Exercise 5.8

Question

\def\eps{\varepsilon} Exercise 8: Suppose $f'$ is continuous on $[a,b]$ and $\varepsilon>0$. Prove that there exists $\delta>0$ such that $$\biggl|\frac{f(t)-f(x)}{t-x}-f'(x)\biggr|<\varepsilon$$ whenever $0<|t-x|<\delta$, $a\le x\le b$, $a\le t\le b$. Does this hold for vector-value functions too?

ghostofgarborg · Accepted Answer

\def\where{\,|\,}
\def\eps{\varepsilon}
(	extit{Matt ``Frito'' Lundy}) \
    Because $f'$ is continuous on the compact set $[a,b]$, $f'$ is uniformly continuous (Theorem 4.19).
    So there exists $\delta >0$ such that for any $x,y$ where $|x-y| < \delta$ we have $|f'(x) - f'(y)| < \varepsilon$.
    But by ``the'' mean value theorem, for any $x,t \in [a,b]$, there exists a $y \in (a,b)$ such that $|x - y| < |x -t|$
    and
    $$f'(y) = \frac{f(t) - f(x)}{t -x }.$$
    So if $\left| x- t ight| < \delta$ we have:
    $$\left| \frac{f(t) - f(x)}{t -x } - f'(x) ight| = \left| f'(y) - f'(x) ight| < \varepsilon$$
    because $|x - y| < |x - t| < \delta$.

Exercise 5.8

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Exercise 5.8

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