# If $\mathrm x\cdot ∇f(\mathrm x)=f(\mathrm x)$ $∀\mathrm x∈\Bbb R^n$, then $f(t\mathrm x) = tf(\mathrm x)$ $∀\mathrm x∈\mathbb{R}^n,∀t∈\mathbb{R}^+$

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?

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

$$\boldsymbol x\cdot (\nabla f)(\boldsymbol x)=f(\boldsymbol x)\tag{1}$$ 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