Axler’s Linear Algebra Done Right, Exercise 1C-24

Problem
A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called even if

$f(-x) = f(x)$

for all $x \in \mathbb{R}$ . A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called odd if

$f(-x) = -f(x)$

for all $x \in \mathbb{R}$ . Let $U_e$ denote the set of real-valued even functions on $\mathbb{R}$ and let $U_o$ denote the set of real-valued odd functions on $\mathbb{R}$ . Show that:

$\mathbb{R}^\mathbb{R}=U_e \oplus U_o$

Proof. To show that $U_e + U_o$ is a direct sum, it is sufficient to prove that $U_e \cap U_o = {0}$ . Let $f \in U_e \cap U_o$ . Hence, $f(x) = f(-x) = -f(-x)$ for all $x \in R$ , which only happens for $f(x) = 0$ . Hence, $U_e + U_o$ is a direct sum.

Let $g \in \mathbb{R}^\mathbb{R}$ . The equation $\mathbb{R}^\mathbb{R}=U_e \oplus U_o$ means that $g$ can be written as the sum of an odd function $o$ and an even function $e$ . Or in other words, for all $x \in \mathbb{R}$ :

(1) $\begin{equation*} g(x) = o(x) + e(x)\end{equation*}$

and also:

(2) $\begin{equation*} g(-x) = o(-x) + e(-x)\end{equation*}$

By the definition of the odd and even functions, (2) can be written as:

(3) $\begin{equation*} g(-x) = -o(x) + e(x)\end{equation*}$

The system of linear equations (1) and (3) has a unique solution $o(x) = (g(x) - g(-x)) / 2$ and $e(x) = (g(x) + g(-x)) / 2$ . Hence, $g$ can always be uniquely written as the sum of an odd function and an even function. $\qed$

Axler’s Linear Algebra Done Right, Exercise 1C-24

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