Suppose that the function $f$ satisfies |$f(x)-f(t)$| $\le (x-t)^2$ for each $x$, $t$ on the reals. Prove that $f$ must be a constant function. [duplicate]

Question

Suppose that the function $f$ satisfies |$f(x)-f(t)$| $\le (x-t)^2$ for each $x$, $t$ on the reals. Prove that $f$ must be a constant function.

I believe I'm close as I can get to the answer without actually being able to put it together, and it's very frustrating.

Answer 1

$$ \left|\frac{f(x)-f(t)}{x-t}\right| \le |x-t|\to 0 \text{ as }x\to t. $$ So $f'(t)=0$ for all values of $t$, so you have a constant function.

PS: The mean value theorem is usually cited in proofs that if $f'=0$ everywhere, then $f$ is constant. If the real line had a gap somewhere, then we could have $f=3$ everywhere to the left of the gap and $f=4$ everywhere to the right of the gap, and $f'=0$ everywhere. However, I wonder if the thing to be proved here might be provable without the gaplessness of the line, and hence without the mean value theorem.