证：
设任意函数f(x)
另g(x) = [ f(x) + f(-x) ] /2
h(x) = [ f(x) - f(-x) ] /2
则：
g(-x) = [ f(-x) + f(x) ] /2 = [ f(x) + f(-x) ] /2 = g(x)
h(-x) = [ f(-x) - f(x) ] /2 = -[ f(x) - f(-x) ] /2 = -h(x)
g(x) + h(x) = f(x)
显然：
g(x)为偶函数
h(x)为奇函数
且g(x) + h(x) = f(x)
证毕