Let $f:\Bbb{R}^n\to \Bbb{R} $ be differentiable such that $$f(\lambda x)=\lambda f(x),\;\forall\;\lambda\in\Bbb{R},\;\forall\;x\in\Bbb{R}^n.$$

  1. Prove that $f(0)=0.$
  2. Prove that $f$ is linear.

Here's what I have done:

  1. $f(0,0,\cdots,0)=[f(0),f(0),\cdots,f(0)]=(0,0,\cdots,0)$

  2. Let $\lambda\in \Bbb{R},\;\;x,y\in\Bbb{R}^n$ where $x=(x_1,x_2,\cdots,x_n)$ and $y=(y_1,y_2,\cdots,y_n)$. Then $$f(\lambda x+y)=f(\lambda x)+f(y)$$ $$=(f(\lambda x_1),f(\lambda x_2),\cdots,f(\lambda x_n))+f(y_1,y_2,\cdots,y_n)$$ $$=(\lambda f(x_1),\lambda f(x_2),\cdots,\lambda f(x_n))+f(y_1,y_2,\cdots,y_n)$$ $$=\lambda[ f(x_1), f(x_2),\cdots, f(x_n)]+f(y_1,y_2,\cdots,y_n)$$ $$=\lambda f(x_1,x_2,\cdots,x_n)+f(y_1,y_2,\cdots,y_n)$$ $$=\lambda f(x)+f(y)$$

Please, I'm I right? If not, could someone show me a proof or reference?

  • $\begingroup$ Why $f(\lambda x +y)= f(\lambda x) + f(y)$? $\endgroup$
    – xbh
    Aug 16, 2018 at 3:06
  • $\begingroup$ @xbh: I feel somehow that, my solution is incorrect! $\endgroup$ Aug 16, 2018 at 3:06

2 Answers 2


It's only necessary the differentiability at $0$.

Proof. Define $$g(x) = f(x) - \sum_{1}^{m}\frac{\partial f}{\partial x_{i}}(0)x_{i}.$$ Note that $g(\lambda x) = \lambda g(x)$. Since $f(0) = 0$ and $f$ is differentiable at $0$, we have $$\lim_{x \to 0}\frac{g(x)}{|x|} = \lim_{x \to 0}\frac{f(x) - f(0) - \sum_{1}^{m}\frac{\partial f}{\partial x_{i}}(0)x_{i}}{|x|} = 0.$$ Now, suppose $B_{\delta}(0) \subset U$ for some $\delta$. If $x \in U\setminus\{0\}$, $tx \in U$ and $$tg(x) = g(tx) \Longrightarrow g(x) = \frac{g(tx)}{t}$$ for some $0 < t < \frac{\delta}{|x|}$. Therefore, $$g(x) = \lim_{t \to 0}\frac{g(tx)}{t} = \lim_{t \to 0}|x|\frac{g(tx)}{|tx|} = 0.$$ and if $x = 0$, $g(0) = f(0) = 0$. Therefore, $g \equiv 0$ and $$f(x) = \sum_{1}^{m}\frac{\partial f}{\partial x_{i}}(0)x_{i}.$$

Alternative (using differentiability):

Definition. A function $f: \mathbb{R}^{n} \to \mathbb{R}$ is $p$-homogeneous if $f(\lambda x) = \lambda^{p}f(x)$ for all $\lambda > 0$ and for all $x \in \mathbb{R}^{n}$.

Proposition. A function differentiable $f:\mathbb{R}^{n} \to \mathbb{R}$ is $p$-homogeneous if only if $\langle x:\nabla f(x) \rangle = pf(x)$.

Proof. Take with the function $\varphi: (0,\infty) \to \mathbb{R}$ defined by $\varphi(s) = f(sx) = s^{p}f(x)$. "$\Longrightarrow$" is direct. For converse, try to show $$s\varphi'(s) - p\varphi(s) = 0$$ and multiplie by $s^{-p-1}$.

In your case, the function is $1$-homogeneous.

  • 1
    $\begingroup$ +1 This is a good answer that avoids requiring that $f$ be continuously differentiable. I will just say I think the argument could be better presented (readability). $\endgroup$ Aug 16, 2018 at 3:25

We have $f(0) = f(0 \cdot 0) = 0 f(0) = 0$ (one of the $0$s is the scalar $0$, the other is the origin).

Pick $x$ and let $\phi(t) = f(tx)$. Note that $\phi'(0) = f'(0) x = f(x)$. Hence $f(x) = f'(0) x$, which is linear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.