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

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.

• Take derivative with respect to t and then put t=1 i.e. define g(t)=f(tx) and compute g'(1). Commented Sep 22, 2018 at 18:27

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.

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?

Moreover, the converse is true.