9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I suppose, we wish to find continous solution $\endgroup$ – Michael Galuza Jun 5 '15 at 19:13
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.