TITLE: Is $f(x) = Cx\log x$ the only solution to $f(xy) = xf(y) + yf(x)$?
QUESTION [9 upvotes]: I was studying $L(x) = x \log x$ function and found that it satisfies the following functional equation for positive $x, y$:
$$
f: \mathbb R^+ \to \mathbb R\\
f(xy) = x f(y) + y f(x)
$$
I have a feeling that $L(x)$ is the only (up to multiplying by a constant term) solution to that equation. How do I show that?

REPLY [3 votes]: In addition to William Huang solution, here's what i tried.
First note that $f(1)=0$ and $f(0)=0.$ 
Now let us set $x=x$ and $y=1+\frac{h}{x}$ 
Which will give us, $f(x+h)=xf\left(1+\frac{h}{x}\right)+\left(1+\frac{h}{x}\right)f(x)$
Which simplifies to, $$\frac{f(x+h)-f(x)}{h}=\frac{x}{h}f\left(1+\frac{h}{x}\right)+\frac{f(x)}{x}$$.
Now, taking limits we will have.
$f'(x)=L+\frac{f(x)}{x}$
Where $$L=\lim_\limits{h\to 0}\frac{f\left(1+\frac{h}{x}\right)}{\frac{h}{x}}$$ , which is of form $\frac{0}{0}$. Hence we can apply lhopitals rule to see that $L=f'(1)$.
Hence $f'(x)=f'(1)+\frac{f(x)}{x}$
Yet the solution to which is, $f(x)=f'(1)x \ln x$