The relative error transformation $T(f)=\dfrac{f^\prime}{f}$ for differentiable functions $f:\mathbb{R}\to\mathbb{R}$ satisfies the properties

  1. $T(fg)=T(f)+T(g)$
  2. $T\left(\dfrac{f}{g}\right)=T(f)-T(g)$
  3. $T\left(f^n\right)=nT(f)$
  4. $(f+g)T(f+g)=f\,T(f)+g\,T(g)$

The first three properties are shared with logarithmic functions on $\mathbb{R}^+$, but not the fourth.

Suppose $f:\mathbb{R}\to\mathbb{R}$ and for $a,\,b\in\mathbb{R}$

\begin{equation} (a+b)f(a+b)=af(a)+bf(b)\tag{1} \end{equation}

Clearly, every constant function defined on $\mathbb{R}$ satisfies this property. If $f(x)=c$ then we have

\begin{eqnarray} (a+b)f(a+b)&=&(a+b)c=ac+bc\\ af(b)+bf(b)&=&ac+bc \end{eqnarray}

Are there other functions $f:\mathbb{R}\to\mathbb{R}$ that satisfy property $(1)$ ?

  • 1
    $\begingroup$ There are other solutions, but these are not continuous anywhere. See e.g. en.wikipedia.org/wiki/Cauchy's_functional_equation $\endgroup$ – Winther Jan 9 '17 at 19:53
  • $\begingroup$ I used to ask my freshmen algebra students to show that if $f,g$ satisfy $(1)$ and $h=c_1f+c_2g$, $c_1,c_2\in\mathbb{R}$ then $h$ satisfies $(1)$. I asked my calculus students to show that if $f$ were differentiable, and if $f$ satisfied $(1)$ then $f$ is a constant function. $\endgroup$ – John Wayland Bales Jan 9 '17 at 20:10

Let $g(x) = xf(x) \implies g(a+b) = g(a) + g(b)$. The general solution for this functional equation is that $g(x) = cx$. Thus $f(x) = c$ is the only function which meets this condition.

  • 1
    $\begingroup$ +1 Nice answer, but I don't think "literature" is the word you want. "Solution" is a good word for this. Or "continuous solution" would be more accurate. $\endgroup$ – Thomas Andrews Jan 9 '17 at 19:37
  • $\begingroup$ Good argument. I was unfamiliar with this argument and could only show it true if $f$ were differentiable since for differntiable $f$ it must be the case that $f^{\prime}(x)=0$. $\endgroup$ – John Wayland Bales Jan 9 '17 at 19:52
  • 2
    $\begingroup$ There are plenty of other solutions to $g(a+b) = g(a) + g(b)$ besides $g(x) = cx$ on $\mathbb R$! en.wikipedia.org/wiki/Cauchy's_functional_equation $\endgroup$ – Joshua Mundinger Jan 9 '17 at 19:53

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.