$f$ Holder continuous with Holder exponent $p>1\implies f \text{ is constant}$

Say I have a function on a an interval $I$ in $\mathbb{R}$ $f: I \to Y $ where $Y$ is any metric space.Say $f$ satisfies $d_{Y}(f(y), f(x)) \leq C\cdot|y-x|^p$, for all $y,x \in I$ where $p \in (1,\infty)$ , i.e. $p$ is bigger than $1$ (so this is much stronger than mere Holder Continuity). I want to show that $f$ is constant on $I$.

Here is what I got so far. $f$ is obviously continuous. Also, it makes intuitive sense for $f$ to be constant, since $p >1$ will make $|y-x|^p$ very small for for $|y-x| \ll 1$.

I also feel like I have to look at the expression $$\frac{d_{Y}(f(y), f(x))}{|y-x|} \leq C\cdot|y-x|^{p-1}$$ (Note: There is no notion of differentiability here). I'm also thinking that splitting up the interval $[x,y]$ (assuming $x<y$) will help me like so:

$$d_{Y}(f(y), f(x))\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p = \sum_{k=1}^{n} (\frac{1}{n})^p = n^{1-p} $$

$ (y = x_n, x = x_0)$.

Is this the answer, that last equality chain?

Edit: That last chain of equations, as suggested in the comments below should be $$d_{Y}(f(y), f(x))\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p $$ $$= C\sum_{k=1}^{n} (\frac{|y-x|}{n})^p =|y-x|^p n^{1-p} $$

Solution 1:

Your approach of splitting the interval is essentially correct. Here it is in detail.

Let $x$ and $y$ be any two distinct points in $I$. Without loss of generality let us suppose $x <y$. Splitting $[x, y]$ in $n$ intervals of length $\frac{|y-x|}{n}$, we have $n+1$ points $x_k$, such that $x=x_0 < x_1 < \dots <x_{n-1} < x_n=y$ and $|x_k -x_{k-1}| = \frac{|y-x|}{n}$. So we have

\begin{align*} d_{Y}(f(y), f(x))&\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p = \\ &=C\sum_{k=1}^{n} \left(\frac{|y-x|}{n} \right)^p =C|y-x|^p n^{1-p} \end{align*}

Now, note that as $n \to \infty$, we have $C|y-x|^p n^{1-p} \to 0$, and so you have that $d_{Y}(f(y)- f(x))=0$. So $f(y)=f(x)$. Since, $y$ and $x$ are arbitrary points in $I$, it follows that $f$ is constant.