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Lemma 6.3 in Rudin's Real and Complex Analysis states that:

For every $n$ and for every $n$-tuple of complex number $z_1,...,z_n$, there exists $S\subseteq \{1,...,n\}$ such that
$$\left| \sum_{k \in S} z_k \right| \geqslant \frac1\pi\sum_{i=1}^{n} | z_i |$$

This can be generalised to $\mathbb R^d$ for all $d$ (see, for example, here) and, exploiting the fact that all norms on a finite dimensional vector space are equivalent, we arrive at the following generalisation

Let $(X,||\cdot||)$ be a finite dimensional normed space. There exists a constant $c>0$, that depends only on $X$ and $||\cdot||$, such that the following holds: for any $n$ and any $n$-tuple $(v_1,v_2,…,v_n)$ of elements of $X$, there exists a subset $J⊆\{1,2,…,n\}$ such that $$\left\| \sum_{j \in J} v_j \right\| \geqslant c \sum_{i=1}^{n} \| v_i \|.$$

This is not always the case for infinite dimensional normed spaces. For example, if $(X,\langle\cdot,\cdot\rangle)$ is an infinite dimensional Hilbert space, pick a countable orthonormal subset $\{e_n\}_{n\in\mathbb N}$. Then for all $n$ and for all subsets $J\subseteq\{0,...,n\}$, we have that $\sum_{i=0}^{n} \| e_i \|=n$ and $\left\| \sum_{j \in J} e_j \right\|=\sqrt{|J|+1}\leqslant\sqrt{n+1}$, but there is no constant $c>0$ such that $\forall n\in\mathbb N:\sqrt{n+1}\geqslant cn$. A similar argument also works for $\ell^p$ for all $p\in(1,\infty]$, however, I wasn't able to apply the same argument for $\ell^1.$

My question is: does there exist an infinite dimensional normed space for which the theorem above holds? I couldn't come up with an example, so i tried to prove that is is never the case. My idea was to use Riesz's lemma, but I got stuck.

1 Answers1

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One argument would be based on Dvoretzky's theorem, i.e. given Banach space $(X,||.||)$ then there are constants $c_{1}$ and $c_{2}$ such that for any $n$ one can find a a subspace $X_{n}$ of $X$ such that $$c_{1}*(\sum a_{k}^{2})^{1/2} <= ||\sum a_{k}*x_{k}|| <= c_{2}*(\sum a_{k}^{2})^{1/2}$$ for $x_{k} \in X$ and $a_{k}$ scalars. In words, $X$ admits uniformly copies of $l_{n}^{2}$. Then one would use your argument for Hilbert space above.
Now, this is too complex argument and there must be a simpler one but I do not see it ... .

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