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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?

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  • $\begingroup$ 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

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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.

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    $\begingroup$ +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
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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.

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