Solution of the logarithmic Cauchy equation $ f ( x y ) = f ( x ) + f ( y ) $

Question

I was reading chapter 1 of the book on functional equation by Titu Anderescu.

I encountered two of the following theorems.

Theorem 1.1. Let $ f : \mathbb R \to \mathbb R $ be an additive function. Then the following four conditions are equivalent:
(i) $ f $ has the form $ f ( x ) = a x $ for some $ a \in \mathbb R $;
(ii) $ f $ is bounded above (below) on an interval;
(iii) $ f $ is increasing (decreasing) on an interval;
(iv) $ f $ is continuous at a point.

Theorem 1.6. Any solution $ f : ( 0 , \infty ) \to \mathbb R $ of the logarithmic Cauchy equation has the form of $ f ( x ) = g ( \ln x ) $, where $ g : \mathbb R \to \mathbb R $ is an additive function.

Then we have the following corollary.

An immediate consequence of Theorems 1.1 and 1.6 is the following characterization of the logarithmic function:

Corollary 1.2. Let $ f : ( 0 , \infty ) \to \mathbb R $ be a solution of the logarithmic Cauchy equation having one of the following properties:
(i) $ f $ is bounded above (below) on an interval;
(ii) $ f $ is increasing (decreasing) on an interval;
(iii) $ f $ is continuous at a point. Then $ f ( x ) = a \lg x $, where $ a \in \mathbb R $.

I am unable to prove Corollary 1.2 and I am also not able to understand that why $ f : ( 0 , \infty ) \to \mathbb R $ (instead of $ g : \mathbb R \to \mathbb R $) has to have any one of the three conditions.

A detailed proof for the same would be highly helpful and will be appreciated.

Answer 1

Proof. Let $f\colon (0,\infty)\to\mathbb R$ be a solution to the logarithmic Cauchy equation having any of the properties (i), (ii) or (iii) from Corollary 1.2. By Theorem 1.6 there is an additive function $g\colon\mathbb R\to\mathbb R$ such that $f(x)=g(\ln x)$. Since $\ln\colon (0,\infty)\to\mathbb R$ is a strictly increasing continuous bijection, we can conclude that $f$ having any of the properties (i), (ii) or (iii) of Corallary 1.2 is equivalent to $g$ having any of the properties (i), (ii) or (iii) of Corollary 1.2. Hence, $g$ has one of those properties and since $g$ is additive we can apply Theorem 1.1 to conclude that $g$ has the form $g(x)=ax$ for some $a\in\mathbb R$. Thus $f(x)=a\ln(x)$. $$\tag*{$\square$}$$

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The important consequences of the natural logarithm being a strictly increasing continuous bijection $(0,\infty)\to\mathbb R$ used here are:

• A subset $I\subseteq (0,\infty)$ is an interval if and only if $\ln(I)\subseteq \mathbb R$ is an interval.
• A function $g\colon\mathbb R\to\mathbb R$ is continuous at $x_0\in\mathbb R$ if and only if the function $f\colon (0,\infty)\to\mathbb R$ given by $f(x)=g(\ln x)$ is continuous at $\exp(x_0)$.
• A function $g\colon\mathbb R\to\mathbb R$ is increasing (decreasing) at $x_0\in\mathbb R$ if and only if the function $f\colon (0,\infty)\to\mathbb R$ given by $f(x)=g(\ln x)$ is increasing (decreasing) at $\exp(x_0)$.
• A function $g\colon\mathbb R\to\mathbb R$ is bounded above (below) on the interval $I\subseteq \mathbb R$ if and only if the function $f\colon (0,\infty)\to\mathbb R$ given by $f(x)=g(\ln x)$ is bounded above (below) on the interval $\exp(I)\subseteq (0,1)$.

Answer 2

By Theorem 1.6 we know a solution $f$ is of the form $f(x)=g(\log(x))$ with $g:\Bbb R\to\Bbb R$ an additive function, so we can't have $f$ defined for any value $x\le0$.

If $f$ is continuous at a point $x0\in(0,\infty)$ then $\lim_{x\to x_0}g(\log(x))=g(\log(\lim_{x\to x_0}x))$. But $\log(x)$ is continuous in $(0,\infty)$, so $\log(\lim_{x\to x_0}x)=\lim_{x\to x_0}\log(x)$. Then we can call $y_0=\log(x_0)$ and thus $\lim_{y\to y_0}g(y)=g(\lim_{y\to y_0}y)$. Therefore $g$ is continuous at a point.

If $f$ is increasing on an interval $I$ then $y>x\Rightarrow f(y)\ge f(x)$ on $I$. Now consider any $s,t\in\log(I)$ such that $t>s$. Since $\log$ is a surjective function there must be some $x,y\in(0,\infty)$ such that $\log(x)=s$ and $\log(y)=t$. Since $\log$ is strictly increasing in $(0,\infty)$ we must have $y>x$, because otherwise $x\ge y\Rightarrow \log(x)\ge\log(y)\Rightarrow s\ge t$, which is a contradiction. Since $y>x$ then $f(y)\ge f(x)\Rightarrow g(\log(y))\ge g(\log(x))\ge g(t)\ge g(s)$, so $g$ is increasing in the interval $\log(I)$. It is an interval because $\log$ is strictly increasing and continuous: continuous functions map connected sets to connected sets, and the only connected sets in $\Bbb R$ are intervals; and strictly increasing functions are open, so the interval must be open. You can prove the case for a decreasing $f$ on an interval similarly.

If $f$ is bounded on an interval $I$, again $\log(I)$ will be an interval. So $g$ is bounded on $\log(I)$.

In any case, the property for $f$ implies the same property for $g$, so by Theorem 1.1 we know $g(y)$ is of the form $g(y)=ay$ for some $a\in\Bbb R$. Therefore $f(x)=a\log(x)$.