Define $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ by $f(u,v)=\displaystyle\sum_{i=1}^n|u_i-v_i|^p$.
Assume $p>1$.
We'd like to maximize $f$ under the following constraints: $$ \left\{\begin{array}{l} g_1(u,v):=f(u,0)=1 \\ g_2(u,v):=f(0,v)=1 \\ g_3(u,v):=\langle u,v\rangle=0\end{array}\right. . $$
The problem is very similar to one posed by Andre Porto here: Help minimizing function
Only I want to maximize, not minimize.
I tried setting up Lagrange multipliers, but I got no further than staring at the system. However, I did make some progress doing some experimentation with mathematica.
It seems that the maximizer $(u,v)$ will satisfy $$u_1 = -v_1$$, $$u_2=v_2$$ $$\dots{}$$ $$u_n=v_n$$
and $$|u_2|=|v_2|=...=|u_n|=|v_n|$$
This implies $$f(u,v)=\frac{2^p}{1+(n-1)^{1-\frac{p}{2}}}$$ which then must be the maximum. I am fairly sure this is right(I tried it with a bunch of $n$ and $p$ values). But how can I prove it? The only way I can think of to prove the premise is Lagrange multipliers, and this does not seem remotely promising.
Partial results are welcome! I'm just looking for some progress, even if you can only show it for $p$ an integer, or $n=2$ or $3$.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$tl;dr The case $n = 2$ can be handled with trigonometry and elementary calculus; the constrained extreme values are $2$ and $2^{p-1}$.
Fix $p > 1$. The "unit circle" in the $p$-norm, a subset of $\Reals^{2}$, can be parametrized as a polar graph: $$ u_{1} = r(t) \cos t,\qquad u_{2} = r(t) \sin t, $$ with $r$ a positive function and $t$ real. Since $1 = |u_{1}|^{p} + |u_{2}|^{p} = r(t)^{p}\bigl(|\cos t|^{p} + |\sin t|^{p}\bigr)$, we have $$ r(t)^{p} = \frac{1}{|\cos t|^{p} + |\sin t|^{p}}. \tag{1} $$ There are two choices of $v$ orthogonal to $u$, given by $$ v_{1} = -r(t) \sin t,\qquad v_{2} = r(t) \cos t $$ and its negative. By the sum formulas for $\cos$ and $\sin$, $$ \sqrt{2} \cos(t - \tfrac{\pi}{4}) = \cos t + \sin t,\qquad \sqrt{2} \sin(t - \tfrac{\pi}{4}) = \sin t - \cos t. $$ For either choice of $v$, equation (1) gives \begin{align*} f(u, v) &= |u_{1} - v_{1}|^{p} + |u_{2} - v_{2}|^{p} \\ &= r(t)^{p} \bigl(|\cos t + \sin t|^{p} + |\sin t - \cos t|^{p}\bigr) \\ &= r(t)^{p} 2^{p/2} \bigl(|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}\bigr) \\ &= 2^{p/2} \frac{|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}}{|\cos t|^{p} + |\sin t|^{p}}. \tag{2} \end{align*} For $p \neq 2$, the function $$ \phi(t) := |\cos t|^{p} + |\sin t|^{p} $$ is even, periodic with period $\frac{\pi}{2}$, strictly monotone on $[0, \frac{\pi}{4}]$, and has extreme values $\phi(0) = 1$ and $\phi(\frac{\pi}{4}) = 2^{1 - (p/2)}$. (If $p = 2$, then $\phi(t) = 1$ for all $t$.) Consequently, when subject to the constraints $$ |u_{1}|^{p} + |u_{2}|^{p} = |v_{1}|^{p} + |v_{2}|^{p} = 1,\qquad u_{1}v_{1} + u_{2}v_{2} = 0, $$ we have:
If $1 < p < 2$, then $\phi$ has a minimum value of $1$ at $0$, a maximum value of $2^{1 - (p/2)}$ at $\frac{\pi}{4}$, and $$ 2^{p-1} \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2. $$
If $2 < p$, then $\phi$ has a minimum value of $2^{1 - (p/2)}$ at $0$, a maximum value of $1$ at $\frac{\pi}{4}$, and $$ 2 \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2^{p-1}. $$
In other words, $$ 1 + 2^{p-2} - |1 - 2^{p-2}| = \min(2, 2^{p-1}) \leq f(u, v) \leq \max(2, 2^{p-1}) = 1 + 2^{p-2} + |1 - 2^{p-2}|. $$
These values can obviously be achieved in higher dimensions by "padding with $0$'s", and the "natural generalizations" do no better: If $n = 2m$ is even, for example, the vectors $$ u = m^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{0, \dots, 0}_{m}),\qquad v = m^{-1/p}(\underbrace{0, \dots, 0}_{m}, \underbrace{1, \dots, 1}_{m}) $$ satisfy $f(u, v) = 2$, and the vectors $$ u = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{1, \dots, 1}_{m}),\qquad v = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{-1, \dots, -1}_{m}) $$ satisfy $f(u, v) = 2^{p-1}$. (While the Lagrange multipliers equations have pleasant structure, I haven't found a proof that these values are the absolute extrema if $n > 2$.)