The function!

We have $\left| f\left(x\right) - f\left(y\right)\right| \le 6\left| x - y\right|^2$ .

We can write it as

$\left| \frac{f\left(x\right) - f\left(y\right)}{x - y}\right| \le 6\left|x-y\right|$

Now let $y$ tends to $x$ where x is any real number .

Therefore,

$\displaystyle\lim_{y\to\ x}\left| \frac{f\left(x\right) - f\left(y\right)}{x - y}\right| \le \displaystyle\lim_{y\to\ x}\left|x - y\right|$

Therefore, $\left| f^{'}\left(x\right)\right| \le 0$ by using LMVT(Lagrange Mean Value Theorem) .

Therefore $f^{'}\left(x\right) = 0$

This means function $f$ is a constant function .

Hence , $f\left(6\right) = 6$

Rewrite the condition as $\left| \frac{f(x) - f(y)}{x-y} \right| \leq 6|x-y|$ We will prove that such a function is complex differentiable:

$0 \leq \lim_{y \to x} \left| \frac{f(x) - f(y)}{x-y} \right| \leq \lim_{y \to x} 6|x-y| = 0 \implies \lim_{y \to x} \left| \frac{f(x) - f(y)}{x-y} \right| = 0$ by the squeeze theorem. This in turn implies that $\lim_{y \to x} \frac{f(x) - f(y)}{x-y} = 0$ by the properties of the absolute value. Thus we have found that $f'(x)$ exists for all $x \in \mathbb{C}$ and $f'(x) = 0$ which implies $f$ is a constant function. And we are already told that $f(x) = 6$ .

Furthermore, as we have found that not only is $f$ differentiable but complex differentiable, this problem could be extended to $\mathbb{C}$ in the sense that this is the only solution, even if we are given a function $f: \mathbb{C} \to \mathbb{C}$ .