# If$|f(x)-f(y)|\le (x-y)^2$, prove that $f$ is constant

(Baby Rudin Chapter 5 Exercise 1)

Let $$f$$ be defined for all real $$x$$, and suppose that $$$$\tag{1} |f(x)-f(y)|\le (x-y)^2$$$$ Prove that $$f$$ is constant.

My attempt:

Let $$f$$ be defined for all real-valued inputs. Let $$x \in \mathbb{R}$$ and $$y \in \mathbb{R} \smallsetminus \{ x \}$$, and suppose that (1) holds. Then, we have: \begin{align*} \left| \dfrac{f(x)-f(y)}{x-y}\right| \le (x-y) \end{align*} As $$x\to y, \lim\limits_{x \to y}\left| \dfrac{f(x)-f(y)}{x-y}\right| \le 0$$. Since it cannot be that $$\left|f'(y)\right| < 0$$, we have that $$\left|f'(y)\right| = 0 \implies f'(y) = 0$$.

Can someone please read over my proof and let me know if it is correct?

• @KaviRamaMurthy Okay, thanks for reviewing it! Commented Jul 19, 2020 at 0:33
• To get this question removed out of "unanswered questions" queue, perhaps you can post your solution in the answer box below. Commented Jul 19, 2020 at 0:34
• sorry to ask, just curious, but why can $|f'(y)|$ not be less than $0$? Commented Jul 19, 2020 at 1:00
• An absolute value of a number can not be a negative number because it contradicts the definition of abs value Commented Jul 19, 2020 at 1:11

Your deduction that $$\frac{|f(x)-f(y)|}{|x-y|}\le x-y$$ is incorrect because it would lead to $$|f(x)-f(y)|\le |x-y|\cdot(x-y)\ne (x-y)^2$$ To make it work, you may want to deduce that $$\frac{|f(x)-f(y)|}{|x-y|}\le |x-y|$$

Technically there is a slight mistake.

You took $$x\in \Bbb{R}$$ and $$y\in \Bbb{R}$$ such that $$y\neq x$$, so you should $$y\rightarrow x$$. Then you arrive $$f'(x)=0$$ for all $$x\in \Bbb{R}$$, which gives $$f$$ is constant.

• Why downvote? Do you have any reason?
– user598858
Commented Jul 19, 2020 at 5:43
• I can't understand your answer. Are you saying that there is a difference between $y\neq x$ and $x\neq y$ Commented Jul 19, 2020 at 6:32
• The OP in the answer said $x\rightarrow y$ but it should be $y\rightarrow x$ because the answer starts with arbitrary $x\in\Bbb{R}$ and $y(\neq x)$ is the limiting variable. And if don't understand then you ask to understand.
– user598858
Commented Jul 19, 2020 at 6:42
• OK. I was asking to understand and I am not the downvote Commented Jul 19, 2020 at 6:43
• Ok, sorry......
– user598858
Commented Jul 19, 2020 at 6:44

Late answer, but here is a slight generalization of this problem.

Let $$X,Y\subseteq\mathbb{R}$$ and suppose $$f:X\to Y$$ is a function. Suppose $$f$$ is $$\alpha$$-Holder continuous with $$\alpha\in\mathbb{R}$$ and $$\alpha>1$$. Then, there exists a $$K\in\mathbb{R}$$ such that for all $$x,y\in X$$, we have $$|f(x)-f(y)|\leq K|x-y|^{\alpha}$$. Now, consider $$0\leq \lim_{x\to y}\bigg|\frac{f(x)-f(y)}{x-y}\bigg|\leq \lim_{x\to y}K|x-y|^{\alpha-1}$$ Since $$\alpha>1$$, then $$\alpha-1>0$$, which implies $$\lim\limits_{x\to y}|x-y|^{\alpha-1}=0$$. By the squeeze theorem, we have $$\lim\limits_{x\to y}\bigg|\frac{f(x)-f(y)}{x-y}\bigg|=0$$ which implies $$f'(x)=0$$. Since the only functions whose derivatives are identically zero are the constant functions, then we must have $$f(x)=c$$ for some $$c\in\mathbb{R}$$.

I stumbled upon this question on Functional Equation book by Christoper G. Small though the slight difference is merely that it is bounded by K(x–y)² for some K>0 instead.

So here's my two cents (from Functional Equation perspective)

\begin{align*} &\forall a,b \in \mathbb{R} \text{ and } a

Obviously, from the construction above, we see that: $$\forall k \in [1,n] \cap \mathbb{Z}, \; \, x_k-x_{k-1}=\frac{b-a}{n}$$

\begin{align*} |f(b)-f(a)| &= \left | \, \sum_{k=1}^n f(x_k)-f(x_{k-1}) \, \right | \\ &\leq \sum_{k=1}^n \left | \, f(x_k)-f(x_{k-1}) \, \right | = \sum_{k=0}^n K \left(x_k-x_{k-1}\right)^2 \\ &= \sum_{k=1}^n K \left( \frac{b-a}{n} \right)^2 = nK \left( \frac{b-a}{n} \right)^2 \\ &= \frac{K(b-a)^2}{n} \end{align*}

$$\text{As } \, n\to \infty, \, f(a)=f(b) \; \text{for all a,b in } \mathbb{R}$$

Therefore, f(x) is constant ∎