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!)
Best Answer
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 $$f(x) = cx$$