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Key reference is the following:

Let's investigate real-valued functions $f(x)$ with the following (additive) property for all $\,a,b$ : $$ f(a+b)=f(a)+f(b) $$ It trivially follows that ($m,n$ are naturals): $$ f(a)=f(a-b+b)=f(a-b)+f(b) \quad \Longrightarrow \quad f(a-b)=f(a)-f(b) \\ f(a-a)=f(a)-f(a) \quad \Longrightarrow \quad f(0)=0 \\ f(n.a)=f(a+a+ ...+a)=f(a)+f(a)+ ... +f(a) \quad \Longrightarrow \quad f(n.a)=n.f(a) \\ f(1)=f(n/n)=n.f(1/n) \quad \Longrightarrow \quad f(1/n)=1/n.f(1) \\ f(m/n)=m.f(1/n)=m/n.f(1) \quad \Longrightarrow \quad f(m/n)=m/n.f(1) \\ f(-m/n)=f(0)-m/n.f(1) \quad \Longrightarrow \quad f(-m/n)=-m/n.f(1) $$ We conclude that $f(x)$ must be of the following form: $$ f(x) = c\cdot x \quad \forall\, x \; \mbox{rational} , \, c \; \mbox{real} $$ But there is another way of deriving this. Let $x$ and $h$ be rational, while assuming that it's possible to differentiate on the rationals (why not): $$ f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{f(h)}{h} = \lim_{h\to 0} \frac{h}{h} f(1) = f(1) $$ With the boundary condition $f(0)=0$ giving the same result as before : $f(x) = c\cdot x$ .
Now suppose that a (real) physicist is looking at this. What would he say? We have just solved an ordinary differential equation, isn't it? So why wouldn't the outcome be valid on the reals as well?
I am somehow at lost why there would be other than linear real-valued functions as the solution.

Q : Can somebody clarify about the supposed alternatives?

This question is related to:

Disclaimer. I'm an absolute nitwit if it comes to measure theory. I've tried to learn something about it via the internet, but it seems that they have caused me running around in circles; I really have no idea where to start or where to end.
Han de Bruijn
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  • I don't really understand what you are asking. If you are asking yourself why there would be other than linear solutions but before you just assumed, that the solution would be differentiable on the rationals, this doesn't make any sense. If you assume it to be differentiable on rationals, then yes, there are only linear solutions. If you drop this assumption you get other solutions aswell. So what's your question exactly? – Lukas Betz Jul 22 '16 at 12:12
  • @LeBtz: As far as I can see, it is not assumed that the solution would be differentiable on the rationals, rather it is proved in the one-liner that $f'(x) = f(1)$. Indeed, that's exactly the source of my confusion. – Han de Bruijn Jul 22 '16 at 18:07
  • You used $f(h) = hf(1)$, this however is only true for rational $h$. You need it for real $h$ even if you only try to show differentiability in rational $x$. The Limit still has to be taken in the reals. Otherwise you just get that $f$ restricted to $\mathbb Q$ has to be $\mathbb Q$-linear. This is a true statement but doesn't say anything about $f$ unrestricted. – Lukas Betz Jul 22 '16 at 18:10
  • Found this : Rational Analysis and this : A dialogue concerning the need for the real number system . – Han de Bruijn Jul 22 '16 at 19:21
  • This is what I meant before. Your question itself is purely mathematical. If you add assumptions which come from physics you get different results than those that you would get without these assumptions. I don't understand why you are confused that the mathematical result differs from the physics one if you have different assumptions. Of course they do. – Lukas Betz Jul 22 '16 at 19:24
  • It seems to me, finally, that the whole thing boils down to the fact that the difference between rationals and irrationals is "measurable" in mathematics but not in (experimental) physics. – Han de Bruijn Jul 22 '16 at 19:25

2 Answers2

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Those other solutions on the reals are not differentiable (in fact, not continuous). So your argument is in a conditional form: IF it is differentiable, THEN it is of the form $c\cdot x$.

Note that for complex numbers $x$, the functional equation $f(x+y) = f(x)+f(y)$ has, in addition to continuous solutions $f(x) = c\cdot x$, also continuous solutions $f(x) = c \cdot \overline{x}$. (Here $\overline{x}$ is the complex conjugate of $x$.) But of course $\overline{x}$ is not (complex) differentiable.

Still, if $f(x)$ is (complex) differentiable, then the complex differential equation shows it is of the form $c \cdot x$.

GEdgar
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If $f$ is additive then $f(x)=cx$ for every rational $x$, this correct. Now for the second part you don't need to ask for derivability : if $f$ is continuous then $f(x)=cx$ for every real number, because $\mathbf Q$ is dense in $\mathbf R$.

But if the function $f$ is not continuous it's not necessarily of the form $f(x)=cx$. In fact one can show that nonlinear additive function do exists, and you do that exactly by using a $\mathbf Q$ hamel basis of $\mathbf R$. Such a function will be pathological, it not continuous, probably not measurable and its construction rely strongly on the axiom of choice.

Renart
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