The norm Hessian as the second derivative of the finite coordinate norm along a line

For a finite index set $\iota$ and $p>4$, define for $v,h:\iota\to\mathbb R$

and

For $v,h:\iota\to\mathbb R$ and $t\in\mathbb R$ such that the point $v+t\cdot h$ is nonzero, this theorem shows that the real function

has derivative $\mathrm{normHessian}_p(v+t\,h,\,h)$ at $s=t$.

The same expression $\mathrm{normSlope}_p(v,h)$ is, for $v\ne0$, the ordinary first derivative at $s=0$ of $s\mapsto\bigl(\sum_i|v_i+sh_i|^p\bigr)^{1/p}$, the finite coordinate $p$-norm of $v+sh$ (differentiating $|v_i+sh_i|^p$ termwise and applying the chain rule for the outer power $1/p$, valid once $v\ne0$ makes the sum inside positive). This theorem supplies the corresponding fact one derivative further, so $\mathrm{normHessian}_p$ is exactly the second derivative — the curvature — of the finite coordinate $p$-norm along a line, at any point away from the origin.

Formalization Note The hypothesis $v+t\cdot h\ne0$ is used only to keep $\mathrm{powerSum}_p(v+th)=\sum_i|v_i+th_i|^p$ positive, so that raising it to the exponent $1/p-1$ behaves as the ordinary reciprocal power. The coordinatewise term $|x|^{p-2}x$ occurring inside $\mathrm{powerPair}$, and inside $\mathrm{normHessian}$'s own residual, is differentiable — with derivative $(p-1)|x|^{p-2}$ — at every real $x$ including $x=0$ (given $p>4$, as assumed here), so no individual coordinate of $v+th$ needs to avoid zero.