Let $f:\mathbb{R}^n\to \mathbb{R}$ be a $C^1$ function such that: $\mathrm{x} ·∇f(\mathrm x) =f(\mathrm x)$, $∀\mathrm x ∈\mathbb{R}^n$.
How to prove that $f(t\mathrm x)=tf(\mathrm x)$, $∀\mathrm x ∈\mathbb{R}^n$, $∀ t ∈\mathbb{R}^+$?

I have considered $\phi(t)=f(t\mathrm x)-tf(\mathrm x)$ and arrived at the conclusion that its derivative equals $\mathrm x\cdot ∇f(t\mathrm x)-f(\mathrm x)$. Is this conclusion valid? How should I proceed from here?

  • $\begingroup$ Yes. Now carefully use the given. If you get stuck, perhaps consider a different $\phi$. $\endgroup$ Jun 30, 2023 at 20:21
  • $\begingroup$ Your space is connected, what can you conclude? $\endgroup$ Jun 30, 2023 at 20:23
  • $\begingroup$ @JulesBesson You’re going way too fast. Are you sure? $\endgroup$ Jun 30, 2023 at 20:29
  • $\begingroup$ I was talking about $\mathbb{R}_+$, so we definitely have $\phi_x$ constant do you think something is missing? $\endgroup$ Jun 30, 2023 at 20:33
  • $\begingroup$ We don’t have $\phi’(t)=0$, do we? $\endgroup$ Jun 30, 2023 at 20:39

1 Answer 1


Note that the equality

$$\boldsymbol x\cdot (\nabla f)(\boldsymbol x)=f(\boldsymbol x)\tag{1}$$ implies also ($\boldsymbol x\to t\boldsymbol x$) $$t\boldsymbol x\cdot (\nabla f)(t \boldsymbol x)=f(t \boldsymbol x)\tag{2}$$

Defining $$\phi(t)=f(t \boldsymbol x)-t~f(\boldsymbol x)$$ This means $$\phi'(t)=(\nabla f)(t \boldsymbol x)\cdot \boldsymbol x-f(\boldsymbol x) \\ =\frac{1}{t}\big(t\boldsymbol x\cdot (\nabla f)(t \boldsymbol x)-t~f(\boldsymbol x)\big)$$ Which by $(2)$ is $$\phi'(t)=\frac{1}{t}\big(f(t \boldsymbol x)-t~f(\boldsymbol x)\big) \\ =\frac{1}{t}\phi(t)$$

This is a simple ODE, with solution $\phi(t)=c~t$. However, noticing that $$\phi(1)=f(1~\boldsymbol x)-1~f(\boldsymbol x)=0$$

we conclude $\phi(1)=0=1c\implies c=0$

Hence $\phi(t)\equiv 0$ and thus $f(t \boldsymbol x)=t~f(\boldsymbol x)$. QED


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