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[Thanh Loan 7012]

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).

[vutuanhien]

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. 

Edited by vutuanhien, 01-03-2017 - 20:08.