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?

  • $\begingroup$ Let $f(x)=x\log{x}$. Do I miss anything: $\sout{f(xy)=xy\log{xy}=xy\log{x}+xy\log{y}\neq x\log{y}+y\log{x}=xf(y)+yf(x)}$. Apologies.. $\endgroup$ – rafforaffo Jun 5 '15 at 8:23
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    $\begingroup$ @rafforaffo Nope. $xy\log x + xy\log y = y(x\log x) + x(y \log y) = yf(x) + xf(y)$. $\endgroup$ – Empiricist Jun 5 '15 at 8:24
  • $\begingroup$ And my previous comment is clearly false...apologies $\endgroup$ – rafforaffo Jun 5 '15 at 8:29
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    $\begingroup$ An operation that satisfies this (or rather $f(xy) = xf(y) + f(x)y$) is called a derivation. If $x$ and $y$ were themselves functions, then differentiation is one such operation. Not that it helps you much. $\endgroup$ – Arthur Jun 5 '15 at 8:31
  • $\begingroup$ @Arthur yes, noticed that too $\endgroup$ – uranix Jun 5 '15 at 8:33

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.

  • $\begingroup$ I suppose, we wish to find continous solution $\endgroup$ – Michael Galuza Jun 5 '15 at 19:13

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 $$f(x)=f'(1)xlnx$$

  • $\begingroup$ I had a similar solution by setting $x=x$ and $y=1+\frac{h}{x}$ and then defining derivative and taking limit. $\endgroup$ – Mann Jun 5 '15 at 9:55

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.

  • $\begingroup$ Both papers you've provided are about arithmetic functions, i.e. defined on $\mathbb{N}$ $\endgroup$ – uranix Jun 5 '15 at 8:56
  • $\begingroup$ @uranix Oops, you're right, I've replaced the reference. $\endgroup$ – Travis Willse Jun 5 '15 at 9:21

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.


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


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