This is an exercise of Analysis III of Amann and Escher:
Note: here $X\subset\Bbb R^n$ is open.
I was trying to follow the hint, however I cant finish a proof. My work so far:
Let $x_0\in C$. Now let $J_0$ the cube centered at $x_0$ with side length $r>0$ such that $J_0\subset X$. Then $\Phi(J_0)$ is compact and convex, and from the mean value theorem we find that $$ \Phi(x)-\Phi(x_0)=\int_0^1\partial\Phi(x_0+t(x-x_0))(x-x_0)\, dt\tag1 $$ Hence $$ |\Phi(x)-\Phi(x_0)|\le |x-x_0|\int_0^1\|\partial\Phi(x_0+t(x-x_0))\|\, dt\tag2 $$ Then setting $\rho(r):=\max_{x\in J_0}\int_0^1\|\partial\Phi(x_0+t(x-x_0))\|\, dt$ we find that $$ \lambda_n(\Phi(J_0))\le \rho(r)^n\lambda_n(J_0)= (r\rho(r))^n\tag3 $$ Now note that $\|\partial \Phi(x_0)\|\le\max_{1\le k\le n}|\lambda_k|$, where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $\partial\Phi(x)_{\Bbb C}$ (where $\partial\Phi(x)_{\Bbb C}$ is the complexification of $\partial\Phi(x)$).
However Im stuck here. I cant see that $\lim_{r\to 0^+}\rho(r)=0$, what I see is that $\lim_{r\to 0^+}\rho(r)=\|\partial\Phi(x_0)\|$, what is not necessarily zero.
Some help will be appreciated, thank you.