Proving that a function satisfying $|f(x)-f(y)| \leq |x-y|^3$ is constant

Question

Let $\mathbb R$ be the set of real numbers and $f: \mathbb R \rightarrow \mathbb R$ be such that for all $x$ and $y$ in $\mathbb{R}$, $$|f(x)-f(y) |\leq |x-y|^3.$$ Prove that $f(x)$ is a constant.

This is a new type of problem for me and I feel I am missing some trick to simplify the given expression. Any help??

The stated condition means that $f$ has a derivative everywhere, which can be found by using the definition of derivative. — Will Jagy, Aug 17 '14 at 05:08

score 4 · Answer 1 · answered Aug 17 '14 at 05:08

4

The limit definition of the derivative says that $f'(y) = \displaystyle\lim_{x \to y}\dfrac{f(x)-f(y)}{x-y}$ (provided the limit exists).

Here, we have that $\left|\dfrac{f(x)-f(y)}{x-y}\right| < |x-y|^2$. What does this tell you about the derivative of $f$?

answered Aug 17 '14 at 05:08

JimmyK4542

54,331

Now, I understand the value of derivative is zero, but I am still confused about the absolute value. – user34304 Aug 17 '14 at 05:13
What about the absolute value is confusing? – JimmyK4542 Aug 17 '14 at 05:23
@JimmyK4542: check the fundamental definition of the derivative at a point $a$. – Mhenni Benghorbal Aug 17 '14 at 07:43
Why the down vote? – copper.hat Aug 17 '14 at 17:55

copper.hat · Answer 2 · 2019-08-07T23:27:16.870

3

Here is a version without derivatives.

Choose $x,y$. Let $t_k = x+{k \over n} (y-x)$, for $k=0,...,n$.

Then $|f(x)-f(y)| \le \sum_{k=0}^{n-1} |f(t_{k+1})-f(t_k)| \le (x-y)^3 \sum_{k=0}^{n-1} {1 \over n^3} = {1 \over n^2}(x-y)^3$.

Since this is true for all $n$, we have $|f(x)-f(y)| = 0$, or in other words, $f(x) = f(y)$.

edited Aug 07 '19 at 23:27

answered Aug 17 '14 at 05:37

copper.hat

172,524

Why the down vote? – copper.hat Aug 17 '14 at 17:54
I was wondering how you get that $(x-y)^3 \sum_{k=0}^{n-1} {1 \over n^3} = {1 \over n^2}(x-y)^3$? – Skm Aug 07 '19 at 19:00
The summand is not a function of $n$. – copper.hat Aug 07 '19 at 19:04
so why don't we have $(x-y)^{3}\sum_{k=0}^{n-1} {1 \over n^3} \le \dfrac{1}{n^{2}}(x-y)^{3}$ instead? – Skm Aug 07 '19 at 19:08
1

I'm not sure what you are asking. $\sum_k {1 \over n^3} = n {1 \over n^3} = {1 \over n^2} $. – copper.hat Aug 07 '19 at 19:11
At first, I thought that since the summand is not a function of $n$, that meant that $\sum_{k} \dfrac{1}{n^{3}} = \dfrac{1}{n^{3}}$ – Skm Aug 07 '19 at 19:25
1

Sorry, I meant not a function of $k$. – copper.hat Aug 07 '19 at 19:26
Should $t_{k}$ should be equal to $x+ \dfrac{k}{n}(y-x)$ instead? – Skm Aug 07 '19 at 20:33
1

@K.M: Yes indeed. Thanks for catching that! – copper.hat Aug 07 '19 at 23:27

Proving that a function satisfying $|f(x)-f(y)| \leq |x-y|^3$ is constant

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