4
$\begingroup$

A function $f:U\subset\mathbb{R}^n \rightarrow \mathbb{R}^m$, $U$ open, is differentiable in $p \in U$ if there exists a linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $f(p+v)=f(p)+T(v)+R(v)$, where $R(v)$ satisfies $lim _{v\rightarrow 0}\dfrac{R(v)}{|v|}=0$, for all $v\in\mathbb{R}^n$ with $p+v\in U$.

That said, if $f$ as above is differentiable and $f'(x)=T$, $\forall x\in\mathbb{R}^n$. I need to show that there is an $a\in\mathbb{R}^n$ such that $f(x)=Tx+a$.

The problem I'm having is, how do I show that $R(v)=0$ for all $v$?

  • 0
    I'm a little confused. Do you mean an $a\in\mathbb{R}^m$ instead of $\in\mathbb{R}^n$?2012-03-17
  • 0
    Yes, I misspelled, it is, in fact, $a\in\mathbb{R}^m$.2012-03-17

2 Answers 2

4

Let $g:\mathbb{R}^n\to\mathbb{R}^m$ be defined by the rule $g(x)=Tx+a$ for all $x\in\mathbb{R}^n$ and let $f:\mathbb{R}^n\to \mathbb{R}^m$ be a differentiable function such that $f'(x)=T$ for all $x\in\mathbb{R}^n$.

If $h=f-g$, then prove that $h'(x)=0$ for all $x\in\mathbb{R}^n$. Finally, prove that this implies $h(x)=c$ for all $x\in\mathbb{R}^n$ and some $c\in\mathbb{R}^m$.

  • 0
    Does $h(x)$ needs to be equal zero? Isn't it sufficient for it to be a constant? I just need to proof existence. Like, if $h(x)=k$, then $f(x)=g(x)+k$ which means $f(x)=Tx+a+k$, and $a+k$ is a constant (in fact, $f(0)=a+k$).2012-03-17
  • 0
    @GustavoMarra You are right! We cannot conclude that $h=0$ identically on $\mathbb{R}^n$; we can only conclude that $h$ is constant on $\mathbb{R}^n$. I have corrected this mistake in my answer above.2012-03-17
  • 0
    Thanks to you, Amitesh! I'm sorry I cannot vote up your answer yet, need 15 rep. but I will definitely do so when I am able to!2012-03-17
  • 0
    @GustavoMarra You now have 16 reputation points (I upvoted your question which gave you five reputation points) ... Of course, if you are satisfied with my answer, then you might wish to accept it.2012-03-17
  • 0
    Just did it! :D2012-03-17
4

Note that $a=f(0)$.

Since the derivative of $x\mapsto Tx+a$ is $T$ and the derivative is linear, all you need to show is that a function with zero derivative is constant. This can be done by using the Mean Value Theorem on the component functions of $f$.