A differentiable function on Euclidean Space compatible with scalar multiplication is a linear map

Question

Here is how the question stated:

Problem $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable and $$f(\lambda x)=\lambda f(x), \forall \lambda\in \mathbb{R}, x\in \mathbb{R}^n.$$ Prove that $f$ is a linear map.

My thoughts The equation $f(\lambda x)=\lambda f(x)$ immediately gives the compatibility of scalar, leaving compatibility of addtion to be verified.
I try to derive the addtion from $f(\lambda x)=\lambda f(x)$. Apart from compatibility of scalar, $f$ is a homogeneous function. Suppose that $x=(x_1,x_2,\cdots,x_n)$, then I get $f(\lambda x_1,\cdots,\lambda x_n)=\lambda f(x_1,x_2,\cdots,x_n)$. Differentiating by $\lambda$, I get $$ f_1x_1+f_2x_2+\cdots +f_nx_n=f\left( x_1,x_2,\cdots ,x_n \right) $$ where $f_i$ is the partial derivative of $f$ about the $i^{\text{th}}$ variable of its domain. What I need now is $$ f\left( x+y \right) =f\left( x \right) +f\left( y \right) ,\forall x,y\in \mathbb{R}^n $$ Similarly, we suppose $y=(y_1,y_2, \cdots ,y_n)$, then we need $$ f\left( x+y \right) =\left( x_1+y_1 \right) f_1\left( x_1+y_1 \right) +\left( x_2+y_2 \right) f_2\left( x_2+y_2 \right) +\left( x_n+y_n \right) f_n\left( x_n+y_n \right) $$ equals to $$ f\left( x \right) +f\left( y \right) =x_1f_1\left( x_1 \right) +x_2f_2\left( x_2 \right) +\cdots +x_nf_n\left( x_n \right) +y_1f_1\left( y_1 \right) +y_2f_2\left( y_2 \right) +y_nf_n\left( y_n \right) . $$ Because $f_i$, as derivative, is linear, we can break the brackets and cancel $x_if_i(x_i)$ and $x_if_i(y_i)$. However, terms of form $x_if_i(y_i)$ and $y_if_i(x_i)$ cannot be cancelled, which puzzles me.

It is possible that my thoughts were totally off the track! Any help or idea would be welcome!

Answer 1

First observe that $f(0)=0$. Now fix $x,y\in \Bbb{R}^n$. For every positive $\lambda$ we know: $$ f(x+y) -f(x)-f(y)=\frac{\lambda(f(x+y) -f(x)-f(y))}{\lambda}=\frac{f(\lambda(x+y)) -f(\lambda x)-f(\lambda y)}{\lambda}$$ This shows the function $\lambda \to\frac{f(\lambda(x+y)) -f(\lambda x)-f(\lambda y)}{\lambda}$ defined for positive $\lambda$ is constant. What about its limit as $\lambda \to 0$?

Answer 2

It's probably phrased in a confusing amount of 'generality'. All that we need is differentiability at $0$. Hence, let $Df(0)$ be the linear map given by the total derivative at $0$. All we need to argue is that $f(x)=Df(0)x$ for all $x$.

Let $x\in \mathbb{R}^n\setminus \{0\}$ and note that for all $\lambda \in \mathbb{R}\setminus \{0\}$

$$ f(x)=\lambda f\left(\frac{x}{\lambda}\right)=\lambda\left( Df(0)\frac{x}{\lambda}+o\left(\left\|\frac{x}{\lambda}\right\|\right)\right)=Df(0)x+\varepsilon\left(x/\lambda\right)\|x\|, $$ where $\varepsilon$ is some function with the property that $\lim_{\|y\|\to 0}\varepsilon(y)=0.$ However, the left-hand side is completely independent of $\lambda$, so we get that

$$ f(x)=Df(0)x+\lim_{\lambda\to \infty}\varepsilon(x/\lambda)\|x\|=Df(0)x $$