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For $1\leq p<\infty$, we denote by $\ell_p^n$ the vector space $\mathbb{R}^n$ endowed with the p-norm

$$\|(a_1,\dots,a_n)\|_p=\left(\sum_{i=1}^n|a_i|^p\right)^{\frac{1}{p}}.$$

For a normed space $X$ we define the Schaffer constant of $X$ as $$\mathfrak{S}(X)=\inf_{u,v\in X,\ ||u||=||v||=1}\max\{\|u+v\|,\|u-v\|\}.$$ It is a fact that $\mathfrak{S}(\ell^n_p)=\min\{2^{1/p},2^{1-1/p}\}$.

We denote the unit sphere of $\ell^n_p$ by $S_{\ell^n_p}$ and we shall see it as a subset of the euclidean space $(\mathbb R^n,\|\cdot\|_2)$ (so we can make use of the usual inner product). Then we can ask

What is the infimum of the p-norm distance between two orthogonal vectors in $S_{\ell^n_p}$? More precisely, we wish to compute $$ \alpha^n_p= \inf\{\|u-v\|_p: u,v\in S_{\ell^n_p} \mbox{ and } u\perp v\}.$$

Since $\|(1,0,\dots,0)-(0,1,\dots,0)\|_p=2^{1/p}$ and

$$\|(2^{-1/p},2^{-1/p},0,\dots,0)-(2^{-1/p},-2^{-1/p},0,\dots,0)\|_p=2^{1-{1/p}},$$ we have that $\alpha^n_p\leq \mathfrak{S}(\ell_p^n)$.

Let us see that the reverse inequality holds when $n=2$, so $\alpha^2_p=\mathfrak{S}(\ell^n_p)$. Indeed, if $n=2$ then $$ a\|\cdot\|_p\leq\|\cdot\|_2\leq b\|\cdot\|_p,$$ where $a=\min\{2^{1/2-1/p},1\}$ and $b=\max\{2^{1/2-1/p},1\}$. Given $u$ and $v$ orthogonal vectors in $S_{\ell^n_p}$, we have that $$\|u-v\|_p^2\geq\frac{1}{b^2}\|u-v\|_2^2= \frac{1}{b^2}(\|u\|_2^2+\|v\|_2^2)\geq\frac{a^2}{b^2}(\|u\|_p^2+\|v\|_p^2)=2\frac{a^2}{b^2}=(\min\{2^{1/p},2^{1-1/p}\})^2,$$ where the last equality is obtained by checking it for the cases $p\geq2$ and $p<2$.

I have no idea whether $\alpha^n_p\geq\min\{2^{1/p},2^{1-1/p}\}$ for $n\geq3$. I can't apply the same technique as above, since for general $n$ the sharpest norm equivalence with the 2-norm is for the constants $a=\min\{n^{1/2-1/p},1\}$ and $b=\max\{n^{1/2-1/p},1\}$.

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