Following is the theorem regarding the solution of the exponential Cauchy equation
$$ f ( x + y ) = f ( x ) f ( y ) \text . $$
Let $ f : D \to \mathbb R $ be a solution of the exponential Cauchy equation, where $ D = \mathbb R $ or $ D = ( 0 , + \infty ) $. Then either $ f \equiv 0 $ or $ f ( x ) = e ^ { g ( x ) } $, where $ g : D \to \mathbb R $ is an additive function.
Below is the corollary which is a direct result of the above theorem.
Let $f:D \to \mathbb R $ be a solution of the exponential Cauchy functional equation, where $D=\mathbb{R}$ or $(0,\infty)$.
(1) If $f$ is monotonic on an interval, then $f(x)=e^{ax}$, where $a\in \mathbb{R}$.
(2) If $f$ is continuous at a point or bounded above or below an interval, then either $f = 0$ or $f(x)=e^{ax}$.
My question is that how can we prove that if $f(x)=e^{g(x)}$ is continuous, monotonic or bounded then $g(x)=\ln{f(x)}$ is also continuous, monotonic or bounded?
Best Answer
If $f(x)=e^{g(x)}$ then $f:D\to(0,+\infty)$, since the exponential function is always positive, so $g(x)=\log(f(x))$ is well-defined in $D$, whatever it may be.
If $f$ is continuous, for any $x_0\in D$ we have $\lim_{x\to x_0}f(x)=f(\lim_{x\to x_0}x)=f(x_0)$. We know $f(x)>0$ and that $\log$ is continuous in $(0,+\infty)$, so $\lim_{x\to x_0}g(x)=\lim_{x\to _0}\log(f(x))=\log(\lim_{x\to x_0}f(x))=\log(f(x_0))=g(x_0)$, so $g$ is continuous.
If $f$ is monotonically decreasing, for $x\le y$ it follows $f(y)\le f(x)$. We know $\log$ is monotonically (strictly) increasing in $(0,+\infty)$ (just check the first derivative), so for $x\le y$ it follows $\log(x)\le \log(y)$. So for $x\le y$ we have $f(y)\le f(x)$, and then $\log(f(y))\le\log(f(x))$, which is the same as $g(y)\le g(x)$. Therefore $g$ is monotonically decreasing. The proof for $f$ monotonically increasing is similar.
If $f$ is bounded above then there is some $M$ such that $f(x)\le M$ for all $x\in D$. Therefore, for all $x\in D$, $f(x)\le M\Rightarrow \log(f(x))\le\log(M)$, so $g$ is bounded above.
The last statement doesn't seem to be true, since $f$ is always bounded below by $0$, but $\log(f(x))$ would approximate $-\infty$ if $f(x)$ approximated $0$. If the statement has to be interpreted as "there is some number in $(0,+\infty)$ such that it works as lower bound", then there is some $M>0$ such that $f(x)\ge M$ for all $x\in D$ and the proof is similar as earlier.