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:

函数\(f: R \rightarrow R\)称为偶函数,如果对所有\(x \in R\)均有\(f(-x) = f(x)\)。函数\(f: R \rightarrow R\)称为奇函数,如果对所有\(x \in R\)均有\(f(-x) = -f(x)\)。用\(U_e\)表示\(R\)上的实值偶函数的集合,用\(U_o\)表示\(R\)上实值奇函数的集合,证明\(R^R = u_e \oplus U_o\).

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