Let $f$ be defined for all real $x$, and suppose $|f(x) – f(y)| \leq (x-y)^2$ for all real $x$ and $y$. Prove $f$ is constant. (From Rudin PMA)
Note: Many proofs are available; this question is to verify and critique this proof.
Proof: For any $x$, choose $y < x$. Then
$$\begin{align*}
f(x) – f(y) &\leq (x-y)^2 \\
\frac {f(x) – f(y)}{x – y} &\leq x-y \\
f'(x) = \lim_{y \to x}\frac {f(x) – f(y)}{x – y} &\leq \lim_{y \to x} x-y = 0.
\end{align*}$$
Thus, $f'(x) = 0$ for all $x$, and $f$ is constant.
Is the proof correct? Can it or the exposition be improved?
Update: Thank you for the invaluable critique. A revised proof is below. Please verify, critique, and suggest improvements.
Revised Proof: For $x \neq y$,
$$\begin{align*}
\left \lvert f(x) – f(y) \right \rvert &\leq (x-y)^2 \\
-|x-y| \leq \left \lvert \frac {f(x) – f(y)}{x-y} \right \rvert &\leq |x-y| \\
\end{align*}$$
and so by the Squeeze Theorem $$\lim_{y \to x} \frac {f(x) – f(y)}{x-y} = 0.$$
Since $f'(x) = 0$ for all $x$, by the Mean Value Theorem $f$ is constant.
Best Answer
There are three main logical gaps in your argument:
Initially, you assume that $y<x$. But when you take a limit as $y\to x$, by definition you are considering values of $y$ which are greater than and less than $x$.
You haven't justified why $f'(x)$ exists.
Even assuming that $f'(x)$ exists and you have managed to establish that it is less than or equal to $0$ for all $x$, you can't automatically conclude from this that $f'=0$.
Congratulations, your revised argument is correct! Here, I give some further comments. The inequality $$ -|x-y| \leq \left \lvert \frac {f(x) - f(y)}{x-y} \right \rvert \leq |x-y| \label{*}\tag{*} $$ is true, but I don't know why you don't instead use the simpler bounds $$ 0\le\left \lvert \frac {f(x) - f(y)}{x-y} \right \rvert \leq |x-y| \label{star}\tag{$\star$} \, . $$ Indeed, the reason that I know that $\eqref{*}$ is true is because of $\eqref{star}$.
Your argument tacitly uses the fact that $\lim_{y\to a}|g(y)|=0\implies \lim_{y\to a}g(y)=0$. This fact is probably elementary enough to go unmentioned, but it should be clear in your own mind why this implication holds (it follows directly from the $\varepsilon$-$\delta$ definition of limits).
In your argument, you fixed an arbitrary $x$ and applied the squeeze theorem to an inequality of the form $p_x(y)\le q_x(y)\le r_x(y)$. It might be slightly easier for the reader to follow if the first sentence of your proof was "Fix $x\in\mathbb R$ and consider any $y\neq x$".
Well done on justifying why $f'=0$ implies that $f$ is constant. This seemingly innocuous result does indeed actually require the full machinery of the mean value theorem.