Proof of a simple property of real, constant functions.

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.

Solution 1:

If we divide through by $|x_1 - x_2|$, we get

$$\left|\frac{f(x_1) - f(x_2)}{x_1 - x_2}\right| \leq |x_1-x_2|,$$

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Hölder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!

Solution 2:

I'll show that $f(0)=f(1)$ and you'll see the trick. For all $n\in \mathbb{N}$, $n > 0$:

\begin{align} |f(1)-f(0)| &= |f(1)-f(\frac{n-1}{n})+f(\frac{n-1}{n})-f(\frac{n-2}{n}) + \dotsc + f(\frac{1}{n}) - f(0)| \newline &\leq \frac{1}{n^2} + \frac{1}{n^2} + \dotsc + \frac{1}{n^2} = \frac{1}{n} \end{align}

(There are $n$ terms in that sum.) Since this holds for all positive $n$ it follows that $f(1) = f(0)$.