C_p \leq 2$, and for $p=2$ we have $C_2 = \sqrt{3}$. Is it true that $C_p \lt 2$? (Can we also hope that lim sup over all $p \gt 1$ of $C_p$ is also less than $2$?)

$C_p$ to me represents the length of the long diagonal of a rhombus (diamond) with short diagonal and sides length $1$. Please retag this post as appropriate.

Gerhard "Still Enjoys His Birthday Presents" Paseman, 2015.11.20

Answer 1

Here is a simple short proof of one of the first main questions.

Claim. $C_p < 2$.

Proof. Let $C_p$ be above. We look at two cases: (i) $1 < p < 2$, and (ii) $p > 2$.

Case (i): From this short note we know that \begin{equation*} \|x+y\|_p^2 \le 2(\|x\|_p^2 + \|y\|_p^2) + (1-p)\|x-y\|_p^2. \end{equation*} Using the hypothesis, $\|x\|_p=\|y\|_p=\|x-y\|_p=1$, we obtain \begin{equation*} \|x+y\|_p^2 \le 5 - p\quad\implies C_p < 2. \end{equation*}

Case (ii): From Hanner's inequalities since $p > 2$ we know that \begin{equation*} \|x+y\|_p^p + \|x-y\|_p^p \le (\|x\|_p+\|y\|_p)^p + | \|x\|_p - \|y\|_p |^p. \end{equation*} Using the hypothesis, we obtain \begin{equation*} \|x+y\|_p \le (2^p - 1)^{1/p},\quad \implies C_p < 2. \end{equation*}

The only case we skipped here is $p=2$, but in that case $C_2=\sqrt{3}$ is already known to the OP.

Answer 2

There is a systematic approach to these type of problems which finds the best possible constants (when $n\to \infty$) but requires plenty (but easy) computations. I will mention the steps and if somebody is interested I can provide more details

What are we looking for? $$ \sup_{x,y \in L^{p}}\{ \|x+y\|_{L^{p}} : \|x\| _{L^p}=1, \|y\| _{L^p}=1, \| x-y\| _{L^{p}}=1 \} $$

Let us consider the following extremal problem in $L^{p}([0,1])$ $$ B(u,v,w) = \sup_{f,g \in L^{p}} \{ \int_{0}^{1}|f+g|^{p} : \int_{0}^{1}|f|^{p}=u, \int_{0}^{1}|g|^{p}=v, \int_{0}^{1}|f-g|^{p}=w \} $$

Then clearly the best possible $C_{p}=(B(1,1,1))^{1/p}$.

Properties of B

Now the advantage of considering the function $B$ is that it satisfies the following properties

1) $B$ is given in the convex cone $\Omega$ such that $(u,v,w)\in \Omega$ iff $w^{1/p}\leq u^{1/p}+v^{1/p}$, $u^{1/p}\leq v^{1/p}+w^{1/p}$ and $v^{1/p}\leq u^{1/p}+w^{1/p}$.

2) $B$ is a concave function in $\Omega$.

3) $B$ has a boundary condition in $\Omega$ i.e, $B(|x|^{p}, |y|^{p}, |x-y|^{p})=|x+y|^{p}$

4) $B$ is minimal among those who satisfy properties 1), 2) and 3).

Now it is clear how to find $B$: $B$ is a minimal concave function in $\Omega$ with a given boundary conditions 3). Notice that $B$ is 1-homogeneous so it is enough to find $B$ just in any section of the convex cone $\Omega$ (say for example $w=1$).

Your initial question reduced to a purely geometrical question: find the concave envelope of the boundary data.

After you find $B$ you can use it for your finite dimensional problem: if $d\mu$ is the uniform counting measure with weights $1/n$ then by Jensen's inequality

$$ \int |x+y|^{p} d\mu = \int B(|x|^{p},|y|^{p},|x-y|^{p})d\mu \leq B\left(\int |x|^{p} d\mu, \int |y|^{p}, \int |x-y|^{p} \right) $$ 1-homogeneity of $B$ and the right hand side gives you an upper bound.