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Question

How to construct a discontinuous $f$ such that $f(xy) = f(x)+f(y)$. Domain of $f$ has to be some subset of $\mathbb{R}$ and range of $f$ is $\mathbb{R}$. Also, try to construct non differentiable $f$ having the above property.

Context

In Apostol's calculus-1, it starts to give the motivation of definition of $\ln x$, where it want to exploit the property of the function that it should satisfy $f(xy) = f(x)+f(y)$ and then it assumes that $f$ is differentiable. I was wondering if we relax the differentiable assumption, how to proceed in finding that function?

Debu
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    Hint: See Cauchy's Functional Equation. Consider properties of the loagarithm. – lulu Jun 15 '24 at 18:05
  • As for your second question, every continuous function with this property from the positive reals to the reals has to be $\log$ in some basis, hence differentiable. The classification of such maps in a much larger context is a fascinating subject. When you're up for it, I would suggest you read about Pontryagin duality. – Uri George Peterzil Jun 15 '24 at 20:42

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Let $B$ be a basis for the reals as a vector space over $\mathbb{Q}$, and fix $v,w\in B$ distinct. Then the function $g_0:B\rightarrow B$ defined by $g_0(v)=w,g_0(w)=v$ and $g_0(u)=u$ otherwise extends uniquely to a $\mathbb{Q}$-linear map $g:\mathbb{R}\rightarrow\mathbb{R}$. Note that it is discontinuous at $v$; you can choose a sequence of $q_n\in\mathbb{Q}$ which both converge to a rational number $q$ and have that $q_n w$ converges to $v$ (this requires some formalizing but can be proved by density of the rational numbers and the fact that $w\neq 0$), which contradicts continuity since $v$ is not a rational multiple of $w$. Define $f:\mathbb{R}_+\rightarrow \mathbb{R}$ by $f=g\circ\log_e$. It is discontinuous at $e^v$ by the above, and by linearity of $g$ and the multiplicativity of the logarithm has:

$$f(xy)=g(\log(x)+\log(y))=g(\log(x))+g(\log(y))=f(x)+f(y)$$

for every pair of real numbers $x,y$. In fact, using a similar argument one can show that this function is nowhere-continuous (start by proving it for basis elements, then for their rational linear combinations, which is all of $\mathbb{R}$).

Uri George Peterzil
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