$f$ is differentiable and satisfies $f(t\vec{x})=t^pf(\vec{x})$ prove $\vec{x} \cdot \nabla f(\vec{x}) = pf(\vec{x})$

Question

let the three-variable function $f$ be differentiable and satisfy $f(t\vec{x})=t^pf(\vec{x})$ for all $,\vec{x} \in \Bbb{R^3}, t \in \Bbb{R}$ and where $p$ is a constant. How would you prove that $\vec{x} \cdot \nabla f(\vec{x}) = pf(\vec{x})$

I am very confused for this question and have no idea on how to use the information given to reach a conclusion.

Answer 1

Note that $f(tx,ty,tz)=t^pf(x,y,z)$ for all $(x,y,z)\in\mathbb{R^3}$,$t\in\mathbb{R}$. So take a specific point $(x,y,z)\in\mathbb{R^3}$ and let's define $g:\mathbb{R}\to\mathbb{R}$ by $g(t)=f(tx,ty,tz)-t^pf(x,y,z)$. Then $g$ is identically zero, so its derivative is identically zero as well. So let's find the derivative.

$0=g'(t)=f_x(tx,ty,tz)\times x+f_y(tx,ty,tz)\times y+f_z(tx,ty,tz)\times z-pt^{p-1}f(x,y,z)$

Now put $t=1$ and you will get the result you need.

Answer 2

Definition. We say that $f: \mathbb{R}^{n}\to \mathbb{R}$ is $p$-homgeneous if $f(tx) = t^{p}f(x)$ for all $t >0$.

Now,

Theorem. If a function $f$ is $p$-homogeneous and differentiable, then $\langle x:\nabla f(x) \rangle = pf(x)$

Proof. Fixed $x \in \mathbb{R}^{n}$, consider the function $\varphi: (0,+\infty) \to \mathbb{R}$ defined by $\varphi(s) = s^{p}f(x) = f(sx)$. Use the Chain Rule and take $s = 1$.

Can you finish the proof?

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Moreover, the converse is true.