因为 f[f(x)] = x,所以 f[f(x)]-f(x) = x-f(x),任取一点 x0,若 f(x0) = x0,则已找到 a = x0,使 f(a) = a；否则设 f(x0) = x1,此时 x0 ≠ x1.于是 f(x1)-x1 = x0-f(x0),也即 f(x1)-x1 与 f(x0)-x0 符号相反且均不为 0.设 g(x) = f(x)-x,则 g(x) 在两点 x0,x1 处反号,于是存在一点 a 属于 (x0,x1),使 g(a) = 0,所以存在一点 a,使 f(a) = a.
取 e^x,e^y 代入 f(xy) = f(x)+f(y),于是有 f(e^(x+y)) = f(e^x)+f(e^y),设 g(x) = f(e^x),那么 g(x+y) = g(x)+g(y)；取 y = dx,那么 g(x+dx) = g(x)+g(dx) = g(x)+g'(0)dx,所以 g'(x) = g'(0),也即 g(x) = g'(0)x+C,所以 f(x) = g(ln(x)) = Aln(x)+C,代入 f'(1) = a,得到 A = a,从而 f(x) = aln(x)+C,又 f(1) = f(1)+f(1),知道 f(1) = 0,所以 C = 0,求得 f(x) = aln(x).