I recently came across the following theorem:

$$ \forall x_1, x_2 \in \mathbb{R},\textrm{function, } f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto y; \ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2 \implies f \textrm{ is constant.}\ \mathbf{(1)} $$

I've been trying for some time, but the proof of $\mathbf{(1)}$ remains as elusive as ever. I've made two major attempts, the second of which I'll outline here. Though, I would be glad to detail the first as well if requested, I won't now since I think it's mostly wrong. But for the second, this is what I have so far:

If, $\forall x_1,\ x_2,\ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2$, then $f$ is continuous. This so as $f$ is defined for all reals, $(\forall x \in \mathbb{R})\ f$ has finite limits, and each of those limits equals $f(x)$. Assume $f$ wasn't constant, then $\exists x_1,\ x_2 \ni x_1 \neq x_2 \implies |f(x_1)- f(x_2)| > 0$. Since $f$ is continuous, there exist an infinity of such pairs, $x_1$ and $x_2$. For all such $x_1$ and $x_2$, we may construct a set, $S$, consitsting of $f(x_1)$ and $f(x_2)$ (not as pairs); since f is defined for all $x,\ S$ is "absolutely" bounded and as such has a least upper bound and and greatest lower bound, which we will denote as $\alpha_1\ = f(a_1)$ and $\alpha_2 = f(a_2)$ respectively. To show $f$ is constant, it will suffice to show that $\alpha_1 = \alpha_2$.

Does anyone see how the proof could be completed? Or even, do you think there might be a better approach? Thank you all in advance.

I recently came across the following theorem:

$$ \forall x_1, x_2 \in \mathbb{R},\textrm{function, } f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto y; \ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2 \implies f \textrm{ is constant.}\ \mathbf{(1)} $$

I've been trying for some time, but the proof of $\mathbf{(1)}$ remains as elusive as ever. I've made two major attempts, the second of which I'll outline here. Though, I would be glad to detail the first as well if requested, I won't now since I think it's mostly wrong. But for the second, this is what I have so far:

If, $\forall x_1,\ x_2,\ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2$, then $f$ is continuous. This is so as $f$ is defined for all reals, $(\forall x \in \mathbb{R})\ f$ has finite limits, and each of those limits equals $f(x)$. Assume $f$ wasn't constant, then $\exists x_1,\ x_2 \ni x_1 \neq x_2 \implies |f(x_1)- f(x_2)| > 0$. Since $f$ is continuous, there exist an infinity of such pairs, $x_1$ and $x_2$. For all such $x_1$ and $x_2$, we may construct a set, $S$, consitsting of $f(x_1)$ and $f(x_2)$ (not as pairs); since f is defined for all $x,\ S$ is "absolutely" bounded and as such has a least upper bound and and greatest lower bound, which we will denote as $\alpha_1\ = f(a_1)$ and $\alpha_2 = f(a_2)$ respectively. To show $f$ is constant, it will suffice to show that $\alpha_1 = \alpha_2$.

Does anyone see how the proof could be completed? Or even, do you think there might be a better approach? Thank you all in advance.