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Apr 6, 2011 at 19:59 comment added Denis Serre @Emil. That was my first solution. But under some external pressure, I switched to my present construction where the $v_k$'s are all positive. Should be more self-confident...
Apr 6, 2011 at 15:56 comment added Emil Jeřábek I hesitate to post this as an answer as all the work has been done by someone else, but the optimal vector for $d=1$ is $v_{2k+1}=\sqrt{k+1}-\sqrt k$, $v_{2k+2}=-v_{2k+1}$. This gives $\|P_ve_I\|\le a\|e_I\|$ for $a=1/\|v\|\sim\sqrt{2/\log n}$. If $w$ is another vector, then one can show the existence of $I$ such that $\|P_we_I\|\ge a\|e_I\|$ by considering the positive and negative entries of $w$ separately, and applying Seva’s argument to both of them.
Apr 6, 2011 at 15:48 comment added Emil Jeřábek $v'=(-1,\sqrt2-1,1)$ achieves $a=1/\sqrt{5-2\sqrt2}\sim0.679$, which is better than $v=(1,\sqrt2-1,\sqrt3-\sqrt2)$ with its $a\sim0.886$.
Apr 6, 2011 at 15:20 comment added Denis Serre What do you mean Aaron with this new vector $(-1,\sqrt2-1,1)$ ?
Apr 6, 2011 at 15:14 comment added Denis Serre @Aaron. One N in Denis in French. Even if there are two A's in Aaron.
Apr 6, 2011 at 14:46 comment added Aaron Meyerowitz @Dennis Aha, so using $\sqrt{2} \approx 7/5.$ in$(-1,\sqrt{2}-1,1).$
Apr 6, 2011 at 13:22 comment added Denis Serre @Aaron. See my edits. Now my $(a_1,a_2,a_3)$ is better than $(-5,2,5)$.
Apr 6, 2011 at 13:22 history edited Denis Serre CC BY-SA 2.5
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Apr 6, 2011 at 11:48 comment added Aaron Meyerowitz At least for $n=3,$ Your example $(-a,0,a)$ is not quite as good as $(-5,2,5)$.
Apr 6, 2011 at 7:59 history edited Denis Serre CC BY-SA 2.5
added 110 characters in body
Apr 6, 2011 at 7:38 history answered Denis Serre CC BY-SA 2.5