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



An extremal property of points on the unit sphere of a 2-dimensional Banach space
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 $y^\*(x) \not= 0$.


