Let $X$ be a Banach space and let $(x_{n})_{n=1}^\infty$ be a (Schauder) basis for $X$. Let $(x^{*}_{n})_{n=1}^{\infty}$ be the biorthogonal functionals associated to the basis $(x_{n})_{n=1}^\infty$. We shall use the notation $\|x^{*}\|_{n}:=\|x^{*}|_{[x_{i}\colon i>n]}\|, \quad (x^{*}\in X^{*}, n\in \mathbb{N}).$ Recall that $(x_{n})_{n=1}^\infty$ is shrinking if and only if $\|x^{*}\|_{n}\to 0$ as $n\to\infty$ for every $x^{*}\in X^{*}$.
Let $X$ be a Banach space with a basis $(x_{n})_{n=1}^\infty$. We set $$\textrm{sh}((x_{n})_{n=1}^\infty)=\sup_{x^{*}\in B_{X^{*}}}\limsup_{n}\|x^{*}\|_{n}.$$ Clearly, $(x_{n})_{n=1}^\infty$ is shrinking if and only if $\textrm{sh}((x_{n})_{n=1}^\infty)=0$.
We consider the $\textrm{sh}((x_{n})_{n=1}^\infty)$ for some familar non-shrinking bases and find that all the $\textrm{sh}$-values are $1$. For example,
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$\textrm{sh}((e^{*}_{n})_{n=1}^\infty)=1$, where $(e^{*}_{n})_{n=1}^\infty$ is the unit vector basis of $\ell_{1}$.
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$\textrm{sh}((s_{n})_{n=1}^\infty)=1$, where $(s_{n})_{n=1}^\infty$ is the summing basis of $c_{0}$.
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$\textrm{sh}((\sum_{i=1}^{n}e_{i})_{n=1}^\infty)=1$, where $(e_{n})_{n=1}^\infty$ is the unit vector basis of the James space $\mathcal{J}$.
Question 1. $\textrm{sh}((x_{n})_{n=1}^\infty)=1$ or $0$ for every basis $(x_{n})_{n=1}^\infty$ ？
Question 2. For each $0<c<1$, does there exist a basis $(x_{n})_{n=1}^\infty$ so that $\textrm{sh}((x_{n})_{n=1}^\infty)=c$ ?
Thank you !
Best Answer
The answer to your first question is "yes". Here is a sketch of an argument that may be clumsier than needed. If $(e_n)$ is not shrinking, take a norm one linear functional $x^*$ that has distance $d$ arbitrarily close to one from the span of $(e_n^*)$ and take a norm one $x^{**}$ with $\langle x^{**}, x^{*}\rangle = d$ and $\langle x^{**}, e_n \rangle =0$ for all $n$. By Goldstine's theorem you get a net $(y_\alpha)$ of unit vectors in $X$ s.t. $\langle x^* , y_\alpha \rangle \to d$ and $\langle e_n^*, y_\alpha \rangle \to 0$ for each $n$. This last condition implies that for every $N$, the distance from $y_\alpha$ to the span of $(e_n)_{n>N}$ goes to zero, and hence $\|x^*\|_N \ge d$.