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.