Assume that exists $T\in\hom\left(\mathbb{R}^{k},\mathbb{R}^{m}\right)$ such that $\left(Df\right)_{a}=T$ forall $a\in\mathbb{R}^{K}$. Prove that exists $T\in\hom\left(\mathbb{R}^{k},\mathbb{R}^{m}\right)$ such that $f\left(x\right)=Tx+f\left(0\right)$ forall $x\in\mathbb{R}^{k}$.
can anyone give me any clue? thanks!
If I understood correctly, you have, by hypothesis, that there exists a linear map $T$ such that, for every $a\in\mathbb R^k, Df_a=T$; and what you have to prove is that, there exists another linear map $S$ such that $f=S+f(0)$, right?
So, for the solution take the function $g=f-T$ which has derivative identically zero, that is, for every $a\in\mathbb R^k, D(f-T)_a=0$. By the connectedness of the domain of $f-T$, we get that $f-t$ is constant. Evaluating at $x=0$ we get $(f-T)(0)=f(0)$. Since $f-T$ is constant, we get, for every $x\in\mathbb R^k$, $f(x)=T(x)+f(0)$. So you have to take $S=T$.