Inequality for Square-Summable Complex Series

Thu, 28 May 2026 00:00:00 +0000

Some inequalities look formidable until the right decomposition makes them transparent. The conjecture below, posed by Zoltan Retkes on the Open Problem Garden in 2012 with a £10 prize attached, is one such case: once the dyadic structure of the positive integers is made explicit, the proof reduces to two classical facts.

Conjecture (Retkes, 2012), now proved

For all $\alpha = (\alpha_1, \alpha_2, \ldots) \in \ell^2(\mathbb{C})$, $$\sum_{n \geq 1} |\alpha_n|^2 \geq \frac{6}{\pi^2} \sum_{k \geq 0} \left|, \sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1} ,\right|^2.$$

The conjecture was confirmed by an anonymous comment on the problem page in November 2013. A self-contained proof and an extension to $\ell^p$ were subsequently published by Ibragimov and Salimova in Elemente der Mathematik 70 (2015), 79–81.

The Dyadic Decomposition #

The index $2^k(2l+1)$ running over $k \geq 0$ and $l \geq 0$ is not arbitrary: it encodes a canonical partition of the positive integers. Every $n \in \mathbb{N}^+$ factors uniquely as $$n = 2^k \cdot r, \qquad k \geq 0,\quad r \text{ odd positive},$$ where $k = v_2(n)$ is the 2-adic valuation of $n$ and $r = n/2^k$ is its odd part. Writing $r = 2l+1$ gives the bijection $\mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}^+$, $(k, l) \mapsto 2^k(2l+1)$. In particular the sets $$A_k = {2^k(2l+1) : l \geq 0} = {2^k, 3 \cdot 2^k, 5 \cdot 2^k, \ldots}$$ form a partition of $\mathbb{N}^+$. Explicitly: $A_0 = {1, 3, 5, 7, \ldots}$ (odd numbers), $A_1 = {2, 6, 10, 14, \ldots}$ (twice an odd number), and so on. This partition is the key structural fact behind the proof.

Proof #

The argument has two ingredients: the Basel sum $\sum_{l \geq 0}(l+1)^{-2} = \pi^2/6$, and the Cauchy–Schwarz inequality in $\ell^2(\mathbb{C})$.

Define two sequences in $\ell^2(\mathbb{C})$: $$x = \left(1,, \tfrac{1}{2},, \tfrac{1}{3},, \ldots\right), \qquad y_k = \left(\alpha_{2^k},, \alpha_{3 \cdot 2^k},, \alpha_{5 \cdot 2^k},, \ldots\right) \quad (k \geq 0).$$

The inner sum in the conjecture is exactly the $\ell^2$ inner product $\langle x, y_k \rangle$: $$\sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1} = \langle x, y_k \rangle.$$

Step 1: Apply Cauchy–Schwarz. For each $k$,

$$|\langle x, y_k \rangle|^2 \leq |x|_2^2 \cdot |y_k|_2^2.$$

Summing over $k \geq 0$,

$$\sum _{k \geq 0} |\langle x, y _k \rangle|^2 \leq |x| _2^2 \sum _{k \geq 0} |y _k| _2^2.$$

Step 2: Evaluate using the Basel problem and the partition. The Basel problem gives $$|x| _2^2 = \sum _{l \geq 0} \frac{1}{(l+1)^2} = \frac{\pi^2}{6}.$$

Since the sets $A_k$ partition $\mathbb{N}^+$, $$\sum _{k \geq 0} |y_k|_2^2 = \sum _{k \geq 0} \sum _{l \geq 0} |\alpha _{2^k(2l+1)}|^2 = \sum _{n \geq 1} |\alpha_n|^2.$$

Combining both steps, $$\sum_{k \geq 0} \left|\sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1}\right|^2 \leq \frac{\pi^2}{6} \sum_{n \geq 1} |\alpha_n|^2,$$ which is the inequality with the $\frac{6}{\pi^2}$ factor moved to the other side.

Sharpness of the Constant #

The constant $6/\pi^2$ is the best possible. To see this, consider the truncated sequence $\alpha^{(N)}$ defined by $\alpha^{(N)}_{2l+1} = 1/(l+1)$ for $l = 0, 1, \ldots, N-1$ and $\alpha^{(N)}_n = 0$ otherwise. Then:

The left-hand side equals $\displaystyle\sum_{l=0}^{N-1} \frac{1}{(l+1)^2} \to \frac{\pi^2}{6}$.
The only non-zero contribution to the right-hand side comes from $k = 0$ (since all non-zero indices are odd, i.e. in $A_0$), giving $\displaystyle\frac{6}{\pi^2}\left(\sum_{l=0}^{N-1} \frac{1}{(l+1)^2}\right)^2 \to \frac{6}{\pi^2} \cdot \frac{\pi^4}{36} = \frac{\pi^2}{6}$.

The ratio of the right-hand side to the left-hand side therefore tends to $1$ as $N \to \infty$, so no larger constant than $6/\pi^2$ can hold universally. Equality is never achieved for $\alpha \in \ell^2(\mathbb{C})\setminus{0}$ with finite norm since the limiting sequence does not belong to $\ell^2(\mathbb{C})$.

Extension to $\ell^p$ #

The Cauchy–Schwarz inequality used above is a special case of Hölder’s inequality, and the proof generalises immediately.

Theorem (Ibragimov–Salimova, 2015)

Let $p, q \in (1,\infty)$ with $\tfrac{1}{p} + \tfrac{1}{q} = 1$. For all $\alpha = (\alpha_1, \alpha_2, \ldots) \in \ell^p(\mathbb{C})$ and $x = (x_0, x_1, \ldots) \in \ell^q(\mathbb{C})$, $$\sum_{n \geq 1} |\alpha_n|^p \geq \left(\sum_{l \geq 0} |x_l|^q\right)^{-p/q} \sum_{k \geq 0} \left|\sum_{l \geq 0} x_l, \alpha_{2^k(2l+1)}\right|^p.$$

Retkes’s original inequality is the case $p = q = 2$ and $x_l = 1/(l+1)$, where $(\sum_{l\geq 0}|x_l|^2)^{-1} = 6/\pi^2$ by the Basel problem.

Remarks on Structure #

The role of the dyadic partition. The sets $A_k$ are the dyadic layers of $\mathbb{N}^+$: each integer sits in exactly one layer determined by its 2-adic valuation. This structure also appears in the theory of Hardy spaces, where the dyadic martingale decomposition underpins the $H^1$–BMO duality, and in wavelets, where the dyadic scaling of the real line organises the multiresolution analysis. The inequality can be read as a norm comparison between the $\ell^2$ norm and a weighted sum over dyadic layers.

Relation to the Basel problem. The constant $6/\pi^2$, the reciprocal of $\zeta(2)$, appears here because the weight sequence $1/(l+1)$ used in the inner sum is precisely the harmonic sequence, whose $\ell^2$ norm squared is $\zeta(2)$. Any other weight sequence $x \in \ell^2(\mathbb{C})$ would produce the analogous inequality with $|x|_2^{-2}$ in place of $6/\pi^2$.

The inequality as a rearrangement estimate. The right-hand side reorganises the entries of $\alpha$ by their dyadic layer and applies a weighted average within each layer. The inequality says the total $\ell^2$ energy cannot be less than $6/\pi^2$ times the energy of this rearranged, averaged version of the sequence, a quantitative statement about how averaging destroys energy.

Further Questions #

While the original conjecture is settled, several natural variants remain.

Question 1

What is the sharp constant in the inequality if the dyadic partition is replaced by the partition induced by a prime $p \neq 2$, i.e. by the sets $A_k^{(p)} = {p^k m : \gcd(m, p) = 1}$? The same argument applies with $x_l = w_l$ for any weight sequence $w \in \ell^2(\mathbb{C})$, but the resulting constant depends on $|w|_2$ and the choice of weight, not on $\pi$.

Question 2

The inner sum $\sum_{l \geq 0} \alpha_{2^k(2l+1)}/(l+1)$ averages the entries in layer $A_k$ with the harmonic weights. What happens if the harmonic weight $1/(l+1)$ is replaced by a weight $w(l)$ depending on the position $l$ within the layer in a more general way, for instance $w(l) = l^{-s}$ for $s > 1/2$? The sharp constant would then involve $\zeta(2s)$ instead of $\zeta(2) = \pi^2/6$.

Question 3

For $p = 1$ the Ibragimov–Salimova theorem requires $q = \infty$, and the Hölder inequality takes a different form. Does an analogue of Retkes’s inequality hold for $\alpha \in \ell^1(\mathbb{C})$, and if so, what is the sharp constant?

References #

Ibragimov, Z. O. & Salimova, D. F. (2015). On an inequality in $\ell_p(\mathbb{C})$ involving Basel problem. Elemente der Mathematik, 70(2), 79–81. https://ems.press/content/serial-article-files/45532
Retkes, Z. (2012). Inequality for square summable complex series. Open Problem Garden. http://www.openproblemgarden.org/op/inequality_for_square_summable_complex_series
Benko, D. & Molokach, J. (2013). The Basel problem as a rearrangement of series. College Mathematics Journal, 44(3), 171–176.
Ritelli, D. (2013). Another proof of $\zeta(2) = \pi^2/6$ using double integrals. American Mathematical Monthly, 120(7), 642–645.

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