**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 $d(x,y)<min\{i_{M}(x),i_{M}(y)\}$, 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)<min\{i_{M}(x),i_{M}(y)\},\\ & \langle(L_{x,y}\zeta \otimes L_{x,y}\zeta)L_{x,y} A .v, \ v \rangle_{y}= \langle (\zeta \otimes \zeta) A .L_{y,x}v, \ L_{y,x}v \rangle_{x}\\ \end{align*} 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.