Is $f$ Constant if Distance Between Points is Raised to a Power Greater than 1?

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Here is this week's POTW:

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If $f$ is a function from $\Bbb R$ into a metric space $(X,d)$ such that for some $\gamma > 1$, $d(f(x),f(y)) \le |x - y|^{\gamma}$ for all $x,y\in \Bbb R$, show that $f$ must be constant.

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Congratulations to castor28 and Opalg for their correct solutions. You can read Opalg’s solution below.

Spoiler

Let $x\ne0$. By the triangle inequality $$\begin{aligned} d(f(x),f(0)) &\leqslant \sum_{k=0}^{n-1}d\bigl(f\bigl(\tfrac{kx}{n}\bigr), f\bigl(\tfrac{(k+1)x}{n}\bigr)\bigr) \\ &\leqslant \sum_{k=0}^{n-1}\Bigl|\frac xn\Bigr|^\gamma = \frac{n|x|^\gamma}{n^\gamma} \to0\ \text{as }n\to\infty \end{aligned}$$ (because $\gamma>1$). Therefore $d(f(x),f(0)) = 0$ and so $f(x) = f(0)$. Since that holds for all $x$, it follows that $f$ is constant.