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

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!

• You might try taking that first expression and differentiate wrt each individual $x_i$ as well; that might get you a little more information. (but this is just a wild guess). – John Hughes Oct 9 '20 at 11:47
• @John Reasonable guess! One of my mates has actually discussed this thought with me but not figured it out yet. Thank you! – atlantic0cean Oct 9 '20 at 12:21
• @trurl thanks! But I think this only implies the addition holds on the line through $0$ and $x$. What about $y\ne kx, \forall k\in \mathbb{R}$? – atlantic0cean Oct 9 '20 at 12:25
• I have a sense that it may have something to do with direction derivatives at $0$, but only a sense... I also figure out an example in $\mathbb{R}^3$: a cone with vertex at $0$. The directrix of cone is possibly not a circle but any figure in $\mathbb{R}^2$. – atlantic0cean Oct 9 '20 at 12:31

## 2 Answers

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 $$\lambda \to 0$$?

• Incredible! It's like a brain teaser to fix $x,y$ and let $\lambda$ varies! The limit as $\lambda\rightarrow 0$ is actually not easy for me haha... Anyway, thank you so much for your help! – atlantic0cean Oct 9 '20 at 14:14
• @atlantic0cean HINT1: as $\lambda\to 0$, $\lambda(x+y)\to ?$, $\lambda x\to ?$, $\lambda y\to ?$ HINT2: we still haven't used the differentiability hypotesis... :-) – FormulaWriter Oct 10 '20 at 15:01

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

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

• Finally I get over this problem with your hint of using $Df(0)$ to evaluate $f(x)$! Many thanks for your help! – atlantic0cean Oct 9 '20 at 14:08