## 设Ue是R上的偶函数集合，Uo是R的奇函数集合，证明R^R = Ue与Uo的直和

Foreword:

We use the operator “$$\oplus$$” to express “direct sum”.

Question:

Solution:

Let

$$U_e = \{f: \mathbb{R} \rightarrow \mathbb{R}$$ : such that f is even$$\}$$,

$$U_o = \{f: \mathbb{R} \rightarrow \mathbb{R}$$ : such that f is odd$$\}$$.

For each function $$f(x) \in \mathbb{R}^\mathbb{R}$$, we can divide it into:

$$f_e(x) = \frac{f(x) + f(-x)}{2}$$,

$$f_o(x) = \frac{f(x) – f(-x)}{2}$$.

Clearly $$f(x) = f_o(x) + f_e(x)$$.

Therefor $$\mathbb{R}^\mathbb{R} = U_e + U_o$$.

Since $$U_e \cap U_o = f_0(x) = 0$$, $$\mathbb{R}^\mathbb{R} = U_e \oplus U_o$$.