Let $\phi :\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $\phi (t) = 0$ if, only if, $t=0$. Let $F:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a map such that for all point $p$ in the compact set $K\subset \mathbb{R}^n$ satisfy:
$$\exists \epsilon>0 \mbox{ e } c>0; \parallel F(x)-F(y)\parallel \leq c\cdot\phi (\parallel x-y\parallel) \forall x,y \in \mathbb{B}(p,\epsilon)$$
show that exists a constant $\lambda$ such that
$$\parallel F(x)-F(y)\parallel \leq \lambda\cdot\phi (\parallel x-y\parallel) \forall x,y \in k$$
All we have to do is prove that $\frac{ ||F(x)-F(y)|| }{\phi(||x-y||)}$ is bounded for $x,y \in K$. Suppose not. Then we have: $\forall M>0 , M\in\Bbb{N}; \exists x_M, y_M \in K$ such that $\frac{ ||F(x_M)-F(y_M)|| }{\phi(||x_M-y_M||)} > M$. However $K$ is compact, and thus we can find a convergent subsequence of $\{x_M\}_M \subset K$, say $x_{M_i} \to x^* \in K (i \to \infty)$. And since the same subsequence $\{y_{M_i}\}_i \subset K$, we can also find a “sub-subsequence” of $\{M_i\}_i$, say $\{M_{i_k}\}$, such that $y_{M_{i_k}} \to y^* \in K (k\to\infty)$. Thus, since a subsequence of a convergent sequence is also convergent, we also have $x_{M_{i_k}} \to x^*(k\to \infty)$.
From now on we write $M_{i_k}$ as $M$ for simplicity.
It is easy to see that $F$ is continuous on $K$, thus by Weiestrass’s theorem, we have that $F$ is also bounded on K. Say $||F|| < R \in \Bbb{R}$
Thus $ M < \frac{ ||F(x_M)-F(y_M)|| }{\phi(||x_M-y_M||)} < \frac{2R}{\phi(||x_M - y _M||)} \implies \phi(||x_M - y_M||) < \frac{2R}{M} \to 0 (M\to\infty)$. Hence $\phi(||x_M - y_M||) \to 0$, and by the continuity of $\phi$ and that $\phi(x) = 0 \iff x=0$ we get that $||x_M - y_M||\to0$, thus $x^* = y^*$. We can now apply the hypothesis for the point $x^*$, and thus we have $\epsilon>0, c>0$ such that $\frac{ ||F(x)-F(y)|| }{\phi(||x-y||)} < c$ if $x,y \in B_{\epsilon}(x^*)$, which is a contradiction since $M > c \implies \frac{ ||F(x_M)-F(y_M)|| }{\phi(||x_M-y_M||)}>c$, and $x_M, y_M \in B_{\epsilon}(x^*)$ for large enough M.