I am having trouble constructing a proof for this preposition. I am not sure if I am misunderstanding the meaning of a homogenous functions, but either way I get stuck in my proof.

Let $k$ be an integer. A function $f : \mathbb{R^n} \to \mathbb{R}$ is called homogenous of degree $k$ if $f(\lambda x) = \lambda^k f(x)$ for all $\lambda \in \mathbb{R}$ and $x \in \mathbb{R^n}$. Prove that if $f$ is homogenous of degree $k$ then $x \cdot \nabla f(x) = kf(x)$.

Here is my go at it.


(=>) Suppose $x \cdot \nabla f(x)$.

Then $$x \cdot \nabla f(x)$$ $$= x \cdot [\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{ \partial x_n}]$$

$$ = [x_1, ..., x_n] \cdot [\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}]$$

$$ = \frac{\partial f}{\partial x_1} x_1 + ... + \frac{\partial f}{\partial x_n} x_n $$

But I feel I've gone off course now.

Any ideas / hints?



1 Answer 1


Suppose $f$ is homogeneous of degree $k$, i.e. $f(\lambda x) = \lambda^kf(x)$ for any $\lambda\in\mathbb{R}$ and $x\in\mathbb{R}^n$. Differentiating this relation with respect to $\lambda$ gives: \begin{align} \nabla f(\lambda x)\cdot x = k\lambda^{k-1}f(x). \tag{1} \end{align} On the other hand, differentiating the same relation with respect to $x$ gives: \begin{align} \nabla f(\lambda x)\lambda = \lambda^k\nabla f(x) \implies \nabla f(\lambda x) = \lambda^{k-1}\nabla f(x). \tag{2} \end{align} Substituting (2) into (1) yields the desired result.

  • $\begingroup$ Thanks. I wasn't too far off. I had used the chain rule incorrectly in (1) because I treated $x \in \mathbb{R}$ as opposed to $\mathbb{R^n}$. Grrr. $\endgroup$ Apr 13, 2017 at 2:08

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