# Let $f:\Bbb{R}^n\to \Bbb{R}$ be differentiable. Prove that $f$ is linear and show that $f(0)=0$

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?

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

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 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). 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.