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Let $f$ be a function $(0;+\infty)\to\mathbb{R}$ with following property: $f(ab) = af(b) + bf(a)$. What can $f$ be?

It can be seen that functions $\delta_p(x) = px\space ln(x)$ work.

Now, let $f_0$ be a solution and $f_0(x_0)=y_0$, $x_0\neq1$. Because $\delta$ is continuous by $p$, there exists such $p_0$ that $\delta_{p_0}(x_0)=y_0$.
It can be proven that $\forall r\in\mathbb{Q}\space f_0(x_0^r)=\delta_{p_0}(x_0^r)$.

In other words, any solution must be a union of several $\delta_{p_S}$ functions restricted to subsets $Q_S\subset\mathbb{R}^+$, which are equivalence classes of relation $a\sim b$ if $a = b^q, q\in\mathbb{Q}$. The question is, can a solution include two or more different $\delta_p$, i.e. can it be non-continuous, i.e. can it be not of form $\delta_p$?

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    $\begingroup$ Potentially useful: If $g(x):=f(e^x)e^{-x}$, then $g(x+y)=g(x)+g(y)$. $\endgroup$
    – πr8
    Commented Sep 27, 2016 at 22:31
  • $\begingroup$ $f(x) = 0$ A polynomial series of $f(x)$ also arrives at this answer. $\endgroup$
    – user186104
    Commented Sep 28, 2016 at 14:17
  • $\begingroup$ @arthur Yes, this is $\delta_0$. $\endgroup$ Commented Sep 28, 2016 at 14:18

3 Answers 3

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Substitute $f(x)=xg(\ln x)$ (so $g(y)=e^{-y}f(e^y)$). Then $g\colon \mathbb R\to\mathbb R$ satisfies the Cauchy's equation $g(a+b)=g(a)+g(b)$, whose solutions are described in a known way (and may well be discontinous).

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As $\pi$r8 pointed out in the comments, this functional equation is really close to Cauchy's.

If $f$ is a solution and that you denote $g:x\rightarrow f(e^x)e^{-x}$, then for all $x,y$ $g(x+y) = e^{-x+y} (e^y f(e^x) + e^x f(e^y))$, $\textit{i.e.}$ $g(x+y) = g(x) + g(y)$, which is Cauchy's functional equation.

Thus, if you suppose for instance that $f$ is continuous at one point at least, you can prove that $g(x) = a x$ for some $a$, and thus $f = \delta_a$ is one of the solutions you identified.

But $f$ can indeed be non-continuous, because you can find $g$ non-continuous satisfying $g(x+y) = g(x) + g(y)$. Classically, in $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$, there exists a subspace $B$ s.t. $\mathbb{R} = \mathbb{Q} \oplus B$. We can consider $g$ the projection along $B$ onto $\mathbb{Q}$.

As $g(\mathbb{R}) = \mathbb{Q}$, you know that $g$ is nowhere continuous. And we define, for $x>0$, $f(x) = xg(\ln (x))$. Then for $x,y>0,\ f(xy) = xy\ g(\ln(x)+\ln(y))$ so $$f(xy) = yx\ g(\ln(x)) + xy\ g(\ln(y)) = yf(x)+xf(y)$$

But $f$ is not one of the $\delta_a$ (f is not continuous, not bounded, ...)

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First, verify that $f(0) = f(1) = 0$

For the functions that have up to third derivative.

$f(xy) = xf(y)+yf(x)$, differentiate by $x$ two times and by $y$ one time: $$ yf'(xy)=f(y)+yf'(x) $$$$ y^2f''(xy)=yf''(x) $$$$ yf''(xy)=f''(x) $$ $$ f''(xy)+xyf'''(xy)=0 $$ Now let $xy=z, f''(z) = \psi(z)$ and we gets differential equation on $\psi$: $\psi(z)+z\psi'(z)=0$ so that $\psi(z) = \frac{C1}{z}$

Then $f'(z)=C1\ln(z)+C2$ and $f(z) = C3 + C2z + C1(z\ln(z) - 1)$

Considering initial condition: $C3-C1=0, C2+C3-C1=0$ so $C2=0$ your initial form is the only correct.

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  • $\begingroup$ The last sentence of the question says that asking whether $f$ may be non-continuous is an alternative way of posing the same problem. So, to assume that it has some derivatives is too much. $\endgroup$ Commented Sep 28, 2016 at 14:56
  • $\begingroup$ Correct. My answer is for differentiable function up to third derivative. It is functional space limitation. Different methods have to be used for different functional spaces... $\endgroup$
    – hOff
    Commented Sep 28, 2016 at 16:29

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