Exercise 9.16

Question

Exercise 16: Show that the continuity of $\v f'$ at the point $\v a$ is needed in the inverse function theorem, even in the case $n=1$: If
    $$
        f(t)=t+2t^2\sin\bigg(\frac{1}{t}\bigg)
    $$
    for $t\ne 0$, and $f(0)=0$, then $f'(0)=1$, $f'$ is bounded in $(-1,1)$, but $f$ is not one-to-one in any neighborhood of 0.

analambanomenos · Accepted Answer

The derivative of $f$ at 0 is
    $$
        f'(0)=\lim_{t\rightarrow 0}\frac{f(t)}{t}=\lim_{t\rightarrow 0}\big(1+2t\sin(t^{-1})\big)=1 \
    $$
    since $\big|2t\sin(t^{-1})\big|\le 2t\rightarrow 0$ as $t\rightarrow 0$.

For $t\in(-1,1)$, $t\ne 0$, we have
    \begin{align*}
        f'(t) &= 1+4t\sin(t^{-1})-2\cos(t^{-1}) \
        \big|f'(t)\big| &\le 1+4\big|t\sin(t^{-1})\big|+2\big|\cos(t^{-1})\big| \
        &\le 3+4|t| \
        &\le 7
    \end{align*}

If $n$ is a positive integer, let $a_n=2n\pi$, $b_n=(2n+1)\pi$. Then, since $\sin a_n=\sin b_n=0$, $\cos a_n=1$, and $\cos b_n=-1$, we have $f'(a_n^{-1})=-1$ and $f'(b_n^{-1})=3$. Hence $f$
    has a local maximum in $(b_n^{-1},a_n^{-1})$ and a local minimum in $(a_n^{-1},b_{n+1}^{-1})$ and so cannot be one-to-one in either interval. Since a neighborhood of 0 contains an infinite
    number of such intervals, $f$ cannot be one-to-one in any neighborhood of 0.

Exercise 9.16

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Exercise 9.16

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