If $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2$ for all $x$ and $y$ in $\mathbb{R}$ then $f$ is a constant function

Question

Assume that $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2$ for all $x,y\in \mathbb{R}$. Prove that $f$ is the constant function.

My first thought was to use continuity of $f$, so that we get that $\forall x\in\mathbb{R},\forall \epsilon>0, \exists \delta>0, \forall y \in \mathbb{R}, \lvert x-y \rvert<\delta \implies \lvert f(x)-f(y) \rvert < \epsilon$.

So choosing $y$ so that the above property holds, we get that $\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2 <\delta^2$

Using the inequality I found (supposing it is correct), how would I continue to prove that $f(x)=f(y)$, and so their difference must be zero.

Answer 1

Use the definition of the derivative.

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

So the function is differentiable with derivative $0$ at all points.

Answer 2

Let $y=x+h$ with $h\ne 0$. Then $$0 \le \left| \frac{f(x+h)-f(x)}{h}\right|<|h|$$ and so $$-|h| \le\frac{f(x+h)-f(x)}{h} \le |h|.$$Take limit for $h$ to zero then $$f^\prime(x)=0.$$

Answer 3

An easy way to understand this is, when $x\neq y$, we can divide by $|x-y|$ to get that: $$\frac{|f(x)-f(y)|}{|x-y|}\leq |x-y|$$ Now, one characterization of the derivative of $f$ is: $$f'(x) = \lim_{y\to x}\frac{f(x)-f(y)}{x-y}$$ As out inequality is valid for all $x,y$, we can pick a sequence $y_k$ with $y_k\to x$, $y_k\neq x$ and look at: $$\lim_{k\to\infty} \left|\frac{f(x)-f(y_k)}{x-y_k}\right|\leq \lim_{k\to\infty} |x-y_k|$$ The left-hand side of this is just the derivative at $x$, so we have that: $$|f'(x)|\leq 0$$ From this, it's clear that $f'(x) = 0$, and so $f(x)$ is constant.