Euler's theorem for homogenic functions states, that an $f:\mathbb{R}^n\to\mathbb{R}$, continuously differentiable function is homogeneous of degree $k$ if and only if for all $r \in \mathbb{R}^n$ the following equation satisfies.
$$\sum_{i=1}^nr_i\partial_if(r)=kf(r)$$
A proof my professor did was fine for the part where we start from the fact that $f$ is homogeneous, so then the equation is satisfied. The other part is where I have problems, when we start from the fact that the upper equation is true.
He did the following:
Define $g:\mathbb{R}\to\mathbb{R}$ to be $g(t) = \frac{f(tr)}{t^k}$ where $t>0$. By the theorem's criterions we know that $g$ is differentiable (since $f$ is continuously differentiable). Taking the derivative of $g$ we get:
$$g'(t) = \frac{\sum_{i=1}^nr_i\partial_if(tr)t^k-kt^{k-1}f(tr)}{t^{2k}} = \frac{\sum_{i=1}^n(tr_i)\partial_if(tr)-kf(tr)}{t^{k+1}} = 0$$
By the criterion, this is zero, and this is where we finished, the theorem is proven. It's not clear how does this finish the proof. Didn't we simlply proved, that $g$ is a constant function, so $f(tr) = ct$ for some $c\in\mathbb{R}$? How can the original statment be deducted from this, that $f$ is homogeneous, by definition, the $f(tr) = t^kf(r)$ equation holds?
Edit: What I thought first is that we can maybe define $g$ to be $g(t) = \frac{f(tr)}{t^kf(r)}$. But then we have no criterion so that $f(r) \neq0$, so it seems incorrect. But even by doing this I'd still have the problem with it simply being a constant function, hence not being able to get rid of the $c$ constant. It seems like I can't notice something.
Best Answer
Thanks Jens Schwaiger for the tip in the comments, the credit is his, I'm just going to post it as an answer.
By showing that $g$ is a constant function, we can plug in any $t > 0$ value to $g$ and that will give us the constant. Plugging in $1$ we get that $g(1) = f(r)$, hence $g(t) = f(r)$ for every $t>0$. $$\frac{f(tr)}{t^k}=f(r)$$ Reordering this we get the desired result, which is: $$f(tr) = t^kf(r)$$