## A bound on linear functionals over cotype 2 spaces

This is a modification of the somewhat naive question that I asked below.

Suppose $X$ is a real Banach space of cotype-2, and $u_1, u_2, ... u_n$ are unit vectors in this space. For $\gamma = ((\gamma_1, \gamma_2, ..., \gamma_n))$ an arbitrary element of $\mathbb{R}^n$, suppose that $\|\gamma_1 u_1 + \gamma_2 u_2 + \cdots \gamma_n u_n\| \geq \|\gamma\|_{\ell^\infty}$. Then is there a bound independent of $n$ (but depending on the cotype constant) of

$$\inf_\epsilon \sup_{\beta \geq 0} \frac{\sum \beta_i\cdot \frac{1}{\sqrt{n}}}{\|\sum \beta_i \epsilon_i u_i \|}$$

where $\epsilon = ((\epsilon_1,\epsilon_2, ..., \epsilon_n))$ varies over all sequences of $\pm 1$, and $\beta \geq 0$ means that $\beta_i \geq 0$ for all $i$.

My suspicion is that this is actually false, but I haven't constructed an explicit counterexample.

The original question I asked was resolved by two answers below. (I had left out a $\sqrt{n}$ term that I meant to include, but even with this, the answer is no.)

Suppose $X$ is a real Banach space of cotype-2, and $u_1, u_2, ... u_n$ are unit vectors in this space. For $\gamma = ((\gamma_1, \gamma_2, ..., \gamma_n))$ an arbitrary element of $\mathbb{R}^n$, suppose that $\|\gamma_1 u_1 + \gamma_2 u_2 + \cdots \gamma_n u_n\| \geq \|\gamma\|_{\ell^\infty}$. (It is important that this inequality be strict, not just an 'up to a constant' sort of bound. What is going on becomes more clear if a picture is drawn.) Is there a bound independent of $n$ on

$\sup_\beta \frac{\sum \beta_i}{\|\beta_1 u_1 + \cdots + \beta_n u_n\|}?$

If not, what if we impose the additional condition that the $\beta_i$ all be non-negative? One might guess the bound is linear with the cotype-2 constant.

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Did you mean to have a factor of ${\sqrt n}$ on the right hand side? Consider $u_i$ orthonormal in a Hilbert space.

In fact, the best constant is of order $n$, but if you restrict the $u_i$ to be C-unconditional, then obviously the sup is bounded by C times the cotype 2 constant times ${\sqrt n}$.

(1) Your inequality says that the $u_i$ are an Auerbach basis for the space they span, which means that there are functionals $u_i^*$ biorthogonal to $u_i$ and having norm one. If you have the inequality but only up to the constant $1/C$, the biorthogonal functionals have norm at most $C$. You can renorm the space with $(1/C\|x\|) \vee \max_i|u_i^*(x)|$ to make the $u_i$ an Auerbach basis--at worst this multiplies the cotype 2 constant by $C$.

(2) You can replace the top of your left hand side with $\sum |\beta_i|$ because your hypotheses on the $u_i$ does not change if you multiply some of them by $-1$. To see that you cannot do better than order $n$ as an upper estimate, consider $u_i :=(1/2)(e_i-e_{i+1})$ in $\ell_1$.

In your modified question, you get a bound independent of $n$ under the stronger hypothesis that $X^*$ has type 2. Indeed, let $u_k^*$ be the (norm one) biorthogonal functionals to $u_k$ and choose signs $\epsilon_k$ so that $y^*:= n^{-1/2}\sum_{k=1}^n \epsilon_k u_k^*$ has norm at most $T_2(X^*)$. Then for any $\beta_k\ge 0$, $\|\sum_{k=1}^n \epsilon_k\beta_k u_k\|$ is at least as large as $T_2(X^*)^{-1} n^{-1/2} \sum_{k=1}^n \beta_k$.

Since for an $n$ dimensional space $E$, $T_2(E^*)\le C \log n C_2(E)$, you do get at worst a $\log n$ growth rate of your ratio.

In your revised question, do you know it is important that the $u_k$ form an Auerbach basis for their span?

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 Thanks for this very helpful response. I tried to post a follow up question in this space, but was held up by character limits, so I've posted it below. – Brad Rodgers Apr 26 2010 at 18:34 Aha! This is great to find out! (Sorry I didn't respond sooner, but I wasn't aware of the update.) In answer to your last question, what I ultimately want to prove is that the above inf-sup quantity is bounded when we have some constant $A$ independent of $n$ so that $\|\gamma_1 u_1 + \gamma_2 u_2 + \cdots \gamma_n u_n\| \leq A\|\gamma\|_{\ell^2}$. This may seem to go in the opposite direction as the criterion I give above, but both impose the constraint that boundary of the unit ball of our Banach space must be somewhat close to meeting each $u_i$ orthogonally. (For large $n$, at any rate.) – Brad Rodgers May 13 2010 at 19:50 Sorry; I had to learn on meta how to make a changed answer move up the thread. So the condition you have is that the formal identity from $\ell_2^n$ into $\ell_\infty^n$ factors through your space. This condition was studied by Pelczynski and somebody (maybe Schutt?). I am traveling and cannot check references or use MathSciNet. – Bill Johnson May 14 2010 at 10:54

No, not even with positive coefficients. Take, for instance, the $u_i$ to be orthonormal vectors in Hilbert space (which has type 2). Then the condition $\| \gamma \|_{\infty} \le \|\gamma_1 u_1 + \gamma_2 u_2 + \cdots + \gamma_n u_n\|$ clearly holds, whereas taking all $\beta_i$ equal to one gives a quotient $\frac{n}{\sqrt{n}} = \sqrt{n}$, definitely not bounded independently of $n$.

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 Sorry to have crossed paths, Alejandro. What are you doing up so early? :) – Bill Johnson Apr 26 2010 at 14:54

Err ... yes. I meant to have normalized this, with each of the $\beta_i$'s divided by $\sqrt{n}$. Thanks for the quick responses!

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A question somewhat less naive than the one I asked above is whether, given the constraints above on the $u_i$, there is a bound independent of $n$ on

$$\inf_\epsilon \sup_{\beta \geq 0} \frac{\sum \beta_i\cdot \frac{1}{\sqrt{n}}}{\|\sum \beta_i \epsilon_i u_i \|}$$

where $\epsilon = ((\epsilon_1,\epsilon_2, ..., \epsilon_n))$ varies over all sequences of $\pm 1$. I suspect now that I've thought a bit more about it that this too is false, but it is at least a bit more reasonable guess.

I should perhaps have mentioned this earlier, but I was motivated to ask the question by some relevance this sort of thing has to a plank conjecture Keith Ball makes at the end of this paper. If what I've guessed here is false, then one could add in an upper-2 bound on the $u_i$'s of the sort Ball uses in the end of his paper.

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Sure, Brad, because the average over $\epsilon_i=\pm 1$ of $\|\sum \beta_i \epsilon_i u_i$ is at least the reciprocal of the cotype 2 constant times $(\sum |\beta_i|^2)^{1/2}$. – Bill Johnson Apr 26 2010 at 18:59
But there is a problem that for each $\epsilon$, the supremum could be achieved for a different $\beta$, no? – Brad Rodgers Apr 26 2010 at 19:04
Hi Brad, it's preferable to edit your original question, rather than clarify and extend in answers. If you want to substantially change a question (in particular, if this would cause existing answers to not make sense), then it's best to just ask a new one, leaving comments on both explaining the relationship between them. – Scott Morrison Apr 26 2010 at 19:46
Right; I misread what you asked. – Bill Johnson Apr 26 2010 at 20:47
Another question, for Scott: ought I uncheck Bill's answer? It (along with Alejandro's) answered my original question, but might otherwise give off the impression that the question I'm asking now has already been answered. I'm guessing the proper etiquette is to not, but perhaps there should be someway to indicate a new question having been asked? – Brad Rodgers Apr 28 2010 at 3:52
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