A Quadratic Modulus of Continuity Forces Constancy

Detailed Solution

Fix arbitrary $x,y\in\mathbb R$ with $x<y$. Divide the interval $[x,y]$ into $n$ equal subintervals. Define \[ x_k=x+\frac{k(y-x)}{n}, \qquad k=0,1,\ldots,n. \] Then, by the triangle inequality, \[ |f(y)-f(x)| \le \sum_{k=1}^{n}|f(x_k)-f(x_{k-1})|. \] Using the given condition on each subinterval, \[ |f(x_k)-f(x_{k-1})| \le |x_k-x_{k-1}|^2 = \left(\frac{y-x}{n}\right)^2. \] Therefore, \[ |f(y)-f(x)| \le n\left(\frac{y-x}{n}\right)^2 = \frac{(y-x)^2}{n}. \] Since this inequality holds for every positive integer $n$, letting $n\to\infty$ gives \[ |f(y)-f(x)|=0. \] Hence \[ f(y)=f(x). \] Because $x$ and $y$ were arbitrary, \[ \boxed{f\text{ is constant on }\mathbb R}. \] Thus the correct answer is \[ \boxed{\mathrm{C}}. \] Why the other options are incorrect: - (A) is false because $f$ cannot be non-constant. In fact, every function satisfying the given inequality is constant. - (B) is false because although every such $f$ is indeed Lipschitz with constant $1$, it must actually be constant. The phrase "need not be constant" is therefore incorrect. - (D) is false because the inequality forces $f$ to be constant. Consequently, $f'(x)=0$ wherever the derivative is considered, so it cannot have $|f'(x)|>0$. Therefore, \[ \boxed{f\text{ must be a constant function}.} \]