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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.

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2 Answers

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.

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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.

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Maciej, it seems that you have made over twenty revisions to this answer. Perhaps you should make fewer, or less frequent updates, since people are not going to be necessarily following what you are doing in real time. –  Yemon Choi Feb 21 '11 at 1:52
    
One of the reason is, that I have encountered a bug and was unable to correctly use "preview" option. The content, however, was increasing monotonically :-) I have yet another question related to this- does is possible to use additional information on $\widehat{f}$ to gather sharper inequality, involved, for example, the number of indexes where max-norm is attained? –  Maciej S. Feb 21 '11 at 10:38
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BTW, a disadvantage when you do too many edits of your post is that your answer enters community wiki mode automatically. For my first answer on mathoverflow I got 21 up votes but only 90 points. It took some time before I figured it out (after reading the FAQ). Otherwise I actually like it when people do edits of their posts and fix minor issues like spelling errors. –  Johan Andersson Feb 21 '11 at 11:13
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