Let $f:\Bbb R^m\to\Bbb R$ be differentiable at $\Bbb R^m\setminus\{0\}$. Show that if $(\nabla f(x)|x)=\alpha f(x)$ for some $\alpha\in\Bbb R$ then $t^\alpha f(x)=f(tx)$ for all $t>0$ and each $x\in\Bbb R^m\setminus\{0\}$.
We want to show that
$$(\nabla f(x)|x)=\alpha f(x),\,\forall t>0,x\in\Bbb R^m\setminus\{0\}\implies f(tx)=t^\alpha f(x)$$
First observe that for $v\neq 0$ and $t>0$ we have that
$$\partial f(tv)v=D_vf(tv)=\lim_{r\to 0}\frac{f(tv+rv)-f(tv)}r=\lim_{r\to 0}\frac{f(v(t+r))-f(tv)}r=\partial_t [f(tv)]$$
Thus $\alpha f(tv)=t \partial_t [f(tv)]$, hence if $f(tv)\neq 0$ for all $t\in[1,s]$ we have that
$$\int_1^s \frac\alpha{t}\mathrm dt=\int_1^s \frac{\partial_t [f(tv)]}{f(tv)}\mathrm dt\implies \alpha \ln s=\ln \left|\frac{f(sv)}{f(v)}\right|\implies s^\alpha f(v)=f(sv)$$
where the cancellation of absolute values is the consequence of the continuity of $f$ in $[1,s]$ together with the fact that $f(tv)\neq 0$ for all $t\in[1,s]$.
Now Im stuck with the possibility that $f(tv)=0$ for arbitrary number of points in $[1,s]$. Some help will be appreciated, thank you.
Well, I solved it using the Gronwall's lemma. Here $\land$ and $\lor$ denote $\min$ and $\max$ respectively, and $s>0$.
If $f(tx)=0$ for some point in $[1\land s,1\lor s]$ then $\partial_t[f(tx)]=0$ in $[1\land s,1\lor s]$. Proof: we can see that if $\partial_t[f(tx)]=\frac{\alpha}t f(tx)$ then
$$\big|\partial_t[f(tx)]\big|\le\left|\frac{\alpha}m\left( f(t_0 x)+ \int_{t_0}^t\partial_h[f(hx)]\mathrm dh\right)\right|\le \frac\alpha{m}|f(t_0 x)|+\frac{\alpha}m \int_{t_0}^t\big|\partial_h[f(hx)]\big|\mathrm dh$$
for $m:=\min\{1,s\}$, what by the Gronwall's lemma imply that $\partial_t[f(tx)]=0$ if $f(t_0x)=0$ for some $t_0\in[1\land s,1\lor s]$.
Then if $f(t_0x)=0$ for some $t_0\in[1\land s,1\lor s]$ due to the result above and the equation $\partial_t[f(tx)]=\frac{\alpha}t f(tx)$ we can conclude that $f(tx)=0$ for all $t\in[1\land s,1\lor s]$.