L1 distance from a trigonometric susbspace How to check, whether the $L^{1}$ distance between a finite exponential sum $S_{F}(x)=\sum\limits_{n\in F} \exp(inx)$ and the $L^{1}$-closure of subspace $\mathrm{span}\left(\exp(inx): n\in \mathbb{Z}\setminus F \right)$ is less or equal than $1$? 
This problem I have found oryginally stated as follows:
For which subset $F\subset\mathbb{Z}$ there exists a function $f\in L^{1}$ such that $\widehat{f}(k)=1$ for $k\in F$ and $\| f \|_1=1$.
I have known some estimates of $L^{1} $ norm of $S_F$, however, such estimates gives an error which leads to many cases left.
 A: I'm not sure about your version of the problem, but the following result of Helson from 1955 partially answers the original problem:
Claim: Let $\mu$ be a measure on $\mathbb{T}$ such that $\hat{\mu}(n)=0$ or $1$ for $n=0,\pm1,\pm2,\ldots.$  Then $\{n\in\mathbb{N}\colon\hat{\mu}(n)=1\}=(F_{1}\cup V)\setminus F_{2}$ with $F_{1},F_{2}$ finite sets and $V$ a periodic set, i.e. a finite sum of arithmetic progressions.
Proof:  We argue by contradiction.  Suppose the conclusion is false.  Then without loss of generality, we can find two increasing sequences $n(t),m(t)$ such that $n(t),m(t)\to\infty$ as $t\to\infty$, $\hat{\mu}(n(t)-j)=\hat{\mu}(m(t)-j)$ for $j=1,\ldots,t-1$, but $\hat{\mu}(n(t))\neq\hat{\mu}(m(t))$.  Write $\mu=fdm+\mu_{s}$ with $dm$ Lebesgue measure and $\mu_{s}$ singular.  By our assumptions (ignoring trivial cases, e.g. when $\mu$ is a finite trigonometric polynomial) $\mu_{s}\neq0$.  (If $\mu_{s}=0$, the Fourier coefficients would vanish at infinity by Riemann-Lebesgue.)  Now, consider the measure
$$\nu_{t}=(e^{-im(t)}-e^{-in(t)})\mu_{s}$$
As $t\to\infty$, we get a weak$^{*}$ limit $\nu\in L_{1}(|\mu_{s}|)$ by Banach-Alaoglu.  Note that $\nu$ is a singular measure, since it is the weak$^{*}$ limit of singular measures.  However, $\hat{\nu}(n)=0$ for $n<0$, so $\nu$ is absolutely continuous, by the F.-M. Riesz Theorem.  This contradiction gives our desired result.
A: Yet another attempt to solve this problem:
Definition An exponential sum is a trigonometric polynomial $S(x)=\sum_{k}c_k\exp(ikx)$ with coefficients $|c_k|=1$.
We will show
Lemma1: $\|S\|_1 \geqslant 1$, provided than $S$ is trigonometric polynomial with one of coeeficients of absolute value greater or equal than $1$, especially when $S$
is an exponential sum.
Proof: We can assume that $0\in \mathrm{supp}{\widehat{F}}$ and $c_0\geqslant 1$, because multiplying $f$ by $\overline{c_n}\exp(inx)$ does not change the $L^{1}$ norm.
We have an integral inequality from Zygmund's book "Trigonometric Series"
$\frac{1}{2\pi}\int\limits_{0}^{2\pi} \log |S(x)| \mbox{d}x \geqslant \log|S(0)|$
which follows from applying the Poisson formula to function $\log|S|$ modified in suitable way in order to be harmonic. Hence, $\frac{1}{2\pi}\int\limits_{0}^{2\pi} \log |S(x)| \mbox{d}x \geqslant 0$, and using the inequality $\log|S| \leqslant |S|-1$ we obtain finally
$\frac{1}{2\pi}\int\limits_{0}^{2\pi} |S(x)| \mbox{d}x \geqslant 1$
with equality if and only if $|S|=1$ a.e., equivalently, if $|\mathrm{supp}(\widehat{S})|=1$. QED
Characterization of closure of the trigonometric subspace spaned by a given subsystem it is also well know. 
Lemma2  In the $L^{1}$ norm we have
$V=\overline{\mathrm{span}\{\exp(inx): n\in \mathbb{Z}\setminus F\}} = \{ f: \widehat(f) = 0 \textrm{ on } \mathbb{Z}\setminus F\}$
Assume next, that $d(S,V)< 1$. Then from Lemma2, there exists trigonometric polynomial $g \in V$, such that $d(S,g) < 1$. Observe however, that Lemma1 holds for $S-g$.
Hence we get contraddiction $d(S,V)\geqslant\|S-g\|_1 \geqslant 1$. Therefore, we have shown
Theorem If $S$ is an exponential sum, then $d(S,V)\geqslant 1$
In view of the above result, it remains to characterize possibility of $d(S_F,V) = 1$.
Next observe, that if we have element $S_F - g$ where $g \in V$, then $S_F*g\equiv 0$ bacuase transforms of $f,g$ have disjont supports. Moreover, $S_F$ is idempotent in a convolution algebra, These remarks allowing us to calculate $(S_F-g)*(S_F-g) = S_F + g*g$. Hence, 
Lemma3 If an element $g\in V$ is the best approximation of $S_F$ and $d(S_F,V)\leqslant 1$ then  also $-g*g \in V$ is optimal.
Remark Suppose, that distance beetween $V$ and $S_F$ is less or equal than $1$.
Note that if $g$ is optimal, then $|\widehat{g}(k)|\leqslant 1$, because of $
\widehat{g-S_F} = \widehat{g}$  on  $\mathrm{supp} (\widehat{g})$. 
......
I do not know, if all of these remarks are valuable. However, finally I have found an argument, which uses some of presented ideas:
Lemma4 Suppose, that $1 = |\widehat{f}(k)|$ for some $k$. Then $\|f\|_1\geqslant 1$ with equality if and only if $f(x) = \epsilon \exp(ikx),\quad |\epsilon|=1$.
Proof: Without lose of generality, we can assume that $k=0$. Let $F_n$ be a Fejer sequence for $f$. It is known, that $F_n$ converges to $f$ almost everywhere and in $L_1$ norm. 
Moreover, $\widehat{F_n}(0) = \widehat{f}(0)$. Inequality from Zygmund's book gives
$\frac{1}{2\pi}\int\limits_{0}^{2\pi} \log |F_n(x)| \mbox{d}x \geqslant \log |\epsilon| = 0$.
Therefore, using $\log t \leqslant t-1$, we obtain
$ \frac{1}{2\pi}\int\limits_{0}^{2\pi} |F_n(x)| \mbox{d}x -1 \geqslant \frac{1}{2\pi}\int\limits_{0}^{2\pi} \log |F_n(x)| \mbox{d}x \geqslant \log |\epsilon| = 0$.
Obviously, we have $\| F_n \|_1 \geqslant 1$ and $\| F \|_1 \geqslant 1$. 
Assuming, that $\| F \|_1 = 1$, we obtain passing to the limit
$\frac{1}{2\pi}\int\limits_{0}^{2\pi} \log |F_n(x)| \mbox{d}x\to 0$, 
hence,
$\frac{1}{2\pi}\int\limits_{0}^{2\pi}| |F_n(x)| - \log |F_n(x)| - 1 |\mbox{d}x  = \frac{1}{2\pi}\int\limits_{0}^{2\pi} (|F_n| - \log |F_n| - 1)\mbox{d}x \to 0$
This can be expressed as
$ |F_n| - \log |F_n|  \to 1$ in $L^1$.
Convergence in $L_1$ implies convergence in the measure. Convergence in the measure implies convergence almost everywhere of some subsequence. We have, however, convergence almost everywhere $F_n\to F$. Therefore,
$|F| - \log |F| = 1$ a.e.
And finally, $|F|=1$, what finishes the proof.
We can rewrite as Young inequality:
Theorem The following estimate holds: $ \| f \|_1  \| \widehat{f} \|\_{\infty}^{-1} \geqslant 1$, with an equality if and only if in case of trigonometric polynomial with support of cardinality $1$.
Remark: Inequality is trivial, however I proved it using other arguments in order to extract some additional information about the equality case.
