An algebra problem by Aaron Jerry Ninan

The condition tells us that $\left|\frac{f(x)-f(y)}{x-y}\right| \; \le \; |x-y|^2$ and so, letting $y$ tend to $x$ , we see that $f$ is differentiable, with $f'(x)=0$ for all $x$ . Thus $f$ is constant, making the answer $\boxed{11}$ .

If $f:\mathbb{R} \to \mathbb{R}$ satisfies: $|f(x) - f(y)| \leq |x - y|^3 \\ f(3) = 11$ then $f(x) = 11, \space \forall x \in \mathbb{R}$ .

Proof .-

$\forall \epsilon > 0$ "very small", $\exists n \in \mathbb{N}$ such that $\frac{1}{n^2} < \epsilon$ (Archimedes property). Then, due to triangular inequality $|f(4) - f(3)| \leq |f(4) - f(4 - 1/n)| + |f(4 - 1/n) - f(4 - 2/n)| + |f(4 - 2/n) - f(4 - 3/n)| +... + |f(4 - (n-1)/n) - f(3)| \leq$ $\leq n \cdot (\frac{1}{n^3}) = \frac{1}{n^2} < \epsilon$ . This implies that $f(4) = f(3)$ . With the same reasoning $f(n) = f(3) = 11, \space \forall n \in \mathbb{Z}$ . And with a similar reasoning, $f(x) = 11, \space \forall x \in \mathbb{R}$ .