A linear functions $f$ is a function that has the following two properties:
But does $c\cdot f(x) = f(cx) \implies f$ is linear? Can we assume property $2$ above, if we have a function $f$ satisfying property $1$?
$\def\norm#1{\lvert\lvert#1\rvert\rvert}\DeclareMathOperator\sign{sign}$ I think the answer is 'no' in a real vector space of dimension $n \geq 2$. For a counterexample, let $f \colon \mathbb{R}^n \to \mathbb{R}$ be the function given by $f(x_1, \dotsc, x_n) = \sign(x_1)\sqrt{x_1^2 + \dotsb + x_n^2}$. This satisfies the scaling condition, because
$$ \begin{align} f( c(x_1, \dotsc, x_n)) &= f(cx_1, \dotsc, cx_n) \\ &= \sign(cx_1)\sqrt{c^2 x_1^2 + \dotsb + c^2x_n^2} \\ &= \sign(c)\sign(x_1)\sqrt{c^2}\sqrt{x_1^2 + \dotsb + x_n^2} \\ &= \sign(c)|c|\sign(x_1)\sqrt{x_1^2 + \dotsb + x_n^2} \\ &= cf(x_1, \dotsc f_n). \end{align} $$ Here $\sign \colon \mathbb{R} \to \mathbb{R}$ is the function defined by $$ \sign(x) = \begin{cases} 1 & \text{if $x>0$} \\ 0 & \text{if $x=0$} \\ -1& \text{if $x<0$} \end{cases}. $$ To see that $f$ does not respect addition, let $e_1 = (1, 0, 0, 0, \dotsc, 0)$ and $e_2 = (0, 1, 0, 0, \dotsc, 0)$, each containing only one entry equal to one. Note that this is possible because we assumed $n \geq 2$. Then
$$ \begin{align} f(e_1 + e_2) &= f(1, 1, 0, 0 , \dotsc, 0) \\ &= \sqrt{1^2 + 1^2 + 0^2 + 0^2 + \dotsc 0^2} \\ &= \sqrt 2. \end{align} $$ But since $f(e_1) + f(e_2) = 1 + 1 = 2 \neq \sqrt 2$, the second condition is not satisfied by $f$.
In dimension $1$ (again over the reals) the answer is yes, as explained by the argument by the poster of this question.
