Derivatives and Linear transformations Let G be a non-empty open connected set in $R^n$, $f$ be a differentiable function from $G$ into $R$, and $A$ be a linear transformation from $R^n$ to $R$. If $f$ '($a$)=$A$ for all $a$ in $G$, find $f$ and prove your answer.
I thought of $f$ as being the same as the linear transformation, i.e. $f(x)$=$A(x)$. Is this true?
 A: It is true, and may be seen so as follows:
Define the function $g(x)$, $g:G \to R$ as
$g(x) = f(x) - A(x); \tag{1}$
then $g(x)$ is clearly differentiable on $G$; also,
$g' = f' - A = A - A = 0. \tag{2}$
Since a connected open subset of $R^n$ is path connected, for any $x_1, x_2 \in G$  there is a differentiable curve $\gamma(t): I = [0, 1] \to G$ with $\gamma(0) = x_1$ and $\gamma(1) = x_2$.  Applying the fundamental theorem of calculus in vector form we have
$g(x_2) - g(x_1) = \int_0^1 (g(\gamma(t))' dt = \int_0^1 g'(\gamma'(t)) dt$
$ = \int_0^1 0(\gamma'(t))dt = \int_0^1 0dt = 0; \tag{3}$
thus
$g(x_2) = g(x_1) \tag{4}$
for any $x_1, x_2 \in G$.  Set
$r = g(x_1) = f(x_1) - A(x_1) \in R; \tag{5}$
(4) now becomes, using (1),
$f(x_2) - A(x_2) = r \tag{6}$
or
$f(x_2) = A(x_2) + r; \tag{7}$
since $f(x)$ is linear, we have for $0 \ne c \in R$,
$cA(x_2) + cr = c(A(x_2) + r) = cf(x_2) = f(cx_2)$
$= A(cx_2) + r = cA(x_2) + r, \tag{8}$
or
$cr = r; \tag{9}$
since (9) binds for all real $c$, we must have $r = 0$; thus
$f(x_2) = A(x_2) \tag{10}$
for all $x_2 \in G$, as was requested to be shown.
