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I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:

Try out some cases:

$x=0:$ $$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$ The same result is for when $y=0$

$x=-y:$ $$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$ I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:

Let $q=1+1+1+...+1$. Then $$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$ I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:

Thus $$f(qx)=qf(x)$$ and it should follow that $$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that $$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that $$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of $f(x+y)=f(x)+f(y)$.

I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of $f(x+y)=f(x)+f(y)$? Or do other solutions exist?

Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!)

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    $\begingroup$ Think about $f(x)=f(1 x)=f(\frac{q}{q} x)$. In order to extend the result to $a$ real you need to use the continuity of $f$. $\endgroup$
    – Quimey
    Apr 10, 2013 at 2:43
  • $\begingroup$ @Quimey What do you mean that I need an extra hypothesis like continuity? I think I mentioned that I am looking for the functions f from $\mathbb R \rightarrow \mathbb R$ that are continuous, is that what you are talking about? $\endgroup$
    – user66807
    Apr 10, 2013 at 2:46
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    $\begingroup$ @user66807, Instead of "functions of the form $f(ax)$," I think you mean "functions of the form $f(x) = ax$". $\endgroup$ Apr 10, 2013 at 2:52
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    $\begingroup$ Once you have proven that $f(rx)=rf(x)$ for every rational number $r$, you have $f(r)=rf(1)$ for every rational number $r$, and by continuity (and the density of the rationals in the reals) for every real number $r$. Ie any solution $f$ is completely determined by its value at $1$. Moreover, every constant $a$ gives rise to a solution of the form $f(x)=ax$ (trivial to check). Thus we have found the full solution set. $\endgroup$
    – goonfiend
    Apr 10, 2013 at 2:59
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    $\begingroup$ Wikipedia: Cauchy functional equation. BTW you mention continuity of $f$ in the title, but I do not see it in the body of your question. $\endgroup$ May 14, 2013 at 13:08

3 Answers 3

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For $f(\frac1q x)$:

$$f(x) = f(q\cdot\frac1q x) = f(\frac1q x+\ldots+\frac1q x) = f(\frac1q x)+\ldots+f(\frac1q x) = qf(\frac1q x)$$

For $f(\frac pqx)$: Set $y=\frac xq$ to get $$f(\frac pqx) = f(p\frac xq) = f(py) = p\,f(y) = p\,f(\frac1q x) = p\cdot\frac1qf(x)$$

So now you know $f(\alpha x)=\alpha f(x)$ for all $x\in \mathbb Q$. Let $(\alpha_n)$ be a sequence of rational numbers converging to the real number $r$. Since $f$ is continuous, we have $\lim_{n\to\infty}f(\alpha_n x)=f(r x)$. On the other hand, since all $\alpha_n$ are rational, we have $\lim_{n\to\infty}f(\alpha_n x) = \lim_{n\to\infty} \alpha_n f(x) = r\,f(x)$. Since for every $r\in R$ there exists a sequence of rational numbers converging to it, we therefore have $f(rx)=r\,f(x)$ for all $r\in \mathbb R$.

Finally, we can get an explicit form by observing that $f(x) = f(x\cdot 1) = x\,f(1)$. Therefore with $f(1)=c$ arbitrary, we get $$f(x) = cx$$

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If you do not assume continuity of $f$, then it depends on the Axiom of Choice, stated as Zorn's Lemma. Consider the family $\cal L$ of linearly independent over $\mathbf Z$ subsets of $\mathbf{R}$. Given a chain of such sets,

$$L_1 \subset L_2 \subset \dotsm \subset L_n \subset \dotsm$$

we can prove $L = \bigcup_n L_n \in \cal L$. Suppose $L$ were not linearly independent over $\mathbf Z$. Then, there would be distinct $x_i \in L$ and nonzero integers $a_i$ such that

$$ a_1 x_1 + \dotsm + a_m x_m = 0 $$

But each $x_i$ must be in $L_n$ for some $n$. Let $N$ be the maximum of all such $n$. Then, each $x_i \in L_N$, and so $L_N$ is a $\mathbf Z$-dependent set, contrary to the assumption.

Since every chain has an upper bound, Zorn's Lemma states that there is a maximal member, $M$, of $\cal L$. Every real number $r$ must be a $\mathbf Z$=linear combination of a finite set of elements of $M$; otherwise, $M \cup \{r\}$ would be $\mathbf Z$-independant, contrary to $M$ being maximal. $M$ is a Hamel basis of $\mathbf R$.

And now, any permutation of $M$ extends by linearity to a discontinuous $f$ such that $f(x+y) = f(x) + f(y)$.

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$$f(x+y)=f(x)+f(x)$$ $$f(0)=0$$ $$f(-x)=-f(x)$$ All these are enough to state a linear function with no constant. Also,$$f'(x)=\dfrac{f(x+h)-f(x)}h$$ if $$f'(x)=\dfrac{f(h)}{h}=c$$

So, The only solution set is $f(x)=cx$

A confirmation too from wolfram. .

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    $\begingroup$ It is not given that $f$ has a derivative, only that $f$ is continuous. $\endgroup$
    – Aryabhata
    Apr 10, 2013 at 19:05
  • $\begingroup$ And any solution must make use of continuity in an essential way. Without that assumption, highly irregular solutions exist (assuming choice). $\endgroup$
    – goonfiend
    Apr 11, 2013 at 2:32
  • $\begingroup$ I hope $f'$ was evaluated because function is continuous.$f(0)=0$ so we can say that $f(h)=0^+$ for $h\to 0^+$ and the $f'$ comes out to be $c$, if it had not been the case then function would have been non-differentiable. $\endgroup$
    – ABC
    Apr 11, 2013 at 2:36

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