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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 $(1)$ ?

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    There are other solutions, but these are not continuous anywhere. See e.g. https://en.wikipedia.org/wiki/Cauchy's_functional_equation – 2017-01-09
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    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. – 2017-01-09

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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.

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    +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. – 2017-01-09
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    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 $f^{\prime}(x)=0$. – 2017-01-09
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    There are plenty of other solutions to $g(a+b) = g(a) + g(b)$ besides $g(x) = cx$ on $\mathbb R$! https://en.wikipedia.org/wiki/Cauchy's_functional_equation – 2017-01-09