EXERCISE 01分析进阶时滞微分方程一例若连续可导函数 fff 满足 ∣f(x)∣≤1|f(x)| \le 1∣f(x)∣≤1, 且 ∀x∈R,f′(x)=f(x+1)−f(x)\forall x \in \mathbb R, f'(x) = f(x + 1) - f(x)∀x∈R,f′(x)=f(x+1)−f(x). 证明: fff 是常数.知乎0 篇解答写解答