aits question

Lemma: If a differentiable function satisfies for all x,y in its domain, the relation |f(x) - f(y) | \le C |x-y|^2 for some C>0, then f(x) is a constant.

Proof: 0 \le \left| \frac{f(x+h) - f(x)}{h} \right| \le Ch

By Sandwich theorem f'(x) = 0.

Here taking logarithms on both sides we get (this is permitted since range of f is the +ve real numbers) |\ln f(x) - \ln f(y)| \le 2 (x-y)^2

So if g(x) = ln f(x) then g(x) and hence f is a constant.

So Option (2) is correct

Dunno abt Option 3