The Polygonal Confinement Theorem can be found in Section 2 of this paper by Rosenthal. I am interested in a generalization of Lemma 3.1 in the paper, which states:

$\textbf{Lemma 3.1:} $ If $v_1,\ldots,v_m \in \mathbb{R}^n$, and $\left\|\displaystyle \sum_{i=1}^m v_i\right\|<\epsilon$, $\|v_j\|<\epsilon$ for all $j$, then there is a constant $C$ which does not depend on $m$ and a permutation $\sigma$ of $\{1,\ldots,m\}$ such that for each $1\leq j\leq m$, $$ \left\|\sum_{i=1}^j v_i\right\| \leq C\epsilon.$$

I am wondering whether the following analogue for entire functions holds. In the question below, $\|f\|_R$ denotes the supremum of $f$ on the disk $|z|\leq R$.

$\textbf{Question:}$ Suppose $R>0$ and $f_i$ are entire functions on $\mathbb{C}$. Let $\epsilon>0$. If $\displaystyle \left\| \sum_{i=1}^m f_i \right\|_R< \epsilon$ and $\displaystyle \|f_j\|_R < \epsilon$ for each $j$, does there exist a constant $C$ independent of $m$ and a permutation $\sigma$ of $\{1,\ldots,m\}$ such that whenever $1\leq j\leq m$, $$ \left\| \sum_{i=1}^j f_{\sigma(i)} \right\|_R< C\epsilon?$$

Lemma 3.1, called the rearrangement theorem by Rosenthal and the Steinitz lemma by others, was the crucial element for proving the Levy-Steinitz theorem (found in the Rosenthal paper). The Levy-Stenitz theorem was generalized to metrizable nuclear topological vector spaces by Banaszczyk in this paper. Since the space of entire functions is a Frechet space, and hence a nuclear space, I am hoping that the analgoue of Lemma 3.1 holds. In fact, it looks like the Corollary in the Banaszczyk paper on page 196 may be what I'm looking for. But the generality and technicality of the paper is well beyond my expertise.

Any insight or references would be most appreciated.


1 Answer 1


The analogue for entire functions does not hold. I will take $R=1$ throughout.

First a paragraph of explanation for why that Banaszczyk paper is not relevant. Your question is equivalent to whether the Polygonal Confinement Theorem holds for some Banach space $X,$ the completion of the space of entire functions under the $\|\cdot\|_R$ norm. (Taking the completion is harmless here: any counterexample in the completion would give a counterexample in the non-completed space.) $X$ is an infinite-dimensional Banach space, so isn't nuclear, and the paper's result don't apply.

There is no constant $C$ independent of $n$ such that PCT holds for $\ell^\infty(n)$ where $\ell^\infty(n)$ is $\mathbb R^n$ with the $\ell^\infty$ norm. To see this consider $m$ even, $n=\binom{m}{m/2},$ enumerate the order $m/2$ subsets of $\{1,\dots,m\}$ as $S_1,\dots,S_n,$ and define $f_i(j)=1$ if $i\in S_j$ and $f_i(j)=-1$ otherwise ($1\leq i\leq m$ and $1\leq j\leq n$). These functions all have norm $1$ and sum to zero, but the sum of any $m/2$ has norm $m/2.$

We can transfer this counterexample to $X$ using an approximate embedding $F:\ell^\infty(n)\to X,$ i.e. $$\tfrac12\|f\|_\infty \leq \|F(f)\|_R\leq 2\|f\|_\infty.\tag{*}$$

where $\|f\|_\infty=\sup\{|f(z)|:|z|<1\}.$ Given such an $F,$ the functions $F(f_i)$ have norm $2$ and sum to zero, but the sum of any $m/2$ has norm at least $m/4.$

It remains to construct $F$ satisfying (*). Take $n$ continuous functions $\phi_1,\dots,\phi_n$ on the unit circle with disjoint support and sup-norm $1.$ By Stone-Weierstrass there are trigonometric polynomials $\phi'_n,$ i.e. polynomials in $z^{-1}$ and $z,$ such that $|\phi_i(z)-\phi'_i(z)|<1/2n$ for $|z|=1.$ Multiplying by a high enough power of $z$ we can assume $\phi'_i$ is in fact a polynomial and hence entire. For any $c\in\ell^\infty(n),$ the norm $\|\sum_{i=1}^n c_i\phi'_i\|_\infty$ is within $\tfrac12 \max_i|c_i|$ of $\|\sum_{i=1}^n c_i\phi_i\|_\infty=\max_i|c_i|,$ which gives (*).

  • $\begingroup$ I don't understand in the $\ell^\infty(n)$ example why the sum of any $m/2$ such functions should have norm $m/2$. E.g., if $\tau(i)$ is the index such that $S_{\tau(i)}$ is the complement of $S_i$, then $f_i + f_{\tau(i)} = 0$. So if you add these up pairing indices $i$ and $\tau(i)$, the partial sums have alternating norms $0$ and $1$. $\endgroup$ Jan 8, 2019 at 22:20
  • $\begingroup$ @LukasGeyer: $i$ doesn't index a subset - note $i\leq m$ not $i\leq n.$ Given distinct $i_1,\dots,i_{m/2},$ there is some $j$ such that $S_j=\{i_1,\dots,i_{m/2}\}$, so $f_{i_1}(j)+\dots+f_{i_{m/2}}(j)=m/2.$ $\endgroup$
    – Dap
    Jan 9, 2019 at 7:45
  • $\begingroup$ Thanks for the clarification, I completely misunderstood it the first time I read it. $\endgroup$ Jan 9, 2019 at 18:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .