Show that $f(tx)=t^pf(x),\;\forall \;t>0,\;\&\;x\in \Bbb{R}^n $ if and only $f'(x)(x)=pf(x),\;\forall \;x\in \Bbb{R}^n$

Question

Let $f:\Bbb{R}^n\to \Bbb{R}$ be a differentiable function such that for some $p>1,$ \begin{align}f(tx)=t^pf(x),\;\forall \;t>0,\;\&\;x\in \Bbb{R}^n \qquad (1)\end{align} $i.$ I want to show that \begin{align}f'(x)(x)=pf(x),\;\forall \;x\in \Bbb{R}^n \qquad (1)\end{align} $ii.$ Is the converse also true?

I believe that we can take \begin{align}\varphi(t)=f(tx),\;\forall \;t>0\end{align} and show that \begin{align}\frac{d}{dt}\big(\frac{\varphi(t)}{t^p}\big)=0,\end{align} but I don't know how to tranform this to a proof. Any help please?

Answer 1

We can assume $n=1$ by restriction to an arbitrary line. Then $$ f'(x) = \lim_{\epsilon\to 0}\frac{f((1+\epsilon)x)-f(x)}{\epsilon x} = \lim_{\epsilon\to 0}\frac{[(1+\epsilon)^p - 1]f(x)}{\epsilon x} = \frac{f(x)}{x}\frac{d}{d\epsilon}(1+\epsilon)^p|_{\epsilon=0} = p\frac{f(x)}{x} $$ as desired. Note differentiability is guaranteed by this condition.

Similarly, if $\frac{df}{dx} = p\frac{f}{x}$ then $\frac{df}{f} = p\frac{dx}{x}$. Integrating both sides says $$ \log f(x) - \log f(1) = p\log x = \log x^p $$ so $$ f(x) = C x^p, $$ which provides the converse.

Answer 2

Let's suppose $$ g(t)=f(tx)-t^pf(x)$$ So $\forall t >0$ $$g(t)=0$$ Therfore $$g'(t)=0$$ Now, by the chain role we have $$g'(t)=xf'(tx)-pt^{p-1}f(x)$$ We can evaluate this at $t=1$ and obtain $$xf'(tx)=pf(x)$$