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Imagine you have the following functional equation: $$ f(x + y) = f(x) + f(y) $$ Obviously the solution is a linear function. Usually when defining a linear application/function it will be done by the following process: $$ f:\Bbb R^n \to \Bbb R^m \\ \; \\ f(x + y) = f(x) + f(y)\\ \; \\ f(\alpha x) = \alpha f(x) $$ But, let consider the following. Using only the first equation, we can conclude that the function obeys the following:

First: $$ f(0 + 0) = f(0) + f(0) \Rightarrow 2f(0) - f(0) = 0 \\ \; \\ f(0) = 0 \\ $$ Then: $$ f(x + x - x) = f(x) + f(x - x) \\ f(x) + f(x) + f(-x) = f(x) + f(0) \\ f(x) + f(-x) = 0 \\ \; \\ f(-x) = -f(x) $$ Also: $$ f(x + x) = f(x) + f(x) \\ \; \\ f(2x) = 2f(x) $$ And also: $$ f(2x) = 2f(x) \\ \frac{1}{2}f(2x) = f(x) \Rightarrow \text{let u = 2x} \; \Rightarrow \frac{1}{2}f(u) = f(u/2) \\ \; \\ f(x/2) = \frac{1}{2}f(x) \\ \; \\ \text{hence:} \\ f(x/4) = \frac{1}{4}f(x) \\ f(x/8) = \frac{1}{8}f(x) \\ \vdots \\ f \left( \frac{x}{2k} \right ) = \frac{1}{2k}f(x), \; \; k \in \Bbb N^n \\ $$ It seems like the first definition of a linear function is "screaming" to you that it obeys the second property (previously definied "by hand"). It is possible to properly demonstrate that: $f(\alpha x) = \alpha f(x)$ as a mere consequence of the first definition?

Álvaro Rodrigo
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  • see also https://math.stackexchange.com/questions/2132215/a-real-function-which-is-additive-but-not-homogenous – Alex Pawelko Nov 12 '22 at 21:54
  • https://math.stackexchange.com/questions/2377638/additive-function-t-mathbbr-rightarrow-mathbbr-that-is-not-linear this also – theGrey Nov 12 '22 at 21:56
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    Using your argument, you can show that if $f$ is continuous and satisfies $f(x+y) = f(x) + f(y)$ then it must be $\mathbb R$ - linear. – paul blart math cop Nov 12 '22 at 23:03
  • A map such that $f(x+y)=f(x)+f(y)$ can be proved to be $\Bbb Q$-linear, with a procedure symbolically (but not semantically) similar to one you've written; it may not be $\Bbb R$-linear, however. It's unclear to me what $f(\frac x{2k})=\frac1{2k} f(x)$ is supposed to mean for $x\in\Bbb R^n$, $k\in\Bbb N^n$ and $f(x),f(\frac x{2k})\in\Bbb R^m$. – Sassatelli Giulio Nov 13 '22 at 08:42

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"Obviously the solution is a linear function"

No.

The equation $\alpha f(x)=f(\alpha x)$ is an axiom on linear functions. In this case you have Cauchy's function equation (I think) and any continuous solution is linear, but there exists discontinuous nonlinear solutions too. For example, any $\Bbb Q$-linear map $\Bbb R\to\Bbb R$ is a solution, but these can be 'wild'. Just fix any basis for $\Bbb R$ over $\Bbb Q$ (you need the axiom of choice to even know this exists) and define any value of $f$ on these basis elements whatsoever. You'll almost surely get a solution that is not $\Bbb R$-linear.

FShrike
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