# Find all functions $f : \mathbb{(0,\infty})\to\mathbb{R}$ such that $f(x+y) = xf(y)+ yf(x)$

Find all functions $$f : \mathbb{(0,\infty)}\to\mathbb{R}$$ such that $$f(x+y) = xf(y)+ yf(x)$$ if $$f$$ is continuous at $$x=1$$

This problem was looking quite easy at first but the domain of positive reals is posing me a problem. I couldn't plug in zero's for $$x$$ and $$y$$. I tried putting $$x=y$$ but the result $$f(2x) = 2xf(x)$$ couldn't be used as a recurrence relation $$\infty$$ times as that would yields $$f(0)$$ again. I've run out of ideas. Please help.

• Also: continuity at $x = 1$ should be a powerful tool, but to use it, you'll need to look at $f(1 + u)$ for small values of $u$. How might you do that? – John Hughes Apr 23 '19 at 17:24
• If I'm correct $\lim_{u\to 0} f(1+u)=f(1)=\lim_{u\to 0} f(u) +uf(1)=\lim_{u\to 0}f(u)$ hence the limit of $f(x)$ near zero exists so by applying $\lim_{y\to 0}$ on both sides we get $f(x) = \lim_{y\to 0}xf(y)=xf(1)$ – kingW3 Apr 23 '19 at 17:42
• I can't get a full answer, but your function is 0 in all integers: Let her be $f$ We know $f(2) = f(1+1) = f(1)+f(1) = 2f(1)$ Also, $f(4) = f(2+2) = 2f(2) + 2f(2) = 8f(1)$ But $f(4) = f(3) + f(1) = f(2+1) + 3f(1) = 2f(1) +f(2) +f(1) = 7f(1)$ Therefore, $8f(1) = 7f(1)$ and f(1) = 0 – David Apr 23 '19 at 17:45

## 2 Answers

This is one possible continuation of your attempted solution.

By letting $$\ y=x=z\$$ you found that $$f(2z) = 2z f(z). \tag{1}$$ Let $$\ y=2z,\ x=z\$$ to get $$f(3z) = 2z f(z) + z f(2z). \tag{2}$$ Let $$\ y=2z,\ x=2z\$$ to get $$f(4z) = 4z f(2z). \tag{3}$$ Let $$\ y=3z,\ x=z\$$ to get $$f(4z) = 3z f(z) + z f(3z). \tag{4}$$ Now eliminate $$\ f(2z),\ f(3z),\ f(4z)\$$ from the four linear equations to get $$(3-6z+2z^2)f(z) = 0. \tag{5}$$ There are two positive roots of the quadratic which implies that $$\ f(z) = 0\$$ for any other values of $$\ z.\$$ Use equation $$(1)$$ to prove $$\ f(z) = 0\$$ for the positive roots also.

Notice that no continuity assumption or $$\ f(0)\$$ was used to prove $$\ f(z)=0\ \forall z.$$

If $$f$$ satisfies the given condition so does $$cf$$ for any constant $$c$$. Unless $$f(1) = 0$$ we might as well assume $$f(1) = 1$$. Some calculations:

$$f(2) = f(1 + 1) = f(1) + f(1) = 2$$ $$f(3) = f(1 + 2) = 2f(1) + f(2) = 4$$ $$f(4) = f(2 + 2) = 2f(2) + 2 f(2) = 8$$ $$f(4) = f(1 + 3) = 3f(1) + f(3) = 3 + 4 = 7$$ The last two equations are incompatible so we must have $$f(1) = 0$$.

Doing calculations like the above gives you $$f(n) = 0$$ and $$f(2^{-n}) = 0$$ for all natural numbers $$n$$. The additivity extends this to $$f(r) = 0$$ for all dyadic rationals $$r$$. This should give you that $$f(x) = 0$$ for all positive $$x$$ if you can exploit the continuity condition.