Let A ⊂ Rn be open. Suppose f : A → Rm is a differentiable mapping, and x, y ∈ A satisfy that the line segment joining x and y lies in A. Prove that for any v ∈ Rm, there exists a convex combination of x and y, denoted as z, such that v · f(y) − f(x) = v · [Df(z) · (y − x)]
what I've done so far is: for $g:[0,1]\rightarrow R$ let $g(\lambda)=f((1-\lambda)x+\lambda y) \therefore g(1)=f(y), g(0)=f(x)$ By the chain rule, $g:(0,1)\rightarrow R^m$ is differentiable as $f$ is differentiable $\therefore Dg(\lambda)=Df((1-\lambda)x+\lambda y)\cdotp (y-x)$ as $(1-\lambda)x+\lambda y$ is a convex combination we show it as z therefore $Dg(\lambda)=Df(z)\cdotp (y-x)$ therefore $v\cdotp Dg(\lambda)=v\cdotp Df(z)\cdot (y-x)$
How do I get $Dg(\lambda)$ to $f(x)-f(y)$ without changing the left hand side of the equation?
By the mean value theorem on one variable,
$$ f(y) - f(x) = g(1) - g(0) = g'(c)(1-0) = Df_z(y-x) $$
with $z = g(c) = (1-c)x + cy$.