1, Given $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying f(0)=2017
$\left | f(x)-f(y) \right |\leq \left | x-y \right |^{2}$
Compute f(2017).
2,1, Given $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying f(0)=0
$\left | f(x)-f(y) \right |\leq \left | x-y \right |^{2}$
Compute f(2017).
Let me prove a general result: If $f:\mathbb{R}\to \mathbb{R}$ satisfies $\left | f(x)-f(y) \right |\leq \left | x-y \right |^2$, then $f$ must be constant.
First, fix $x_{0}\in \mathbb{R}$ and $\forall \epsilon>0$, consider $\delta=\sqrt{\epsilon}$. Then for all $x\in (x_{0}-\delta, x_{0}+\delta)$, we have $\left | f(x)-f(x_{0}) \right |\leq \left | x-x_{0} \right |^2<\epsilon$ and we get $f$ is continuous, which follows from definition.
Now fix $y\in \mathbb{R}$. Since $\left | f(x)-f(y) \right |\leq \left | x-y \right |^2$ for all $x\in \mathbb{R}$, we have $0\leq \dfrac{\left | f(x)-f(y) \right |}{ \left | x-y \right |}\leq \left | x-y \right |$ (for all $x\neq y$). Let $x\to y$ and by the squeezing theorem, $\lim_{x\to y}\dfrac{\left | f(x)-f(y) \right |}{\left | x-y \right |}=0$. So $\lim_{x\to y}\dfrac{f(x)-f(y)}{x-y}=0$, which means $f$ is differentiable and $f'(y)=0$ (for all $y\in \mathbb{R}$). Therefore, $f$ must be constant.
It follows from the above result that $f(x)\equiv 2017$ in your first problem and $f(x)\equiv 0$ in your second problem.
Bài viết đã được chỉnh sửa nội dung bởi vutuanhien: 01-03-2017 - 20:08
"The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!" - Grothendieck