Estimate $\sup_n\left|\int_0^1 \left(1-e^{i\alpha_n x}\right) f(x)d\mu(x)\right|$

Question

Let $\mu$ be a singular Borel probability measure on $[0, 1)$, and $f\in L^2(\mu)$. Estimate $$\sup_n\left|\int_0^1 \left(1-e^{i\alpha_n x}\right) f(x)d\mu(x)\right|$$ where $\alpha_n\in\mathbb R$ and $|\alpha_n|\leq L<1$.

This is my approach: $$\sup_n\left|\int_0^1 \left(1-e^{i\alpha_n x}\right) f(x)d\mu(x)\right|\leq \left(\sup_n \left|1-e^{i\alpha_n}\right|\right)\left|\int_0^1 f(x)d\mu(x)\right|=\left(2-2\cos L\right)\left|\int_0^1 f(x)d\mu(x)\right|$$ where the equality is thanks to $|\alpha_n|\leq L<1$.

Probably it is wrong, because it should be

$$\sup_n\left|\int_0^1 \left(1-e^{i\alpha_n x}\right) f(x)d\mu(x)\right|\leq \left(\sup_{n,\mu(x)} \left|1-e^{i\alpha_n x}\right|\right)\left|\int_0^1 f(x)d\mu(x)\right|$$ But here I do not know how to continue. Any suggestions please?

Answer 1

This is more like comment, but too long.

1. You probably want to estimate, not to calculate.

2. $\mu$ is singular --- is this relevant?

3. You're doing almost right, although there should be absolute value under the integral: \begin{gathered} \left|\int_0^1 \left(1-e^{i\alpha_n x}\right) f(x)d\mu(x)\right|\leq \left(\sup_{x\in[0,1]} \left|1-e^{i\alpha_nx}\right|\right)\int_0^1 |f(x)|d\mu(x)\\ \le \left(2-2\cos L\right)\int_0^1 |f(x)|d\mu(x). \end{gathered}

4. Since $f\in L^2(\mu)$, you can also use Cauchy--Schwarz: \begin{gathered} \left|\int (1-e^{i\alpha_n x})f(x) d\mu(x)\right|^2 \le \int |1-e^{i\alpha_n x}|^2 d\mu(x) \int |f(x)|^2 d\mu(x)\\ = 2(1-\operatorname{Re}\varphi(\alpha_n)) \int |f(x)|^2 d\mu(x), \end{gathered} where $\varphi$ is the characteristic function of $\mu$.