Exercise 5.1

Question

Exercise 1: Let $f$ be defined for all real $x$, and suppose that $\big|f(x)-f(y)\big|\le(x-y)^2$ for all real $x$ and $y$. Prove that $f$ is constant.

ghostofgarborg · Accepted Answer

(	extit{Matt ``Frito'' Lundy}) \
    For any $x$, we have:
    \begin{align*}
    \left| \frac{f(t) - f(x)}{t - x} ight| &\leq \left| \frac{(t - x)^2}{t - x} ight| \
    &=\left| t - x ight| 	o 0
    \end{align*}
    as $t 	o x$. Because $f'(x) = 0$ for all $x$, $f(x)$ is a constant by Theorem 5.11(b).

Exercise 5.1

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Exercise 5.1

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