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Given $n$ vectors $v_1,\dots,v_n\in\Bbb R^N$ with $\|v_i\|_2^2\leq1$ at every $i\in\{1,\dots,n\}$, Komlos conjecture states that, there is a $c\in\Bbb R$ (independent of $n,N$) such that at some $\epsilon\in\{-1,+1\}^n$, $$\Big\|\sum_{i=1}^n\epsilon_iv_i\Big\|_\infty<c.$$ What is best lower bound known for $c$?
Best upper bound is $c=O(\sqrt{\log n})$.
Is there a set of examples for $c=1+\delta$ for any $\delta>0$ where Komlos conjecture fails?
A simple way of obtaining a lower bound $c\ge\sqrt{2}$ is to consider pairs of vectors $u,v\in\mathbb{R}$. First of all, it makes sense to focus on pairs of unit vectors for which all $\{-1,1\}$-linear combinations are as long as possible (note that this is just an interesting special case, I'm not saying that it is opotimal in any way). This is achieved when $u,v$ are orthogonal, and by checking the possible rotations we find that $u=\frac{1}{\sqrt{2}}(1,1), v=\frac{1}{\sqrt{2}}(1,-1)$ show that $c$ must be at least $\sqrt{2}$.
This example can be generalized to the sets of vectors $V_k=\{2^{-\frac{k}{2}}v_{i,k}\ |\ 0\le i\le k\}\subseteq\mathbb{R}^{2^k}$, where the $j$-th coefficient $(v_{i,k})_j$ of $v_{i,k}$ is $1$ if the $i-th$ binary digit in $j$ is $0$, and $-1$ otherwise.
The $\infty$-norm of any $\{-1,1\}$-linear combination of the vectors in $V_k$ is $(k+1)2^{-\frac{k}{2}}$, which reaches its maximum $\frac{3}{2}$ at $k=2$, with the set of vectors
$V_2=\{\frac{1}{2}(1,1,1,1),\frac{1}{2}(1,1,-1,-1),\frac{1}{2}(1,-1,1,-1)\}$.
There may be better lower bounds, but this is a start.
Taking the $v_i$ to be the columns of this matrix shows $c \geq \frac{4}{\sqrt{6}} \approx 1.633$ (I found and verified the matrix by computer experiment):
$$M = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 & 1 & 1 & -1 & 1 & -1\\ 1 & 1 & -1 & 1 & -1 & -1\\ -1 & 1 & 1 & 1 & 1 & 1\\ -1 & 1 & 1 & -1 & -1 & 1\\ -1 & -1 & 1 & 1 & -1 & -1 \\ -1 & 1 & -1 & -1 & 1 & -1 \end{pmatrix}$$
There is no best lower bound, because c is an upper bound. The best known upper bound is $O(\sqrt{\log n})$, as proved by Banaszczyk in 1998.
