Question:
Let $E$ a vector field ( $\mathbb{R}$ or $\mathbb{C}$) of dimension $n$ and $K$ a compact convex set of $E$ s.t. $Vect(K)=E$. Moreover for the set $End(E)$ we define the following norm: $\left \|u \right \|_\infty = max\left \{ ||u(e_i)||, i=1,...,n \right \}$ for $e_i$ a basis of E.
We know that the following set $F = \left \{ u \in End(E) | u(K) \subset K \right \}$ is bounded, now proved that $F$ is closed.
I try different ways: by avsurd approach, by matrix representation of linear mapping approach, by definition... but I am very stuck here. Any help will be very nice.
I think that I ve found a simple answer and I hope that it is correct I hope to read your feedback on it. take in count that I am just a student that is trying to answer a question he was stuck on it :-) .
My solution:
-First of all let remind that the set of all linears mapping on the vector field $E=E( \mathbb{F} )$ (with $\mathbb{F}$ the mathematic field equal to $\mathbb{R}$ or $\mathbb{C}$) of finite dimension $N$ is equivalent to the set of all matrix $\mathbb{F}^{N \times N}$ as we know to every linear mapping corresponds a matrix representation. So we can writte that the set $End(E)$ is equivalent to $\mathbb{F}^{N \times N}$. This help us because we know that the set $\mathbb{F}^{N \times N}$ is a vector space of dimension $N^2$.
-Now as we know that $||.||_{\infty} $ is a metric we get that $(\mathbb{F}^{N \times N}, ||.||_{\infty})$ is a finite vector space normed space.
-It is very useful because we know that any subspace of a finite dimentional normed vector space is closed. And of course $F = \left \{ u \in End(E) | u(K) \subset K \right \}$ is a subspace of $\mathbb{F}^{N \times N}$ (as it concentrates only on a specific linear mapping) hence $F$ is closed.
EDIT 1 : I get an answer from my professor and despite the down vote the answer is correct.
EDIT 2 : In the website we used the following property "all finite sub space of a vectorial normed space is closed."