Properties of Continuous Functions with Additivity
Theorem
- [1] If a continuous function $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$ $$ f(x) = f(1) x $$
- [2] If a continuous function $g : \mathbb{R} \to ( 0 , \infty )$ satisfies $g(x + y) = g(x) g(y)$ for all $x, y \in \mathbb{R}$ $$ g(x) = \left( g(1) \right)^x $$
Explanation
The property where addition is preserved across functions as in $f(x + y) = f(x) + f(y)$ is called additivity, and the property where multiplication is preserved is called multiplicativity. $g$ feels like a mix of additivity and multiplicativity, which is called homomorphism.
Proof
Strategy: First, it’s shown that rational numbers can move in and out of functions, then a sequence of rational numbers converging to $x$ is created. Since continuity is guaranteed, $\lim$ is also utilized as a point that moves in and out of the function.
[1]
Part 1. $f(-x) = - f(x)$
By additivity $$ f( 0 ) = f( 0 + 0 ) = f(0) + f(0) \implies f(0) = 0 $$ Similarly, by additivity $$ 0 = f( 0 ) = f( x + (-x) ) = f(x) + f( -x ) \implies f(-x) = - f(x) $$
Part 2. $f(qx) = q f(x)$
For $n \in \mathbb{N}$ $$ f(nx) = f \left( \underbrace{x + \cdots + x}_{n} \right) = \underbrace{f(x) + \cdots + f(x)}_{n} = n f(x) \implies f(nx) = nf(x) $$ If we set $\displaystyle y := {{ x } \over {n}}$ $$ f(x) = f(ny) = n f(y) = n f \left( {{x} \over {n}} \right) \implies f \left( {{x} \over {n}} \right) = {{1} \over {n}} f(x) $$ And according to Part 1, these properties also hold for negative numbers, so for all $q \in \mathbb{Q}$ $$ f(qx) = q f(x) $$
Part 3. $f(x) = mx$
Defining a sequence of rational numbers converging to $x \in \mathbb{R}$, $\left\{ q_{n} \right\}_{n \in \mathbb{N}}$, based on Part 2 and the continuity in $f$
$$ \begin{align*} f(x) =& f( x \cdot 1 ) \\ =& f( \lim_{n \to \infty} q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} f( q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} q_{n} f( 1 ) = f(1) x \end{align*} $$
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[2]
Part 1. $\displaystyle g(-x) = \left( g (x) \right)^{-1}$
Since $g(x) \ne 0$ $$ g( 0 ) = g( 0 + 0 ) = g(0) g(0) \implies g(0) = 1 $$ Similarly, $$ 1 = g( 0 ) = g( x + (-x) ) = g(x) g( -x ) \implies g(-x) = {{1} \over {g(x)}} $$
Part 2. $g(qx) = \left( g(x) \right)^{q}$
For $n \in \mathbb{N}$ $$ g(nx) = g \left( \underbrace{x + \cdots + x}_{n} \right) = \underbrace{ g(x) \times \cdots \times g(x)}_{n} = \left( g(x) \right)^{n} \implies g(nx) = \left( g(x) \right)^{n} $$ If we set $\displaystyle y := {{ x } \over {n}}$ $$ g(x) = g(ny) = \left( g(y) \right)^{n} = \left( g \left( {{x} \over {n}} \right) \right)^{n} \implies g \left( {{x} \over {n}} \right) = \left( g(x) \right)^{{1} \over {n}} $$ And according to Part 1, these properties also hold for negative numbers, so for all $q \in \mathbb{Q}$ $$ g(qx) = \left( g(x) \right)^{q} $$
Part 3.
With $g(x) = a^x$ and defining a sequence of rational numbers converging to $x \in \mathbb{R}$, $\left\{ q_{n} \right\}_{n \in \mathbb{N}}$, based on Part 2 and the continuity in $g$ $$ \begin{align*} g(x) =& g( x \cdot 1 ) \\ =& g( \lim_{n \to \infty} q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} g( q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} g( 1 )^{q_{n}} \\ =& g( 1 )^{ \displaystyle \lim_{n \to \infty} q_{n}} \\ =& \left( g(1) \right)^{x} \end{align*} $$
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