I have been working on a math problem which I devised myself but am unable to fully solve. It is based on the "additive function" let
f:R-->R such, as f(x+y)=f(x)+f(y). Trivially, the linear function having y-intercept at zero are additive functions, namely, y=ax.
However, this does not PROVE that all real valued additive function must be a linear function of form y=ax.

Thus, I attacked the problem like this, let f be a real valued function having the additive property AND a DIFFRENCIABLE function. Then we have the following f'(x)=lim f(x+h)-f(x)/h and based on our initial hypothesis f(x) is diffrencial thus f"(x) exist everywhere. But, since f is additive we have
f'(x)= lim f(x)+f(h)-f(x)/h thus we are left with lim f(h)/h. But since f is a diffrencialbe function then lim f(h)/h exists futhermore it is the same value regradless of x. Thus, f'(x)=L. Thus, f(x)=ax+b. If b is not zero than by basic observation we could conclude the f is not additive however if b is zero than it is additive. Thus, we have proved a additive function that is diffrenciable must be a linear function of the form y=ax.

Now the problem remains what happens if f is not diffrenciable? Then must it be a linear function of the form y=ax? So I decided to attack the problem as follows. I already worked it out for diffrenciable functions. But what about countinous functions? Must it be that a countinous function having the additive property be a linear funtion of the form y=ax? Try as I might I have not been able to prove it, nor find a counterexample.

And finally, the biggest challenge to prove all additive functions are linear functions. Which I completely doubt that it is true. Interestingly enough I was not able to find a counterexample of a non-linear function. I am guessing there are probably Infinitely many such functions, non of which are elementary.

2. Originally Posted by ThePerfectHacker
I have been working on a math problem which I devised myself but am unable to fully solve. It is based on the "additive function" let
f:R-->R such, as f(x+y)=f(x)+f(y). Trivially, the linear function having y-intercept at zero are additive functions, namely, y=ax.
However, this does not PROVE that all real valued additive function must be a linear function of form y=ax.
This is an interesting problem, which has however already been treated elsewhere (Try to get your hands on Natanson's book "Theory of Functions of a Real Variable" or on Kolmogorov/Fomin). You can prove that any additive function that is also continuous at 0 must be of the form f(x) = ax, as follows.

Firstly, f(0) = f(0+0) = f(0) + f(0), so f(0) = 0. If f is continuous at x = 0, then $\lim_{h \to 0}f(x+h) = \lim_{h \to 0}(f(x)+f(h)) = f(x) + 0$, so f is continuous everywhere.

Next set a = f(1). By induction you can prove that f(n) = an for all positive integers. Also, 0 = f(0) = f(n-n) = an + f(-n), so this is true for all integers n.

Then consider a rational number x = m/n, with n > 0. Then nx = m = x + ... + x, so by induction $f(nx) = f(x+\dots+x) = nf(x) = f(m) = am$, thus $f(x) = \frac{an}{m} = ax$.

Finally for general x, $f(x) = \lim_{y \to x, \, y \in Q}f(y) = \lim_{y \to x, \, y \in Q}ay = ax$.

Now if f is not continuous at 0, things are different. Your instinct that an additive function need not be of the form f(x) = ax is indeed correct. Counterexamples are tricky and require the Axiom of Choice. Namely, consider a Hamel basis H, that is a basis for theset of real numbers viewed as a vector space over the rationals. For each $h \in H$, define $f(h) = a_h$, where $a_h$ is an arbitrary real number. Then if $x = \sum_i \alpha_i h_i$ where the $\alpha_i$ are rational and the $h_i \in H$ , one extends f as $f(x) = \sum_i \alpha_i a_{h_i}$. This is an additive function that is not of the simple form f(x) = ax (and not continuous at 0).

3. Thank you very much for a good explaination. It could easily be shown that this additive function cannot be a piecewise elementary function this is why it is so difficult to find a counterexample.