Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $v$?

share|improve this question
I'm a little confused. Do you mean an $a\in\mathbb{R}^m$ instead of $\in\mathbb{R}^n$? –  Neal Mar 17 '12 at 0:18
Yes, I misspelled, it is, in fact, $a\in\mathbb{R}^m$. –  Marra Mar 17 '12 at 0:24

2 Answers 2

up vote 4 down vote accepted

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 $c\in\mathbb{R}^m$.

share|improve this answer
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, $f(0)=a+k$). –  Marra Mar 17 '12 at 0:36
@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. –  Amitesh Datta Mar 17 '12 at 0:58
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! –  Marra Mar 17 '12 at 1:08
@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. –  Amitesh Datta Mar 17 '12 at 2:39
Just did it! :D –  Marra Mar 17 '12 at 11:20

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 $f$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.