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 $f(x)=ax$?
How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable
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0http://www.math.rutgers.edu/~useminar/cauchy.pdf – 2012-01-18
3 Answers
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.
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0@John Dawkins: Brilliant!!! – 2012-01-18
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.
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0@Jonas Meyer: I have known your idea, thanks a lot – 2012-01-18
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: $f(x)=ax$