Is $x^a$sin$(x^{-a})$ Hölder continuous?

I'm currently trying the following Exercise #11 of Chapter 3 in Stein's Real Analysis.

Exercise 11. If $$a, b>0$$, let $$f(x)=\begin{cases} x^a \sin \left(x^{-b}\right) & \text { for } 0 Prove that $$f$$ is of bounded variation in $$[0,1]$$ if and only if $$a>b$$. Then, by taking $$a=b$$, construct (for each $$0<\alpha<1$$ ) a function that satisfies the Lipschitz condition of exponent $$\alpha$$ $$\begin{equation*} |f(x)-f(y)| \leq A|x-y|^\alpha \end{equation*}$$ but which is not of bounded variation.

[Hint: Note that if $$h>0$$, the difference $$|f(x+h)-f(x)|$$ can be estimated by $$C(x+h)^a$$, or $$C^{\prime} h / x$$ by the mean value theorem. Then, consider two cases, whether $$x^{a+1} \geq h$$ or $$x^{a+1}. What is the relationship between $$\alpha$$ and $$a$$ ?]

I've managed to solve the former problem; $$f$$ is of bounded variation iff $$a > b$$. The thing is the latter one; when $$b = a$$, $$f$$ satisfies $$|f(x) - f(y)| \leq A|x-y|^{\alpha}$$ for each $$0 < \alpha < 1$$.

After searching for some materials, I've found a similar question in here, and I tried similar approach based on this answer as follows.

Let $$q = 1/\alpha, p = (1-1/q)^{-1}$$, and $$g(t) = 1$$ for $$t \in [0,1]$$.

Using Holder's Inequality with $$0 < x \le y < 1$$ (wlog),

$$|f(y)-f(x)| = \left|\int_x^y f'g\right| \le \left\{\int_x^y |f'|^p\right\}^{1/p}\left\{\int_x^y |g|^q\right\}^{1/q}$$ $$= \left\{\int_x^y |f'|^p\right\}^{1/p}|y-x|^{\alpha}$$

If this is correct, it remains to show that $$\int_0^1 |f'|^p < \infty$$, where I'm stuck.

(Do I have to use $$|f'| \le ax^{a-1}|$$sin$$(x^{-a})| + ax^{-1}|$$cos$$(x^{-a})|$$?)

So, any comments about my trial, and (if exists) other proper approach to solve the problem would be appreciated.

(I ignored the given hint because I found it difficult to apply, but any explanation about the hint is also welcome.)

Your approach: The problem with your approach is that what remains to prove in order to conclude is false. For any $$p \geq 1$$, there holds $$\int_0^1 |f'|^p = + \infty$$. As you correctly computed, $$f'(x)$$ contains a term of amplitude $$x^{-1}$$, which is not integrable near $$0$$, even with $$p = 1$$. Hence, you cannot conclude as you planned.

Let us prove the following statement, using the given hint.

For any $$0 < a < 1$$, $$x \mapsto x^a \sin x^{-a}$$ is $$\alpha$$-Hölder continuous on $$[0,1]$$ with $$\alpha = \frac{a}{a+1} \in (0,1)$$.

Claimed bounds: The claimed bounds are easy to derive. For the first one, write, for $$x,h\geq0$$, $$|f(x+h)-f(x)| \leq |f(x+h)|+|f(x)| \leq (x+h)^a + x^a \leq 2 (x+h)^a.$$ The second one follows from the median value inequality as claimed, and from your computation of $$f'$$ which proves that $$f'(x) \leq 2 a x^{-1}$$ on $$(0,1)$$ so indeed, $$|f(x+h)-f(x)| \leq 2 a \frac{h}{x}.$$

Intuition: The considered function oscillates increasingly fast as $$x \to 0$$. For a fixed $$x$$, when we consider an increment $$h$$ which is sufficiently small, then the local approximation of $$f$$ by its tangent is good. On the contrary, when $$h$$ is small but not enough, the first order approximation is not relevant, and the best we can do is bound the difference using the amplitude of the envelope of $$f$$ at $$x$$. This is why we must distinguish cases.

Proof: Take $$x,h \geq 0$$. We need to bound $$|f(x+h)-f(x)|$$ by some constant times $$h^{\alpha} = h^{\frac{a}{a+1}}$$.

1. First case: $$h \leq x^{a+1}$$. We use the tangent approximation and $$x^{-1} \leq h^{-\frac{1}{1+a}}$$. $$|f(x+h)-f(x)| \leq 2a\frac{h}{x} \leq 2a h^{1-\frac{1}{1+a}} = 2 a h^{\alpha}.$$
2. Second case: $$x^{a+1} < h$$. We use the envelope bound and $$x < h^{\frac{1}{1+a}}$$. $$|f(x+h)-f(x)| \leq 2 (x+h)^a \leq 2 (h^{\frac{1}{1+a}}+h)^a \leq 2 (2 h^{\frac{1}{1+a}})^a = 2^{1+a} h^\alpha.$$
• Ah, now I clearly understand the meaning of the given hint. Thank you so much! Commented May 15, 2023 at 23:43
• Yes it is like the spirit of good kernel one care about far and the other care about near
– lee
Commented Dec 27, 2023 at 17:01