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I wanna determine whether it is true that if $|f(x)− f(y)| ≤ |x−y|$ for all real $x$ and $y$, then $f$ is a constant function. How can this be proved?

I know the solution for $|f(x)− f(y)| ≤ |x−y|^2$, then you simply divide by $(x-y)$ and take the limit, but what about this problem?

In addition, for which $p ∈ \Bbb R$ is it true that if $|f(x)− f(y)| ≤ |x−y|^p$ for all real x and y, then f is a constant function?

Sebastiano
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John_Average
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4 Answers4

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There are lots of function $f$ which satisfy $|f(x)-f(y)|\leq|x-y|$ and are not constant. One example is the sine function: $$|\sin(x)-\sin(y)|=|\cos(\xi)|\cdot|x-y|\leq|x-y|$$ for some $\xi$ between $x$ and $y$ by the mean value theorem. Functions like $\sin$ which satisfy $|f(x)-f(y)|\leq M|x-y|$ for some constant $M$ are known as Lipschitz-continuous functions.

Functions which satisfy the more general property $|f(x)-f(y)|\leq M|x-y|^\alpha$ are known as $\alpha$-Hölder continuous functions. Note that $\alpha=1$ corresponds to Lipschitz functions, so this is a generalization of the concept of Lipschitz continuity. All $\alpha$-Hölder continuous functions with $\alpha>1$ are constant by the same argument that you gave: $$\lim_{x\to y}\left|\frac{f(x)-f(y)}{x-y}\right|\leq\lim_{x\to y}\frac{M|x-y|^\alpha}{|x-y|}=\lim_{x\to y}M|x-y|^{\alpha-1}=0,$$ so every $\alpha$-Hölder continuous function with $\alpha>1$ is differentiable and has $f'=0$ everywhere, and is hence constant.

csch2
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  • So how should I think about the more general, i.e. for which $p ∈ R$ is it true that if $|f(x)− f(y)| ≤ |x−y|^p$ for all real x and y, then f is a constant function? – John_Average Mar 31 '21 at 20:02
  • From the argument I gave you have that if $p>1$, then $f$ must be constant. I gave an explicit counterexample (the sine function) in the case $p=1$. Can you construct counterexamples for all $p<1$? – csch2 Mar 31 '21 at 20:04
  • @John_Average Adding to comment forall $p > 1$, $f$ would be a constant; wheras for all $p \leq 1$, you can find a counterexample. – Anon Mar 31 '21 at 20:06
  • Is it possible to find some general sufficient conditions on the function $c_p : R×R → R$ such that $c$ is not constant function, and if $|f(x)− f(y)| ≤ c_p(x, y)(x−y)^p$ for all real x and y, then f is a constant function? – John_Average Mar 31 '21 at 20:11
  • Yeah $c_p(x,y) = |x - y|^{|p| + 2}$ is a function which makes $f$ always constant, lol. – Anon Mar 31 '21 at 20:14
  • Can this answer be proved, and for those p for which such functions c exist, can you give example of some nonconstant function $c_p(·,·)$ with that property? – John_Average Mar 31 '21 at 20:17
  • Proof is trivial (try it yourself), and the above function holds for all $p$. – Anon Mar 31 '21 at 20:19
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To get a counterexample, simply consider the sine function. The inequality is satisfied by the Mean value inequality. Yet, is $\sin x$ constant?

Edit:

The condition means the function is $1$-Lipschitz. Another simple example of a non-constant $1$-Lipschitz function is the absolute value, since, by the triangle inequality ($2^\text{nd}$ form), $$\bigl||x|-|y|\bigr|\le |x-y| .$$

Bernard
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Take $f(x)=\sin(x)$.

By MVT,

$$(\forall x,y\in\Bbb R)\;\;|f(x)-f(y)|=|(x-y)\cos(c)|$$ $$\le |x-y|$$

but $ f $, as i think, is not constant.

hamam_Abdallah
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A trivial non-constant $f$ such that $|f(x) - f(y)| \leq |x - y|$ is $f(x) = x$.

In fact, consider $f(x) = x$ if $0 \leq x \leq 1$, $f(x) = 0$ if $x \leq 0$, and $f(x) = 1$ if $x \geq 1$. In this case, we see that for all $p \leq 1$, for all $x, y$, $|f(x) - f(y)| \leq |x - y|^p$.

Now suppose that $p > 1$. In this case, we see that for all $n$, we have $|f(0) - f(x)| = |f(0) - f(\frac{1}{n} x) + f(\frac{1}{n} x) - f(\frac{2}{n} x) + ... + f(\frac{(n - 1)}{n} x) - f(x)| \leq |f(0) - f(\frac{1}{n} x)| + |f(\frac{1}{n} x) - f(\frac{2}{n} x)| + ... + |f(\frac{(n - 1)}{n} x) - f(x)| \leq |0 - \frac{1}{n} x|^p + |\frac{1}{n} x - \frac{2}{n} x|^p + ... + |\frac{n - 1}{n}x - x| = n \cdot (\frac{1}{n})^p = \frac{1}{n^{p - 1}}$.

But since $p > 1$, we can make $n^{p - 1}$ arbitrarily large. Then $|f(0) - f(x)| = 0$. Then $f(0) = f(x)$. Then $f$ is constant.

Mark Saving
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