Given $v_i∈B^n$, bounding $\sum b_iv_i$ for $b_i= \pm 1$

Question

Let $(v_i)_{i∈ℕ}$ be a vector sequence. Say $(v_i)$ is boundable (under $M$) if there exists a sequence $(b_i)_{i∈ℕ}$ taking values in $\{-1,1\}$ such that $(|\sum^N_i b_iv_i|)_{N∈ℕ}$ is bounded (under $M$). If $(v_i)$ takes values in the unit ball $B^n⊆ℝ^n$, does it follow $(v_i)$ is boundable?

With the definition formulated analogously for finite sequences, I see I can utilize the (weak) Konig's lemma to show that a vector sequence is boundable under $M$ if and only if all its finite starting sequences are boundable under $M$. As such, it follows that if every such vector sequence is boundable, then there is a minimal such $α_n$ such that every vector sequence is boundable under $α_n$ (for example, $α_1=1$). Based on some randomized computations of mine it appears the hypothesis holds for $n=2$ and $1.6≤α_2≤1.7$, however I'm not sure how to continue.

Edit: It appears actually that should the hypothesis be true, then $α_2 ≥ \sqrt{3}$. This diagram indicates how to construct a finite sequence unboundable under $r$ when $r < \sqrt{3}$: Start with the resultant vector $w_1$ (initial $v_1=(1,0)$) of magnitude $1$ or greater; choose the next vector $w_2$ such that $|w_1+w_2| > r$ yet the angle between $w_1,w_2$ is less than $2π/3$. Then $|w_1-w_2| > |w_1|$ and we can repeat the process with resultant vector $w_1-w_2$ to at least constantly greater effect each time. (Note the mirror situation of $b_1=-1$ is taken care of by symmetry)

Answer 1

According to [$K_2$, p.22], $\alpha_2=\sqrt{3}$, $\alpha_n\le n$, and the problem if $\sup\frac{\alpha_n}{\sqrt{n}}$ is finite is open. In [B] is proved that $\alpha_n(N)\le C_1\sqrt{n} + C_2\log N$ and is considered the counterpart of the problem for the maximum norm $\|v\|_\infty$.

References

[B] Wojciech Banaszczyk, On series of signed vectors and their rearrangements, Random Struct. Algorithms 40:3 (2012), 301-316.

[$K_1$] Vladimir Kadets, Review of [B], Zbl 1250.46011.

[$K_2$] Vladimir Kadets, Rearrangements of series in Banach spaces: A short survey, Palermo, July 2013.

Answer 2

Here is a sketch of a proof showing that $\alpha_2 \le \sqrt 3$.

We choose the signs one at a time as we look at the sequence, except that we allow ourselve to leave one sign undecided somewhere.

So at each step, we are either at one single point, or at two points at a distance at most $2$ from each other. We are given the next step $v_i$ and we have to decide what to do with them.

The following procedure should work :

Always try to go for one single point $P$ where $|P| \le 1$.
If that's impossible, we can go for two points $P,Q$ where $|P+Q| \le 2$. Either by splitting from the previous point ; by translating the previous two points and keeping the midle point in the circle of radius $1$ ; or by moving $P$ and $Q$ in opposite directions, keeping their distance small and not moving the middle point.

The farthest point we can get from a pair of point configuration is when the middle point is at $(1,0)$ and the two points are a small perturbation of $(1 \pm 1/2, \pm \sqrt 3/2)$, one of which is at distance $\sqrt 3$

Answer 3

This answer is based on my misuderstanding of the question, see the second Feryll's comment below.

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According to Wojciech Banaszczyk [B1], S. Sevastyanov [S] and, independently I. Bárány (unpublished) proved that $\alpha_n=\sqrt{n}$. Also there is a lot of the generalizations of this inequality for different norms, see, for instance, the references.

References

[B1] Wojciech Banaszczyk, A Beck-Fiala-type Theorem for Euclidean Norms, Europ. J . Combinatorics 11 (1990), 497-500.

[B2] Wojciech Banaszczyk, Balancing vectors and convex bodies, Studia mathematica, 106 (1993), 93-100.

[G] Apostolos A. Giannopoulos, On some vector balancing problems, Studia mathematica, 122:3 (1997), 225-234.

[S] S. V. Sevastyanov, On the approximate solution of the problem of calendar planning, Upravlyaemye Systemy, 20 (1980), 49–63 (in Russian).