Logarithmic and exponential functions

Question

This is an exercise from the book "Logaritmos" by Elon Lages Lima.

A bijection $E:\mathbb{R} \rightarrow \mathbb{R}^+$ is called an exponential funtion when its inverse $F:\mathbb{R}^+ \rightarrow \mathbb{R}$ is a logarithmic function. Prove that the bijetion $E:\mathbb{R} \rightarrow \mathbb{R}^+$ is an exponential function if, and only if:

a) $E$ is increasing, i.e., $x < y \Rightarrow E(x) < E(y)$;

b) $E(x+y) = E(x)\cdot E(y)$.

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Additional info: in this book, the logarithmic function is defined as a function $L: \mathbb{R}^+ \rightarrow \mathbb{R}$ with the following properties:

i) $x< y \Rightarrow L(x) < L(y)$;

ii) $L(xy) = L(x) + L(y), \forall x, y \in \mathbb{R}^+$

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I managed to prove the first implication, i.e., that $E = F^{-1} \Rightarrow E$ has properties a) and b).

I've been stuck with the second implication.

Answer 1

The inverse of any strictly increasing function exists and is strictly increasing; this general fact shows that (a) implies (i).

If (b) holds, and $F=E^{-1}$, then applying (b) with $F(x)$ and $F(y)$ in place of $x$ and $y$ yields $$ E(F(x)+F(y)) = E(F(x))\cdot E(F(y)) = x\cdot y. $$ Now taking $F$ of both sides gives $$ F(x) + F(y) = F(E(F(x)+F(y))) = F(x\cdot y) $$ which is (ii).

Answer 2

$(\Rightarrow)$ We assume that $E$ is exponential, i.e., $E = F^{-1}$. We must show that a) and b) hold.

Let's show a), i.e., $E$ is increasing. Take $x_1, x_2, \in \mathbb{R}$ with $x_1 < x_2$. Assume that $E(x_1) \geq E(x_2)$.

Since $F$ is increasing and $E(x_1), E(x_2) \in \mathbb{R}^+$, $F(E(x_1)) \geq F(E(x_2))$. Since $E= F^{-1}$ by assumption, it follows that $x_1 \geq x_2$, which is a contradiction. Therefore, $E(x_1) < E(x_2)$, i.e., a) holds.

Now let's prove that b) hold.

Let $x, y \in \mathbb{R} = Im (F)$. Thus, there exists $x_0$ and $y_0 \in \mathbb{R}^+$ such that $F(x_0) = x$ and $F(y_0)= y$ (by surjectivity of $F$).

Adding the two equations we have $F(x_0) +F(y_0) = x + y$. Since $F$ is logarithmic, $F(x_0) +F(y_0) = F(x_0 \cdot y_0) = x + y$

Applying $E$, it follows that $E(x+y) = E(F(x_0 \cdot y_0)) = x_0 \cdot y_0$.

However, $F(x_0) = x \Leftrightarrow E(x) = x_0 $ and $F(y_0) = y \Leftrightarrow E(y) = y_0$.

Thus, $E(x+y)= x_0 \cdot y_0 \Rightarrow E(x + y) = E(x) \cdot E(y)$.

Therefore , b) holds.