Suppose that $f, g : R → R$ are functions such that
$$|f(x) − f(y)| ≤ |g(x) − g(y)| \sqrt{|x − y|}$$ for any $x, y ∈ R$. If $g$ is differentiable with bounded derivative on all $R$, show that $f$ is constant.
I know I am meant to be using MVT for this question, I attempted to use $g(x)$ as the function because we know it is differentiable (condition required for MVT): $$\frac{g(b)-g(a)}{b-a}=g'(c)$$ $$|g(b)-g(a)|=|g'(c)|b-a|$$ I am lost, not sure where to start the question.
If the derivative of $g$ is bounded as $|g'(c)| \le M$ for all $c$, then $$\left| \frac{f(x)-f(y)}{x-y} \right| \le \frac{|g(x)-g(y)|}{\sqrt{|x-y|}} \le M \sqrt{|x-y|}$$ for any $x,y$, where the last inequality is from your application of the mean value theorem. Can you take it from here?