# An extremal property of points on the unit sphere of a 2-dimensional Banach space

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 $$y^*(x)=0$$?

• Note that the condition $y^*(x)=0$ is called Birkhoff Orthogonality
– erz
Oct 26, 2019 at 6:14
• 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 $y^*(x) \not= 0$. Oct 26, 2019 at 10:36
• 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. Oct 26, 2019 at 11:07
• Do you know a single non-Hilbertian case where it holds (for all $x,y$)?
– YCor
Oct 26, 2019 at 11:14

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 $$y^*(x) \not= 0$$.
• When something is true for exponent $2$, and fishy for larger exponents, $p=4$ is the first case to look at! Oct 26, 2019 at 11:09