Problem:
We say $f: R^n - \{ 0\} \rightarrow R$ is homogenous of degree $k$ if $f(tx) = t^k f(x)$ for all $t > 0$. Prove that $f$ is homogenous of degree $k$ if and only if $Df(x)x = kf(x)$ for all nonzero x $\in R^n$. (Hint: fix x and consider $h(t) = t^{-k}f(tx$).)
Explanation:
This problem is from the book Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds and my question is concerned with the proof of the second implication $Df(x)x = kf(x)$ implies that the function $f$ is homogenous of degree $k$, i.e., $f(tx) = t^kf(x)$ for all $t > 0$.
My Proof:
$h'(t) = t^{-k}Df(tx)x - kt^{-(k+1)}f(tx)$
I would like to use the fact that $Df(x)x = kf(x)$, so since $t^{-k} = t^{-k-1}t$ $\hspace{0.1cm}$we have:
$h'(t) = t^{-(k + 1)}Df(tx)tx - kt^{-(k+1)}f(tx) = t^{-(k + 1)}kf(tx) - kt^{-(k+1)}f(tx) = 0$
Therefore $h(t)$ is a constant function.
My question is how to deduce that since $h(t)$ is a constant function it must be that $f(tx) = t^kf(x)$? taking $f(tx) = t^kf(x)$ would cause $h(t)$ to be constant but is that the only way for it to be constant?
is there a way to directly prove that $f(tx) = t^kf(x)$?