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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.

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  • $\begingroup$ 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? $\endgroup$ – John Hughes Apr 23 '19 at 17:24
  • $\begingroup$ 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 $f(x) = \lim_{y\to 0}xf(y)=xf(1)$ $\endgroup$ – kingW3 Apr 23 '19 at 17:42
  • $\begingroup$ 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 $\endgroup$ – David Apr 23 '19 at 17:45
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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 $\ f(z)=0\ \forall z.$

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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.

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