# A proof about the relation between differentiable map and linear transformation

#### ianchenmu

##### Member
Let $V⊂\mathbb{R}^n$ be an open, connected set and let $f: V→\mathbb{R}^m$ be a differentiable map. Suppose
that there exists a linear transformation $T∈L(\mathbb{R}^n;\mathbb{R}^m)$ such that $Df(x)=T$ for all $x∈V$. Prove
that there is a $c∈\mathbb{R}^m$ such that $f(x)=c+T(x)$ for all $x∈V$.

(I found $Df(x)=DT=T$, is that correct and what this means?)

#### jakncoke

##### Active member
Re: a proof about the relation between differentiable map and linear transformation

since f is differentiable at all points x.

if means that for an $\epsilon$ neighborhood of a $\in V$. f(x) = f(a) + T(x-a) for all points x in that $\epsilon$ neighborhood

so f(x) = f(a) + Tx - Ta or f(x) = f(a) - Ta + Tx
Now say G(x) = f(a) -Ta + Tx is a linear map over all $x \in V$, since it is linear, it is continous.

If we take another point b $\in V$, again for a suitable neighborhood around v. f(x) = f(b) + T(x-b) , with x being points in that neighborhood.

Since V is connected it is path connected, so we have a continous function $$r:[0,1] \to V$$ such that r(1) = b, r(0) = a.

Since parametrizations dont change continuity of functions G(r(t)) is also continuous.
if we vary t starting from 0 to 1, then G(r(0)) $\to$ G(r(1)), or G(a) $\to$ G(b)
since G is continuous. f(a) - Ta + Tx $\to$ f(b)-Tb + Tx, which imples f(a)-Ta = f(b) - Tb=c.

thus there is a constant c such that f(x) = c + Tx, for all x.