Mean Value Theorem for derivatives Suppose $f$ is differentiable at every $x$ and $|f(x) - f(y)| \leq C|x-y|^2$ Use the definition of the derivative and the MVT to show that $f$ is constant.


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*So, $$|f(x) - f(y)| \leq C|x-y||x-y|$$
$$\frac{|f(x) - f(y)|}{|x-y|} \leq C|x-y|$$

*Now that looks a lot like the computation for the tangent line and since $f$ is a constant, that entire left-hand side is $0$?

*I'm confused as to whether or not this is the correct step that needs to be taken, and if it is I'm stuck as to where to go after this fact. Any direction would be great thank you. 

Fix a real number $y$. As you've shown, for all $x\neq y$, $\frac{\lvert f(x) - f(y)\rvert}{\lvert x - y\rvert} \le C\lvert x - y\rvert$, that is, 
$$\biggl\lvert \frac{f(x)-f(y)}{x-y}\biggr\rvert \le C\lvert x - y\rvert$$
or
$$-C\lvert x - y \rvert \le \frac{f(x)-f(y)}{x-y}\le C\lvert x - y\rvert$$
Since $\lim_{x\to y} C\lvert x - y\rvert = 0$, the squeeze theorem gives
$$\lim_{x\to y} \frac{f(x)-f(y)}{x-y} = 0$$
that is, $f'(y) = 0$. Since $y$ is arbitrary, $f$ is constant.
