Verify MCQ options $|f(x)-f(y)|<7|x-y|^{201}$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for any reals $x$ and $y$, $$|f(x)-f(y)|\le7|x-y|^{201}$$

We have options $$a)f(101)=f(202)+8\\b)f(101)=f(201)+1\\c)f(101)=f(200)+2\\d)\text{None of the above}$$

My attempt:

$$\left|{f(x)-f(y)\over x-y}\right|\le7(x-y)^{200}\\\text{let $x=y+h$}\\\implies |f'(x)|\le0\\\implies f'(x)=0\\\implies f(x)=k, k\in\mathbb{N}$$

Thus d is the correct answer. Is my reasoning correct or am I missing something?

Answer 1

Your reasoning seems fine.

Alternatively, rather than proving $f$ must be constant. Notice that $f$ being a constant satisfies the condition, hence immediately ruling out the other options.

Answer 2

Note that a $\alpha$-Hölder continuous functions with $\alpha > 1$ is constant. Indeed $$ \frac{|f(x)-f(y)|}{|x-y|} \leq C|x-y|^{\alpha-1} \to 0 $$ as $x\to y$, so $f'(y) = 0$ for every $y$.