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Suppose $f:(a,b) \to \mathbb{R} $ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some $\alpha >1$ and all $x,y \in (a,b) $. Prove that $f$ is constant on $(a,b)$.

I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?

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yes, indeed! (to your last question) – Avitus Dec 11 '13 at 20:41
What does the value $M$ mean? – John Dec 11 '13 at 20:41
the condition should be $\alpha>1$ otherwise it just an holder function and they are everything but constant (in general). In case $\alpha>1$ you just show $f$ is differentiable with $0$ has a derivative. (like the answer below). Can you correct the question ? – user42070 Dec 11 '13 at 21:08
up vote 5 down vote accepted

divide by $|x-y| $ both members. you get

$$\frac{|f(x) - f(y)|}{|x-y|} \le M |x-y|^{\alpha - 1} \ \ (1)$$

now, since $(1)$ has to hold $\forall \ x, y$ then set $y = x + h$, with $h \to 0$

it becomes

$$|f'(x)| \le M \ |h|^{\alpha-1} = 0$$ ($\alpha - 1 > 0$ so there's no problem there)

So $|f'(x)| \le 0$, but of course also $|f'(x)| \ge 0$, it implies $|f'(x)| = 0$. Hence $f'(x) = 0$


We can formalize it more. Let's do it.

$$\frac{|f(x) - f(y)|}{|x-y|} \le M |x-y|^{\alpha - 1} \ \ (1)$$

We can always set $x = y + h, h > 0$ since $(1)$ has to hold $\forall x, y$


$$\frac{|f(y+h) - f(y)|}{h} \le M h^{\alpha - 1} \ \ (1)$$

(note $|h| = h$)

Now, let's suppose $f(y+h) - f(y) \ge 0 \ \ \ \ \ (2a) $

It implies that $f'(y) \ge 0$ (it suffice to divide $(2a)$ by $h$ and then taking the limit for $h \to 0$ to show it)

But it also implies, recalling (1) that

$$\frac{f(y+h) - f(y)}{h} \le 0 \Rightarrow f'(y) \le 0$$

(Again by taking the limit of both parts.)

But these last two results imply that $f'(y) = 0 $

We can do the exact same reasoning in the case that $f(y+h) - f(y) \le 0$

So again, in this case we find $f'(y) = 0$

Thus we have demonstrated that in both cases ($f(y+h) > f(y)$ and $f(y+h) < f(y)$) we have $f'(y) = 0$, so this has to hold $\forall y$

This implies $f = const$

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this is good one – james Miler Dec 11 '13 at 20:47
Are you using this for your second equation? $\lim_{x \rightarrow a}{|f(x)|}=|\lim_{x \rightarrow a}{f(x)}|$, I thought this equality does not hold? – Idonknow Dec 11 '13 at 21:06
edited.. it should be better now :-) – Ant Dec 11 '13 at 23:56

Hint: Show that $f'(y)$ exists and is equal to $0$ for all $y$. Then as usual by the Mean Value Theorem our function is constant.

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+1 for speed :-) – Avitus Dec 11 '13 at 20:41
so what is first conditio refer to? – james Miler Dec 11 '13 at 20:42
Divide by $|x-y|$. We are left on the right with $\le M|x-y|^{\alpha-1}$, which approaches $0$ as $x\to y$. – André Nicolas Dec 11 '13 at 20:43

Without any derivative...

Choose $y\gt x$ in the interval $(a,b)$. For every $n\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\leqslant M(x_{i+1}-x_i)^\alpha=M(y-x)^\alpha n^{-\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\leqslant \sum\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\leqslant M(y-x)^\alpha n^{1-\alpha}.$$ If $\alpha\gt1$, the RHS goes to zero when $n\to\infty$ because $n^{1-\alpha}\to0$, hence $f(x)=f(y)$, QED.

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Interesting answer. Do you think we can adapt it in order to prove this:… – Tomás Dec 11 '13 at 21:42
Hadn't seen this way before--I appreciate it. – nayrb Dec 12 '13 at 0:00
Nicely done! (+1) – leo Dec 12 '13 at 3:58

HINT: Your idea is a good one. What happens when you divide the inequality by $|x-y|$?

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It suffices to show that $f$ is differentiable and its derivative vanishes everywhere.

The hypothesis implies that, for every $x\in(a,b)$ and $h$, such that $x+h\in(a,b)$, we have that

$$ \frac{|f(x+h)-f(x)-0\cdot h|}{|h|} \le M\,|h|^{\alpha-1}. $$

The right hand side of the above tends to zero, as $h\to 0$, and therefore $f'(x)=0$!

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