How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable

Question

Suppose $f$ is integrable on any bounded interval in $\mathbb R$, and it satisfies the equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R$. How to prove that there exists a constant $a\in\mathbb R$ such that $f(x)=ax$ for all $x\in\mathbb R$?

Answer 1

Integrate the functional equation with respect to $x$ between $0$ and $1$. The result is the equation $$ \int_y^{y+1} f(u) du = \int_0^1 f(x) dx + f(y) \text . $$ The integral on the left side exists and is continuous in $y$ because $f$ is locally integrable. Therefore the right side is also continuous in $y$; that is, $f$ is continuous! The rest is clear sailing.

Answer 2

This is known as Cauchy's functional equation. It is easy to see that $f(0)=0$, as $f(x)=f(x+0)=f(x)+f(0)$ for all $x$. If we let $a=f(1)$, we get that $f(n)=f(\underbrace{1+1+\dots+1}_{n\text{ times}})=an$ for any $n\in\mathbb Z$ by induction, and similarly we see that $$bf\left(\frac n b\right)=\underbrace{f\left(\frac n b\right)+\dots+f\left(\frac n b\right)}_{b\text{ times}}=f\Big(\underbrace{\frac n b+\dots+\frac n b}_{b\text{ times}}\Big)=f(n)=an$$ so $f(x)=ax$ for any rational $x$. To extend this to all real $x$ under your restrictions, I suggest you look at other solutions than $ax$ and examine how they differ near $0$, then observe that these solutions also get scaled by addition and so blow up to be very different from $ax$ away from the origin. I will let you work this part out on your own.