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Let $(X,\|\cdot\|)$ be a 2-dimensional real Banach space and $S=\{x\in X:\|x\|=1\}$ be its unit sphere. Assume that $S$ is smooth in the sense that for any $y\in S$ there exists a unique functional $y^*:X\to\mathbb R$ such that $y^*(y)=1=\|y^*\|$. This unique functional $y^*$ will be called the supporting functional at $y$.

Let $x,y\in S$ be points such that $$\|y-x\|+\|y+x\|=\max\{\|s-x\|+\|s+x\|:s\in S\}.$$

Question. Is $y^*(x)=0$?

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  • $\begingroup$ Note that the condition $y^*(x)=0$ is called Birkhoff Orthogonality $\endgroup$
    – erz
    Commented Oct 26, 2019 at 6:14
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    $\begingroup$ Numerical computations suggest that this is not true for general $\ell^p$-norms on $\mathbb{R}^2$. For instance, for $p = 4$ and $x = (1,0)$, I wrote an Octave script which yields that $y \approx (0.8204, 0.8600)$ maximizes the expression you are interested in. But $y^*$ is the pointwise product of $y$ and $|y|^{p-2}$, so $y^*(x) \not= 0$. $\endgroup$
    – Jochen Glueck
    Commented Oct 26, 2019 at 10:36
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    $\begingroup$ Such a norm $\|u\|$ is Fréchet differentiable for $u\neq0$, with $d\|u\|=u^*$, and the Lagrange equation for the maximizer $y\in S$ of $\|s+x\|+\|s-x\|$ for $s\in S$ is $$(y-x)^*+(y-x)^*=\lambda y^*$$ I can't imagine how this could impliy $\langle y^*,x\rangle=0$ without special assumptions on the norm. $\endgroup$
    – Pietro Majer
    Commented Oct 26, 2019 at 11:07
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    $\begingroup$ Do you know a single non-Hilbertian case where it holds (for all $x,y$)? $\endgroup$
    – YCor
    Commented Oct 26, 2019 at 11:14

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The answer is no, in general.

For a counterexample, consider the $\ell^p$-norm on $\mathbb{R}^2$ with $p=4$, and let $x = e_1 = (1,0)$.

We first note that the vectors $e_2 = (0,1)$ and $-e_2$ do not maximize the function \begin{align*} f: S \ni s \mapsto \|s-x\|+\|s+x\| \in [0,\infty). \end{align*}

Indeed, we have $f(e_2) = f(-e_2) = 2\cdot\|(1,1)\| = 2\cdot 2^{1/4}$. On the other hand, consider the vector $s_0 = 2^{-1/4} \cdot (1,1) \in S$. Then \begin{align*} f(s_0) = \|(2^{-1/4} - 1, 2^{-1/4})\| + \|(2^{-1/4}+1, 2^{-1/4})\| > 2\cdot 2^{1/4} \end{align*} (where I, admittedly, used a computer to check the latter inequality). So $f$ is neither maximized at $e_2$ nor at $-e_2$.

Hence if $y$ maximizes $f$, then the first entry of $y$ is non-zero. But $y^* \in \mathbb{R}^2$ is the pointwise product of $y$ and $|y|^{p-2}$. Thus, the first entry of $y^*$ is also non-zero, which shows that $y^*(x) \not= 0$.

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    $\begingroup$ When something is true for exponent $2$, and fishy for larger exponents, $p=4$ is the first case to look at! $\endgroup$
    – Pietro Majer
    Commented Oct 26, 2019 at 11:09

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