# Show $\langle x,\nabla f \rangle = pf(x)$

I'm trying to figure out this problem. Perhaps Someone could give me some hints/solve it for me? It would be much appreciated.

Let $U$ be an open subset of $R^n$ and suppose $f:U\rightarrow R$. Then $f$ is homogeneous of degree $p$ if $f(\lambda x) = \lambda^p f(x)$ for all $x\in U$, $\lambda \in R$,and $\lambda x \in U$

prove that

$\langle x,\nabla f(x) \rangle = pf(x)$

thanks a bunch guys

• What happens if you take the derivative of $f(\lambda x) = \lambda^p f(x)$ with respect to $\lambda$? Mar 17, 2013 at 19:57

$$f(\lambda x)=\lambda^pf(x)\Longrightarrow \frac{d}{d\lambda}f(\lambda x)=xf'(\lambda x)=p\lambda^{p-1}f(x)=\frac{d}{d\lambda}(\lambda^p f(x))\Longrightarrow$$

$$\Longrightarrow \;\;\text{for}\;\;\lambda=1\;,\;\;xf'(x)=pf(x)\ldots$$

• Then I would just have to related $pf(x)$ to the directional derivative ?
– Neo
Mar 17, 2013 at 20:09
• Well...if you want, or else just accept the sheer fact as it is. Mar 17, 2013 at 20:26
• I see why you chose $\lambda = 1$, but that means this would not hold if it were = to 2 ?
– Neo
Mar 17, 2013 at 20:29
• What is "this" in your question, @Neo? If you choose $\,\lambda=2\,$ then you get another identity that is not the one you wanted...! Mar 17, 2013 at 20:35
• @DonAntonio: You always left some of the way of proofing for the OP by writing "..." at the end. I like this manner Don. +1 Mar 18, 2013 at 2:58

$$\langle x,\nabla f(x)\rangle=\lim_{h \to 0}\frac{f(x+hx)-f(x)}{h}=\lim_{h \to 0}\frac{(1+h)^p-1}{h}f(x)=pf(x)$$

The above formula holds because $\langle x,\nabla f(x)\rangle$ is the directional derivative of $f$ in the direction of $x$.