On the $L^1$-norm of certain exponential sums I am stuck with an elementary-looking problem, which does not belong to my usual field of research so I eventually decided to ask it on MO.
Let $S$ be a finite set of integers. For $P$ a subset of $S$, I note
$$s_P = \sum_{s \in P} s$$
the sum of elements in $P$.
I assume that $S$ satisfies the following property
$$ (*) \ \ \ \ \ \  \forall P,Q \subset S, \ \ \ s_P=s_Q  \Longrightarrow P=Q$$
In other words, $(*)$ means that the set $R=$ {$ s_P,\  P \subset S$} has
maximal cardinality:
$$ (*) \ \ \ \ \ \  |R| = 2^ {|S|}$$
(Is there a name for the property $(*)$? additively free maybe?)
Now for $x \in [0,1]$ a real number, I consider the exponential sum
$$f(x) = \prod_{s \in S} (1+e^{2 i \pi s x}) = \sum_{r \in R} e^{2 i \pi r x}.$$
(Seen as a function of $e^{2i \pi x}$ on the unit circle of $\mathbb C$, $f$ is the Fourier
transform $\widehat{\chi_R}$ of the chatacteristic function $\chi_R$ of $R \subset \mathbb Z$.) I am interested in upper bounds for the $L^1$ norm of $f$,
$$||f||_1 = \int _{x=0}^1 |f(x)| dx.$$
A very simple application of Cauchy-Schwarz gives
$$||f||_1 < \sqrt{|R|} = \sqrt{2}^{|S|}$$
My question is:

When $|S|$ goes to $\infty$, can we prove an asymptotically better upper bound for $||f||_1$ than the Cauchy-Schwarz estimate above?
For example, is there a positive real
$\alpha < \sqrt{2}$ such that $||f||_1 < \alpha^{|S|}$ for all $S$ satisfying $(*)$ ? (Edit: same question for sets $S$ satisfying the stronger property $(*2)$ below).


Example: when $S=${$ 1,2,4,\dots,2^{n-1}$}, then $|S|=n$ and $R=${$0,1,2,3,\dots,2^n-1$}, so $S$ satisfies $(*)$, then $f(x)$ is essentially (up to multiplication by a complex of modulus 1)
the  Dirichlet kernel $D_{2^{n-1}}(x)$ and a famous result (of Dirichlet?) says that $$||f||_1 \sim \log (2^{n-1}) \sim n = |S|,$$ so in this case we get a much better bound than the Cauchy-Schwarz bound, and which is essentially optimal by the Littlewood conjecture (now a theorem, saying that $||\widehat{\chi_R}||_1 >> \log |R|$).
Edit: This suggests to restrain ourselves at first to sets $S$ satisfying the stronger property
$(*2)  \ \ \ $ every element of $S$ is a power of $2$.
Non-Example: removed as pointless after Noam's comment; replaced with this Remark:
Without the hypothesis $(*)$, (for $f$ defined as a product over $S$ as above, or as a sum over $R$ if $R$ is interpreted as a multiset),
the Cauchy-Schwarz bound $||f||_1 \leq \sqrt{2}^{|S|}$ does not hold any more, and one can not in general improve on the trivial bound $||f||_1 \leq 2^{|S|}$ as shown in Noam's comment
below.

But in general I am stuck. I thought that the product expression of $f(s)$ might help,
but I was not able to use it in  a clever way.
Numerical evidence is not very conclusive but does not seem to point toward a polynomial bounds in $|S|$. Any advice, reference, intuition, conjecture, solution (proof or disproof) welcome. I am not even sure that my tags are right, please feel free to modify them. PS: my motivation comes form Galois representations.
 A: As Noam correctly mentioned, the upper bound has to be exponential. However, the exponent can, indeed, be improved. 
One (fairly cheap) trick is to look at $|f(x)|^2=2^{|S|}\prod_S(1+\cos 2\pi sx)$. If we can show that the product is bounded from above by $e^{-c|S|}$ outside a set of measure $e^{-c|S|}$, we can get away with Cauchy-Schwarz. Now, $\log(1+y)\le y-cy^2$ for $y\in[-1,1]$ with some $c>0$. Since $2\cos^2 y=1+\cos 2y$, we see that the square term will give us a fixed linear in $|S|$ push down, so it remains to show that for every $a>0$, the set of $x$ for which $\left|\sum_{s\in S} \cos 2\pi sx \right|$ is at least $a|S|$ has exponentially small measure (we'll need to use that twice: once for $x$ and once for $2x$). 
This is actually pretty easy if we recall the reverse inequality $\log(1+x)\ge x-Cx^2$ for $x\in[-\frac 12,\frac 12]$, say. Note that the unique representation as a sum property implies that
$$
\int_0^1 \prod(1+t\cos 2\pi sx)\,dx=1
$$
for every $t\in\mathbb R$. Assuming that $\sum_{s\in S} \cos 2\pi sx>a|S|$ on a set $E$, choose $t\in(0,\frac 12)$ so small that $b=at-Ct^2>0$. Then, clearly, the product is at least $e^{b|S|}$ on $E$ and the desired exponential bound on the measure of $E$ follows at once. The other set where $\sum_{s\in S} \cos 2\pi sx<-a|S|$ is treated the same way using small negative $t$.
This all is extremely crude, of course. However, it answers the original question in the affirmative, so I'll stop here :).  
