Inequality for Fourier Transform of Indicator Function If $A\subset [0,1]$ is measurable, then is it true that $|\int_0^11_A(x)e^{2\pi i x}dx| \le 1/\pi$, where $1_A(x)$ is the indicator for $A$?  Can someone provide a reference (and/or correction)?
Motivation/Background: For $x_1,\ldots,x_n \in \mathbb{C}$ there is always a subset $A\subset [n]$ for which $\frac{1}{\pi}\sum_{k=1}^{n}|x_{k}|\le |\sum_{k\in A}^{}x_k|$.  I think that the integral inequality above can be used to show that $1/\pi$ is the best possible constant that makes the above statement true for all subsets $\{x_1,\ldots,x_n\}\subset \mathbb{C}$, by taking $\frac{1}{n}\sum_{k=1}^n\delta_{x_k}$ to be approximately the uniform distribution on $S^1$.  
 A: Here is a proof of the stronger estimate
  $$ \Big |\int_A e^{2\pi ix} \,dx \Big| \le\frac1\pi\sin(\pi\mu), $$
where $\mu$ is the measure of $A$. 
It is convenient to work with the symmetric interval $[-1/2,1/2]$. We have 
\begin{align*} 
  \Big |\int_A e^{2\pi ix} \,dx \Big|^2 
    &= \Big| \int_A \cos(2\pi x)\,dx + i\int_A \sin(2\pi x)\,dx\Big|^2 \\ 
    &=\Big( \int_A \cos(2\pi x)\,dx \Big)^2 + \Big( \int_A \sin(2\pi x)\,dx\Big)^2.
\end{align*}
Write $\mathrm{mes}(A)=2c$.
Since $\cos(2\pi x)>\cos(2\pi c)$ if $|x|<c$, and $\cos(2\pi x)<\cos(2\pi c)$ if $c<|x|<1/2$, we have
  $$ 1_A(x)(\cos(2\pi x)-\cos(2\pi c)) \le 1_{[-c,c]}(x) (\cos(2\pi x)-\cos(2\pi c)) $$
for all $x\in[-1/2,1/2]$. Integrating, we get
\begin{align*} 
  \int_A \cos(2\pi x)\,dx - 2c\cos(2\pi c) 
    &\le \int_{-c}^c\cos(2\pi x)\,dx - 2c\cos(2\pi c) \\
    &= \frac1{\pi} \sin(2\pi c)  - 2c\cos(2\pi c);
\end{align*}
that is, 
  $$ \int_A \cos(2\pi x)\,dx \le \frac1{\pi} \sin(2\pi c). $$
Indeed, this inequality is true for any set $A$ of measure $2c$. Applying it to the set $A+\alpha$ with a fixed parameter $\alpha$, we get
  $$ \frac1\pi\sin(2\pi c) \ge \int_A \cos(2\pi x-2\pi\alpha)\,dx = C\cos(2\pi\alpha) + S\sin(2\pi\alpha), $$
where $C=\int_A \cos(2\pi x)\,dx$ and $S:=\int_A \sin(2\pi x)\,dx$. To optimize, we choose $\alpha$ so as to have $\cos(2\pi\alpha)=C/\sqrt{C^2+S^2}$ and $\sin(2\pi\alpha)=S/\sqrt{C^2+S^2}$. This gives
  $$ \Big |\int_A e^{2\pi ix} \,dx \Big| = \sqrt{C^2+S^2} \le\frac1\pi\sin(2\pi c), $$
as wanted.
