If $f$ is holder continuous for $\alpha >1$ then $f$ is constant. [duplicate]

Solution 1:

Look at this beautiful proof I learned from my advisor that avoids differentiability:

For fixed $x<y$ take, for a natural number $N$, the points $$x_o=x, x_1=x_0+\frac{y-x}{N}, x_2=x+2\frac{y-x}{N}, \cdots, x_N=x+N\frac{y-x}{N}=y \ . $$

Now observe that $$|f(y)-f(x)|\leq \sum|f(x_{i+1})-f(x_i)| \leq \sum|x_{i+1}-x_i|^\alpha = N \times(\frac{y-x}{N})^\alpha = \frac{y-x}{N^{\alpha-1}}.$$

The LHS is independent of $N$, so, by letting $N \rightarrow \infty$, we get $$|f(y)-f(x)|=0.$$

Solution 2:

Given $x \in \Bbb R$, we have for any $y \neq x$,

$$\left| \frac{f(y) - f(x)}{y-x} \right| \le C |x - y|^{\epsilon}$$

Taking $y \to x$, we get that:

$$\lim_{y \to x} \left| \frac{f(y)-f(x)}{y-x} \right| = 0 $$

Which gives:

$$\lim_{y \to x} \frac{f(y)-f(x)}{y-x} = 0$$

So, $f$ is differentiable at $x$ and $f'(x) = 0$. Therefore $f$ is constant.

Solution 3:

The idea is that the derivative is $0$ over an interval.

$f'(x_0)=0\space \forall x_0 \in A \subset \mathbb R$

$|f'(x_0)|=\lim_{h\to 0}\frac{|f(x_0+h)-f(x_0)|}{|h|}\le\lim_{h\to 0}\frac{C|h|^{\alpha}}{|h|}=0$ if $\alpha \gt 1$ .