Let $A$ be a set of $n$ integers and consider the quantity:
$$\int\_{0}^1 \left| \sum\_{a \in A} e^{2\pi i a x} \right|dx. $$
The (now solved) Littlewood conjecture is the claim that this quantity is lower bounded by $c \log n$ for some universal constant $c>0$.


The (still open) strong Littlewood conjecture is the claim that
$$\min\_{A; |A|=n} \int\_{0}^1 \left| \sum\_{a \in A} e^{2\pi i a x} \right|dx = \int\_{0}^1 \left| \sum\_{a=1}^{n} e^{2\pi i a x} \right|dx$$
Or in, other words, the exponential sum above is minimized when the set $A$ is an arithmetic progression. This seems to be a hard open problem not near resolution.


My question is:



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> Can one prove the "first case" of the strong Littlewood conjecture. Or, in other words, prove that a three-term arithmetic progression has the smallest $L^1$ norm of any three-term idempotent trigonometric polynomial?
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The first case of the strong Littlewood conjecture
The following human-verifiable proof is in collaboration\* with Fedja.


**Lemma** $1$: We have the following for $0 \le x \le 3$: $$\frac{6204}{6750}x^2-\frac{8429}{60750}x^4+\frac{4475}{546750}x^6\le x.$$



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> Proof: The above is equivalent to $$\frac{6204}{750}\left(\frac{x}{3}\right)^2-\frac{8429}{750}\left(\frac{x}{3}\right)^4+\frac{4475}{750}\left(\frac{x}3\right)^6\le x,$$ so letting $x=3y$, multiplying through by $750$, and dividing by $y$, we wish to show $$2250-6204y+8429y^3-4475y^5 \ge 0$$ for $0 \le y \le 1$. The above polynomial is divisible by $1-y$, and dividing yields $$4475y^4+4475y^3-3954y^2-3954y+2250.$$ But $$25\cdot (4475y^4+4475y^3-3954y^2-3954y+2250)$$ $$= 179(5y-3)^4 + (15215y+4638)(5y-3)^2 + 15y+9.$$ $\hspace{130mm}$ $\square$
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**Lemma** $2$: For any $a,b \in \mathbb{N}$ with $b > a$ and $b \not = 2a$, we have $$\int\_0^1 \left|1+e^{2\pi i a \theta}+e^{2\pi i b \theta}\right|^4 d\theta \le 15.$$



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> Proof: A quick proof is that when $b=2a$, the integral is $19$ and there are only $4$ nontrivial 4-tuples $(c\_1,c\_2,d\_1,d\_2) \in \{0,a,b\}$ with $c\_1+c\_2 = d\_1+d\_2$, namely when one is $0$, one is $b$, and the other two are $a$. $\square$
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**Lemma** $3$: For any $a,b \in \mathbb{N}$ with $b > a$, we have $$\int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|^6d\theta \ge 93.$$



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> Proof: We wish to lower-bound the number of 6-tuples $(c\_1,c\_2,c\_3,d\_1,d\_2,d\_3) \in \{0,a,b\}$ with $c\_1+c\_2+c\_3 = d\_1+d\_2+d\_3$. Having $|\{c\_1,c\_2,c\_3\}| = 3$ gives $6\cdot 6 = 36$ 6-tuples, having $|\{c\_1,c\_2,c\_3\}| = 2$ gives $6\cdot 3\cdot 3 = 54$ 6-tuples, and having $|\{c\_1,c\_2,c\_3\}| = 1$ gives $3$ 6-tuples. $\square$
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**Proposition** $4$: For any $a,b \in \mathbb{N}$ with $b > a$, we have $$\int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|d\theta \ge \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}\right|d\theta.$$ Consequently, the first (nontrivial) case of the strong Littlewood conjecture is true.



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> Proof: Suppose $b \not = 2a$. By Lemmas $1$, $2$, and $3$, we have \begin{align\*} \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|d\theta & \ge \frac{6204}{6750}\cdot 3 - \frac{8429}{60750}\cdot 15 + \frac{4475}{546750}\cdot 93 \\ &= \frac{130972}{91125} \\ &> 1.437 \\ &> \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}\right|d\theta,\end{align\*} where we used \begin{align\*} \int\_0^1 |1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}|d\theta &= \int\_0^1 |1+e^{2\pi i \theta}+e^{2\pi i 2\theta}| d\theta \\ &= \frac{2\sqrt{3}}{\pi}+\frac{1}{3} \\ &< \frac{2\cdot 1.733}{3.141}+\frac{1}{3} \\ &< 1.437.\end{align\*}
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\*If this were a paper, Fedja would be an author $99$ times and I once (if lucky).


