Let $f$ be a function such that $f(xy)=xf(y)+yf(x)$ for all $x,y \in \mathbb{R}$. Prove that $f(1)=0$ and that $f(u^n)=nu^{n-1}f(u)$

Question

$u \in \mathbb{R}$ and $n \in \mathbb{N}$.

To prove that $f(1)=0$, am I on the right track by letting $x=\frac{1}{y}$ to get $f(xy)=f(\frac{1}{y}y)=\frac{1}{y}f(y)+yf(\frac{1}{y})$? If yes, how do I get $0$ from that?

To prove $f(u^n)=nu^{n-1}f(u)$.

There was a hint that said to use $f(u)$ to get $f(u^2)$ and up.

From that I got

$f(u)=f(u)$

$f(u^2)=f(u \cdot u)=2uf(u)$

$f(u^3)=f(u^2 \cdot u)=u^2f(u)+u(2uf(u))=3u^2f(u)$

$f(u^4)=f(u^3 \cdot u)=u^3f(u)+u(3u^2f(u))=4u^3f(u)$

Continuing this to n, I see that it is how you would calculate a derivative.

To figure out $n$, I'm letting $x=u^{n-1}$ and $y=u$. After that I get $f(u^n)=u^{n-1}f(u)+uf(u^{n-1})$. After looking at $u,u^2,u^3$, and $u^4$, I see that $uf(u^{n-1})=u(n-1)u^{n-2}f(u)$. How do I draw that conclusion in the proof and how does that become $nu^{n-1}f(u)$?

Edit: I know that $f(u)=f(u)$ only because I assumed that $f(1)=0$, but I don't how how to get that $f(1)=0$.

You're indeed on the right track. What happens when $y = 1$? As for the other part, are you familiar with mathematical induction? — eyeballfrog, Jul 06 '22 at 20:18
This section is about induction but I'm unsure how I'm supposed to use it in this one. — cypherfry, Jul 06 '22 at 20:23
For letting x=y=1 wouldn't that be me assuming at f(1)=0 to prove that f(1)=0 though? — cypherfry, Jul 06 '22 at 20:24
No, it wouldn't. Plugging in values assumes nothing about that. The equation you get is ultimately just $2z=z$ (for $z=f(1)$). Subtract $z$ from both sides to get $z=0$. No assumptions are made here. — PrincessEev, Jul 06 '22 at 20:25
Does this answer your question? Induction with multiple variables - found using an Approach0 search. There are also several closely related questions on this site, e.g., Proving $f'(1)$ exist for $f$ satisfying $f(xy)=xf(y)+yf(x)$, Is $f(x) = Cx\log x$ the only solution to $f(xy) = xf(y) + yf(x)$? and Continuous Functions Problems. — John Omielan, Jul 06 '22 at 22:53
@cypherfry Welcome to Math SE. FYI, there are also quite a few related AoPS threads, e.g., Probably an easy functional equation f(xy)=xf(y)+yf(x), Functional Equations and Functions Question. — John Omielan, Jul 06 '22 at 22:59

score 1 · Accepted Answer · answered Jul 06 '22 at 20:23

To prove that $f(1) = 0$ just let $x=y=1$. Then

$$f(1 \cdot 1) = f(1) \qquad 1 \cdot f(1) + 1 \cdot f(1) = 2 f(1)$$

Of course, if $f(1) = 2 f(1)$, then $f(1) = 0$

For the other, let $x = u, y = 1$. Then

$$f(u) = u f(1) + f(u) = f(u) = 1 \cdot u^{1-1} f(u)$$

Now work by induction.

Suppose that $f(u^n) = nu^{n-1} f(u)$ for all $n \in \{1,\cdots,k\}$. We wish to validate the claim for $k+1$:

$$f(u^{k+1}) = (k+1) u^k f(u)$$

Well,

$$f(u^{k+1}) = f(u \cdot u^k) = u f(u^k) + u^k f(u) = u^k f(u)$$

by using the definition of $f$. Using the induction hypothesis,

$$u f(u^k) = ku^k f(u)$$

so

$$f(u^{k+1}) = ku^k f(u) + u^k f(u) = (k+1) u^k f(u)$$

as desired.

Let $f$ be a function such that $f(xy)=xf(y)+yf(x)$ for all $x,y \in \mathbb{R}$. Prove that $f(1)=0$ and that $f(u^n)=nu^{n-1}f(u)$

1 Answers1