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,


1. $\textrm{sh}((e^{\*}\_{n})\_{n=1}^\infty)=1$, where $(e^{\*}\_{n})\_{n=1}^\infty$ is the unit vector basis of $\ell\_{1}$.
2. $\textrm{sh}((s\_{n})\_{n=1}^\infty)=1$, where $(s\_{n})\_{n=1}^\infty$ is the summing basis of $c\_{0}$.
3. $\textrm{sh}((\sum\_{i=1}^{n}e\_{i})\_{n=1}^\infty)=1$, 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 !



Quantifying shrinking bases
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
$\|x^\*\|\_N \ge d$.


