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

Suppose $f(x)$ is integrable in any bounded interval on $\mathbb R$, and it satisfies the equation $f(x+y)=f(x)+f(y)$ on $\mathbb R$. How to prove $f(x)=ax$?

Integrate the functional equation with respect to $x$ between 0 and 1. The result is the equation $$\int_y^{y+1} f(u) du = \int_0^1 f(x) dx + f(y).$$ 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.

• Neat! ${}{}{}{}$ – Jonas Meyer Jan 18 '12 at 7:49
• @John Dawkins: Brilliant!!! – 89085731 Jan 18 '12 at 12:54

This is known as Cauchy's functional equation. It is easy to see that $f(0)=0$, as $f(x)=f(x+0)=f(x)$ for all $x$. If we left $a=f(1)$, we get that $f(n)=f(1+1+\cdots+1)=an$ for any $n\in\mathbb Z$ by induction, and similarly we see that $$bf(n/b)=f(n/b)+\cdots+f(n/b)=f(n/b+\cdots+n/b)=f(n)=an$$ 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.

• I have already known your solution, but I just don't know how to use the condition f(x) is integrable in any bounded interval on R – 89085731 Jan 18 '12 at 3:15
• @gingerjin: So you know how to handle the case where $f$ is continuous. You can show that if $f$ is discontinuous at a single point, then it is discontinuous everywhere. This is enough to imply that $f$ is Riemann integrable on no interval. If you are referring to Lebesgue integrability, with more work it can be shown that discontinuity of such $f$ implies that $f$ is not measurable. – Jonas Meyer Jan 18 '12 at 3:35
• @Jonas Meyer: Consider f(x)=x^(-0.5), it is not continuous when x=0, but it is integrable. – 89085731 Jan 18 '12 at 3:43
• @gingerjin: That function does not satisfy $f(x+y)=f(x)+f(y)$. (It also has the wrong domain.) – Jonas Meyer Jan 18 '12 at 3:50
• @Jonas Meyer: I have known your idea, thanks a lot – 89085731 Jan 18 '12 at 4:06

Define: $$g(x):=\int_0^xf(t)\,dt$$ It's easy to see that: $$g(x+y)=\int_0^{x+y}f(t)\,dt=\int_x^{x+y}f(t)\,dt+\int_0^xf(t)\,dt=g(x)+\int_0^yf(x+t)\,dt=g(x)+\int_0^yf(t)\,dt+\int_0^yf(x)\,dt=g(x)+g(y)+yf(x)$$ By symmetry, we get $g(x+y)=g(x)+g(y)+xf(y)$ so: $$yf(x)=xf(y)$$ Letting $a:=f(1)$ and putting $y=1$ in the last equation, we have: $$f(x)=ax$$