A basic question on bound of a function

Question

Is the function $f(x) = -\frac{a}{x} cos (1/x^a) + ax^{a-1} \sin(x^{-a})$ bounded in $x \in [0,1]$ for some values of $a >0$ ? I don't think so.

Answer 1

Let's consider $x_{n} = \left( \frac{1}{2 n \pi} \right)^{1/a}$

As $a > 0$, $\lim\limits_{n \to + \infty} x_{n} = 0$, and one has $\cos (x_{n}^{-a}) = 1$ and $\sin(x_{n}^{-a}) = 0$, so that we have

$$ f(x_{n}) = -a \, (2 n \pi)^{1/a} \to + \infty \;\;\; \text{for } \;\;n \to + \infty $$

As a conclusion, $\forall a > 0$, f is not bounded on $[0,1]$

Answer 2

Call your function $f(\cdot; a)$, just to put in evidence the dependence on the parameter $a>0$.

Obviously, your function is continuous in $]0,1]$ for every $a>0$, hence it is bounded in every left neighbourhood $[\varepsilon ,1]$ of $1$ (here $\varepsilon \in ]0,1[$); therefore you have to pay attention only to the behaviour of $f(\cdot ;a)$ in a small right neighbourhood of $0$.

You have: $$f(x;a)= \frac{a}{x}\ \left(- \cos\left(\frac{1}{x^a}\right) + x^a\ \sin\left(\frac{1}{x^a}\right) \right)\; ;$$ roughly speaking, the second summand is negligible w.r.t. the first when $x$ is small enough (because $\sin x^{-a}$ is bounded and $x^a\to 0$ as $x\to 0$), thus: $$\tag{1} f(x;a)\sim -\frac{a}{x}\ \cos\left(\frac{1}{x^a}\right) \quad \text{as } x\to 0^+\; .$$ The RH side of (1) is awfully obscillating at the right of $0$, because: $$\begin{split} \limsup_{x\to 0^+} \frac{a}{x}\ \cos\left(\frac{1}{x^a}\right) &= +\infty\\ \liminf_{x\to 0^+} \frac{a}{x}\ \cos\left(\frac{1}{x^a}\right) &= -\infty\; , \end{split}$$ therefore $f(\cdot ;a)$ exhibits the same obscillating behaviour when $x\to 0^+$ and so it cannot be bounded in any interval $]0,\varepsilon]$, no matter how small you take $\varepsilon$.

This is only a rough picture, but you can make it precise.