I'm hoping someone can verify my answers. I have that for $x\in\mathbb{Q}$ we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{f(x+h)-1}{h}$$ and since $\mathbb{Q}$ is dense in $\mathbb{R}$, $(x+h)$ is rational as $h\to 0^+$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$ Similarly for $D^-f(x)$ we have $$D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$And for $D_+f(x)$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\liminf_{h\to 0^+}\frac{0-1}{h}=-\infty$$ Finally for $D_-f(x)$ we have $$D_-f(x)=\liminf_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\liminf_{h\to 0^+}\frac{1-0}{h}=\infty$$ Now for $x\in\mathbb{R}\setminus\mathbb{Q}$ we have that $f(x)=0$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-0}{h}=\infty$$ $$D^-f(x)=\limsup_{h\to 0^+}\frac{0-1}{h}=-\infty$$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ $$D_-f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ I am having trouble convincing myself of the effect (and proper explanation) of $\liminf$ vs $\limsup$ and their impact on the derivates. Is this correct? Could someone help me better conceptually understand what is going on?
Thanks
Best Answer
First comment
Your definition for the various $D^+, D_+, \dots$ are not the right ones. For example $D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x-h)}{h}$ and not $D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x+h)}{h}$. See Dini derivatives for the correct definitions.
Second comment
You got some right results in term of values. However, the way you did it is not the best one and can lead you to wrong results when applied to others functions.
Let's take the example of $D^+f(x)$ when $x \in \mathbb Q$ where you wrote $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{1-1}{h}.$$
This is not correct as $f(x+h)$ depends on whether $x+h$ is rational or not. What you should say is that for any $\epsilon \gt 0$, $\{f(x+h) \mid 0 \lt h \lt \epsilon\} = \{0,1\}$ and therefore
$$\left\{\frac{f(x+h)-f(x)}{h} \mid 0 \lt h \lt \epsilon\right\} = \{0,-1/h\}.$$
As for $h \gt 0$ we have $\sup \{0,-1/h\} = 0$, we can conclude that
$$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=0.$$