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Let $$D^+f(x)=\lim\limits_{h \to 0}\left[\sup\limits_{0<|t|\leq h}\frac{f(x+t)-f(x)}{t}\right] \text{ and } D^-f(x)=\lim\limits_{h \to 0}\left[\inf\limits_{0<|t|\leq h}\frac{f(x+t)-f(x)}{t}\right]$$ represent the upper and lower derivative respectively.

Let $f(x)=\begin{cases} x\sin(\frac{1}{x}) \text{ if } x \neq 0 \\ 0 \text{ if } x =0\end{cases}$ and $g(x)=\chi_{\mathbb{Q}}$.

My attempt at a solution:

If $x=0$, $D^+f(x)=0$ and $D^-f(x)=-\infty$. If $x \neq 0$, then $D^+f(x)=\infty$ and $D^-f(x)=0$.

I believe the answer is similar to $g(x)$.

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You seem to be missing some $t$s somewhere in the bracketed terms. Perhaps you mean to write: $$\sup_{0 < |t| \leq h} \frac{f(x+t) - f(t) }{t} ? $$ – JavaMan Dec 8 '12 at 7:48
Then what is $t$? – Nameless Dec 8 '12 at 8:11
I take it back. My notes were wrong and the book was correct. – emka Dec 8 '12 at 8:16
Now are you sure your calculations are correct? I have $D^+f(0)=1$ while $D^-f(0)=-1$ – Nameless Dec 8 '12 at 8:18
Would it just be zero? Since its either 1 or -1 at those points, taking the derivative would be 0. – emka Dec 8 '12 at 8:26
up vote 3 down vote accepted

$$D^+f(0)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{f(t)-f(0)}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{t\sin \frac1t}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\sin\frac{1}{t}$$ Now no matter how small $h$ gets, there always exist a point $t$ in $(0,h]$ so that $\sin\frac{1}{t}=+1$ (take $t=\frac{1}{n\pi +\frac{\pi}{2}}$ for sufficiently large $n$) Therefore, $D^+f(0)=+1$. Similarly $D^{-}f(0)=-1$.

At all points $x\neq 0$, $x\sin\frac{1}{x}$ is differentiable and as thus $$D^+f(x)=D^-f(x)=f^{\prime}(x)=\sin \frac{1}{x}-\frac{1}{x}\cos\frac1x$$

For $g$: If $x\in \mathbb{Q}$,$$D^+g(x)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{g(x+t)-g(x)}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}$$ Now $$\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}=\begin{cases} 0&\mbox{if } t\in \mathbb{Q}\\ \frac{-1}{t}&\mbox{if } t\notin \mathbb{Q}\end{cases}$$ The supremum is achieved when $t<0$ and $t\notin \mathbb{Q}$ which means $0<-t\le h\Rightarrow -h\le t<0$. No matter how small $h$ gets there always exists a point $t\in [-h,0)\cap \mathbb{Q}^c$ (density argument) and so $$D^+g(x)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}=\lim_{h \to 0}\sup\limits_{-h\le t<0}\frac{-1}{t}=+\infty$$ All other cases are done in the same spirit

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As a sidenote, the case $x\notin \mathbb{Q}$ for $D^+g$ is harder because you need to examine whether or not for arbitrary $h>0$, $\exists t\in (0,h):x+t\in \mathbb{Q}$ – Nameless Dec 8 '12 at 9:20

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