connection between affine map and differential

Question

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!

What is your definition of the differential of a function, when it exists? — Bernard, Nov 23 '20 at 16:42
$Df$ such that $\underset{x\rightarrow0}{\lim}\frac{f\left(a+x\right)-f\left(a\right)-Dfx}{\left\Vert x\right\Vert _{\mathbb{R}^{k}}}=0$ — Yechiel Merzbach, Nov 23 '20 at 16:44
The definition of a differentiable map $f$ at the point $x$ of its domain is that there exists a linear map $\ell$ such that $f(x+h)=f(x)+\ell(h)+o\bigl(|h|\bigr)$. — Bernard, Nov 23 '20 at 18:46

score 0 · Accepted Answer · edited Nov 23 '20 at 20:19

0

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 $S=T$.

edited Nov 23 '20 at 20:19

amWhy

209,954

answered Nov 23 '20 at 19:00

Emmanuel C.

144

why the derivative is identically zero? – Yechiel Merzbach Nov 23 '20 at 19:57
The linearity of the derivative implies that $D(f-T)_a=Df_a-DT_a$, also we have, by hypothesis, $Df_a=T$, and, since $T$ is linear, for every $a\in\mathbb R^k$, $DT_a=T$. So $D f_a-DT_a= T-T=0$. Better? :) – Emmanuel C. Nov 23 '20 at 20:57
yes, thanks. and why $(f-T)(0)=f(0)$? – Yechiel Merzbach Nov 23 '20 at 21:16
Great! Oh, that's because linear functions map $0_k\in\mathbb R^k$ to $0_m\in\mathbb R^m$ ($T(0_k)=T(0\cdot 0_k)=0\cdot T(0_k)=0_m$, here $0$ denotes the escalar $0\in\mathbb R$). So $(f-T)(0_k)=f(0_k)-T(0_k)=f(0_k)-0_m=f(0_k)$. – Emmanuel C. Nov 23 '20 at 22:16
Thank you! Its good – Yechiel Merzbach Nov 24 '20 at 10:23

connection between affine map and differential

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