(Baby Rudin Chapter 5 Exercise 1)

Let $f$ be defined for all real $x$, and suppose that \begin{equation}\tag{1} |f(x)-f(y)|\le (x-y)^2 \end{equation} 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?

  • $\begingroup$ @KaviRamaMurthy Okay, thanks for reviewing it! $\endgroup$
    – User31415
    Commented Jul 19, 2020 at 0:33
  • 2
    $\begingroup$ To get this question removed out of "unanswered questions" queue, perhaps you can post your solution in the answer box below. $\endgroup$
    – Prism
    Commented Jul 19, 2020 at 0:34
  • $\begingroup$ sorry to ask, just curious, but why can $|f'(y)|$ not be less than $0$? $\endgroup$
    – C Squared
    Commented Jul 19, 2020 at 1:00
  • 2
    $\begingroup$ An absolute value of a number can not be a negative number because it contradicts the definition of abs value $\endgroup$
    – WindSoul
    Commented Jul 19, 2020 at 1:11

4 Answers 4


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 $$\frac{|f(x)-f(y)|}{|x-y|}\le |x-y|$$

Your solution is otherwise correct.


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.

  • $\begingroup$ Why downvote? Do you have any reason? $\endgroup$
    – user598858
    Commented Jul 19, 2020 at 5:43
  • $\begingroup$ I can't understand your answer. Are you saying that there is a difference between $y\neq x$ and $x\neq y$ $\endgroup$ Commented Jul 19, 2020 at 6:32
  • $\begingroup$ 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. $\endgroup$
    – user598858
    Commented Jul 19, 2020 at 6:42
  • $\begingroup$ OK. I was asking to understand and I am not the downvote $\endgroup$ Commented Jul 19, 2020 at 6:43
  • $\begingroup$ Ok, sorry...... $\endgroup$
    – 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<b, \; \text{let } x_0=a; \, x_n=b \\ &x_k=x_0 + k \, \frac{b-a}{n} \end{align*}

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 ∎


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