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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 $v$?

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    Yes, I misspelled, it is, in fact, $a\in\mathbb{R}^m$.2012-03-17

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

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    Just did it! :D2012-03-17
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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 $f$.