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 $f'(x) = 0$?
Derivatives are not needed here (which is a good thing since proofs based on them can lead to erroneous arguments, see the accepted answer for an example)...
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.
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.
HINT: Your idea is a good one. What happens when you divide the inequality by $|x-y|$?
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 $f'(x)=0$!
Caveat (by @Did, 2018-09-12): This accepted answer is wrong. For unfathomable reasons, for years now, neither its author not the OP seem interested in correcting the situation -- hence the present warning.
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 $f'(x) = 0$
EDIT:
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 $\forall x, y$
Then
$$\frac{|f(y+h) - f(y)|}{h} \le M h^{\alpha - 1} \ \ (1)$$
(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 $f = const$
