Villaretq0

2022-06-22

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by

Find the values of the four Dini derivatives of $f$ at $x=0.$
I could easily find out the values of the upper right ${D}^{+}f$ and lower right ${D}_{+}f$ Dini derivatives at $x=0,$ values being 2 and 0 respectively. But, in case of upper left Dini derivative of $f$ at $x=0,$ we have:
$\begin{array}{rl}{D}^{-}f\left(0\right)& =\underset{h\to 0-}{lim sup}\frac{f\left(h\right)-f\left(0\right)}{h}\\ \\ & =1+\underset{h\to 0-}{lim sup}\frac{\sqrt{-h{\mathrm{sin}}^{2}\left(\mathrm{ln}|h|\right)}}{h}.\end{array}$
Next putting: $-h=p$ so that $p\to 0+,$, I obtained:
${D}^{-}f\left(0\right)=1-\underset{p\to 0+}{lim inf}\frac{|\mathrm{sin}\left(\mathrm{ln}|p|\right)|}{\sqrt{p}}.$
Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term $\sqrt{p}$ present in the denominator. I am confused. Please give some insights. Thanks in advance.

Savanah Hernandez

We have
$\begin{array}{rl}{D}^{-}f\left(0\right)& =\underset{h\to {0}^{-}}{lim sup}\frac{h+\sqrt{-h{\mathrm{sin}}^{2}\left(\mathrm{ln}\left(-h\right)\right)}}{h}\\ & =\underset{h\to {0}^{-}}{lim sup}\left(1-\sqrt{-\frac{{\mathrm{sin}}^{2}\left(\mathrm{ln}\left(-h\right)\right)}{h}}\right)\end{array}$
In the last line, we used $\frac{1}{h}=-\sqrt{\frac{1}{{h}^{2}}}$ for $h<0$. Let
$g\left(h\right)=1-\sqrt{-\frac{{\mathrm{sin}}^{2}\left(\mathrm{ln}\left(-h\right)\right)}{h}}$
Clearly, $g\left(h\right)\le 1$ for all $h<0$. The limit superior would then be equal to 1 if, for example, we show g attains the value of 1 in every left neighborhood of 0. Consider the sequence
${x}_{n}=-{e}^{-n\pi }$
Note ${x}_{n}<0$, $\underset{n\to \mathrm{\infty }}{lim}{x}_{n}=0$, and
$g\left({x}_{n}\right)=1-\sqrt{{e}^{n\pi }{\mathrm{sin}}^{2}\left(-n\pi \right)}=1-0=1$
Thus, ${D}^{-}f\left(0\right)=1$.

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