Hi Im stuck on this exercise : for which $a \in (0,1]$ is $f(x)=x^{2}\sin(\frac{1}{x^{3}})$ in $C^{a}((0,1])$
This is my attempt so far :
$|f(x)| \leq x^{2} $
$|f'(x)| = 2x\sin(1/x^{3})-\frac{3}{x^{2}}\cos(1/x^{3}) \leq 2x+ \frac{3}{x^{2}} $
Then for $ 0<y<x \leq 1$ we have
$|f(x)-f(y)| \leq x^{2} +y^{2} \leq 2x^{2}$
I would like now to get this in terms of $|x-y|$ and then use the mean value theorem to find $a$ for which $f(x)$ is in $C^{a}((0,1])$.
I believe $f\in C^{1/2}$ and $f\notin C^{\alpha}$ for $\alpha >1/2.$ I'll consider $0<x<y\le 1/(2\pi),$ and be somewhat brief. Ask if you have questions.
Let $a_n = 1/(2\pi n)^{1/3}, n=1,2\dots $ and let $I_n=[a_{n+1},a_n].$ Verify the following:
$ i)\,\, f(I_{n+1})\subset f(I_n), n=1,2\dots$
$ii)\,\,\sup_{I_n} |f'| \le C_1n^{2/3}$
$iii)\,\, a_n-a_{n+1} \le C_2n^{-4/3}.$
In $ii),iii)$ the constants $C_1,C_2$ are independent of $n.$
Suppose $x,y\in I_n.$ Then the MVT shows
$$\tag 1 \frac{|f(y)-f(x)|}{|y-x|^{1/2}} \le \frac{(\sup_{I_n} |f'|)|y-x|}{|y-x|^{1/2}}=(\sup_{I_n} |f'|)|y-x|^{1/2}.$$
Now $|y-x|\le a_n-a_{n+1}.$ So by $ii),iii)$ above the right side of $(1)$ is bounded above by $ C_1n^{2/3}\cdot [C_2 n^{-4/3}]^{1/2}.$ The last expression is bounded, independent of $n,$ which implies $f$ is Holder $1/2$ on each $I_n$ with a uniform bound on the Holder constants.
Now consider the case where $n>m, x\in I_n, y\in I_m.$ Because of $i)$ above, we have $f(I_n) \subset f(I_m).$ Thus there exists $x'\in I_m$ such that $f(x')=f(x).$ Therefore
$$\frac{|f(y)-f(x)|}{|y-x|^{1/2}} = \frac{|f(y)-f(x')|}{|y-x|^{1/2}} \le \frac{|f(y)-f(x')|}{|y-x'|^{1/2}}.$$
The last expression is bounded independent of $m,n.$ So we have shown there exists $C>0$ such that
$$\frac{|f(y)-f(x)|}{|y-x|^{1/2}}\, \le C\,\,\text { for all } 0<x<y\le1/(2\pi).$$
Perhaps you would like to try showing $f\notin C^{\alpha}$ for $\alpha >1/2?$
