# p-Laplacian is intrinsically uniformly continuous

Definition- The operator $$G$$ is intrinsically uniformly continuous(IUC) with respect to $$x$$ in $$TM-\{0\}\times Sym TM$$ if there exists a modulus of continuity $$w_{G}:[0,\infty)\to [0,\infty)$$ with $$w_{G}(0^{+})=0$$ such that

$$G(\zeta , A) - G(L_{x,y} \zeta , L_{x,y} A) \leq w_{G} (d(x,y))$$

for any $$(\zeta,A) \in (T_{x}M-\{0\})\times Sym TM_{x}$$, and $$x,y \in M$$ with $$d(x,y), where $$i_{M}(x)$$ denotes the injectivity radius at $$x$$ of $$M$$.

Affirmation : the $$p$$-laplacian for $$p \geq 2$$ is IUC. Here the $$p$$-laplacian is

$$\Delta_{p}u=|\nabla u|^{p-2}tr\Big [ \Big ( I +(p-2)\frac{\Delta u}{|\nabla u|}\otimes \frac{\nabla u}{|\nabla u|} \Big ) D^{2}u \Big ].$$

The author asserts that for any $$(\zeta,A) \in (T_{x}M-\{0\})\times Sym TM_{x}$$, and $$x,y \in M$$ with \begin{align*} & d(x,y) for any $$v \in T_{y}M$$.

Thus the p-Laplacian is IUC with $$w_{G}=0$$.

I confess I have no idea how to prove it, i.e, what is the relation between the p-Laplacian is IUC and the identity above?. The unique idea is that one piece is invariant by parallel transport which is the trace.