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

Question

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

Exercise 11. If $a, b>0$, let $$ f(x)=\begin{cases} x^a \sin \left(x^{-b}\right) & \text { for } 0<x \leq 1 \\ 0 & \text { if } x=0 \end{cases} $$ 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 $x^{a+1}<h$. 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.

Thank you in advance.

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

Answer 1

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 $x < h^{\frac{1}{1+a}}$. $$|f(x+h)-f(x)| \leq 2 (x+h)^a \leq 2 (h^{\frac{1}{1+a}}+h)^a \leq 2 (2 h^{\frac{1}{1+a}})^a = 2^{1+a} h^\alpha. $$