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?



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


