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 $$f(\lambda x)=\lambda f(x),\;\forall\;\lambda\in\Bbb{R},\;\forall\;x\in\Bbb{R}^n.$$


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*Prove that $f(0)=0.$

*Prove that $f$ is linear.


Here's what I have done:


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*$f(0,0,\cdots,0)=[f(0),f(0),\cdots,f(0)]=(0,0,\cdots,0)$

*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?

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.
