Is $f(x) = Cx\log x$ the only solution to $f(xy) = xf(y) + yf(x)$?

Question

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?

Answer 1

Introduce a new function $g : \Bbb{R} \to \Bbb{R}$ by

$$g(x) = e^{-x} f(e^x).$$

Then satisfies the Cauchy's functional equation

$$g(x+y) = g(x) + g(y).$$

This equation is extensively studied, and even a mild regularity condition will force the solution to be of the form $g(x) = cx$. On the other hand, under the Axiom of Choice we can construct a solution which is not of this form.

Answer 2

Since the domain of $f$ is $\Bbb R_+$, we get an equivalent condition by dividing both sides by $xy$, namely $$\frac{f(xy)}{xy} = \frac{f(x)}{x} + \frac{f(y)}{y}.$$ By definition, this holds iff the function $$g(x) := \frac{f(x)}{x}$$ satisfies $$g(xy) = g(x) + g(y);$$ a function that satisfies this latter property is called additive.

Now, any monotonic additive function on $\Bbb R_+$ is a multiple of $\log x$ (this can be found at Dieudonné, Jean (1969). Foundations of Modern Analysis 1. Academic Press. p. 84.). So, if we require this condition on $g$, we have $$g(x) = C \log x$$ and hence $$f(x) = C x \log x$$ as desired.

On the other hand, not all additive functions are monotonic, so there are functions $f$ that satisfy the function equation but which are not of the indicated form.

Answer 3

I suppose the function $f$ is differentiable. Let $y=1$, it is easy to check $f(1)=0$.

Then, $$\lim_{y\rightarrow 1}\frac{f(xy)-f(x)}{xy-x}=\lim_{y\rightarrow 1}\frac{xf(y)+yf(x)-f(x)}{xy-x}=\frac{f(x)}{x}+\lim_{y\rightarrow 1}\frac{f(y)-f(1)}{y-1}$$ We obtain a differential equation, $$f'(x)=\frac{f(x)}{x}+f'(1)$$ And it is easy to find the solution is $$f(x)=f'(1)xlnx$$

Answer 4

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, $f(x)=f'(1)x \ln x$