= Solution
A <Schauder basis> of a real <Banach space> $X$ is a sequence $(e_n)$ for which every $x\in X$ has a unique norm-convergent expansion $x=\sum_na_ne_n$. Its <basis projection> is
$$
P_nx=\sum_{k=1}^na_ke_k,
$$
and its <basis constant> is $K_b=\sup_n\lVert P_n\rVert<\infty$.
For the sequence space $Y$ in the question, define
$$
J:Y\longrightarrow X,
\qquad
J(a)=\sum_{n=1}^\infty a_ne_n.
$$
Convergence in the definition of $Y$ makes $J$ well defined, and uniqueness of basis coefficients makes it a <linear isomorphism>[linear bijection]. Moreover,
$$
\lVert J(a)\rVert
=\lim_n\left\lVert\sum_{k=1}^na_ke_k\right\rVert
\leq\lVert a\rVert_Y,
$$
whereas
$$
\lVert a\rVert_Y
=\sup_n\lVert P_nJ(a)\rVert
\leq K_b\lVert J(a)\rVert.
$$
Thus $J$ is an <isomorphism of Banach spaces>[isomorphism] and, in particular, the displayed supremum really is a complete norm on $Y$.
The <coordinate functional of a Schauder basis> is
$$
e_n^*\left(\sum_ka_ke_k\right)=a_n.
$$
It is bounded because $a_ne_n=(P_n-P_{n-1})x$. On finite linear combinations of the $e_n^*$, the nth partial-sum operator is the restriction of $P_n^*$, since
$$
P_n^*e_j^*=\begin{cases}e_j^*,&j\leq n,\\0,&j>n.\end{cases}
$$
The operators have norms at most $K_b$, so the standard basis criterion shows that the <dual sequence of a Schauder basis> $(e_n^*)$ is a <basic sequence> in $X^*$. For every $x^*\in X^*$ and $x\in X$,
$$
\langle P_n^*x^*,x\rangle
=\langle x^*,P_nx\rangle
\longrightarrow\langle x^*,x\rangle,
$$
which is precisely $P_n^*x^*\to x^*$ in the <weak-star topology>.
Suppose now that $X$ is a <reflexive Banach space>. If $\lVert x^*-P_n^*x^*\rVert$ did not tend to zero, approximation by finite basis blocks would give a bounded block sequence $(u_j)$ and an $\varepsilon>0$ such that $|x^*(u_j)|\geq\varepsilon$. Reflexivity gives a weakly convergent subsequence. Every fixed coordinate functional is eventually zero on a block sequence, so its weak limit has every basis coordinate zero and is therefore zero. This contradicts $|x^*(u_j)|\geq\varepsilon$. Hence $P_n^*x^*\to x^*$ in norm for every $x^*$, so the basis is <shrinking Schauder basis>[shrinking] and $(e_n^*)$ is a basis of $X^*$.
The converse fails. The standard basis of $c_0$ is shrinking because its dual sequence is the standard basis of $\ell^1=(c_0)^*$, but $c_0$ is not <reflexive Banach space>[reflexive].
Finally assume $(e_n^*)$ is a basis of $X^*$. Map $x^{**}\in X^{**}$ to
$$
a_n=x^{**}(e_n^*).
$$
For $s_n=\sum_{k=1}^na_ke_k$,
$$
\lVert s_n\rVert
=\sup_{\lVert x^*\rVert\leq1}
|x^{**}(P_n^*x^*)|
\leq K_b\lVert x^{**}\rVert,
$$
so $(a_n)\in Z$. Conversely, if $(a_n)\in Z$ and $M=\sup_n\lVert s_n\rVert$, define
$$
F(x^*)=\lim_nx^*(s_n).
$$
The limit exists: for $m>n$,
$$
|x^*(s_m-s_n)|
\leq2M\lVert x^*-P_n^*x^*\rVert\longrightarrow0
$$
because $(e_n^*)$ is a basis. Also $|F(x^*)|\leq M\lVert x^*\rVert$, so $F\in X^{**}$ and $F(e_n^*)=a_n$. These two constructions are inverse and satisfy
$$
\lVert F\rVert\leq\lVert(a_n)\rVert_Z
\leq K_b\lVert F\rVert.
$$
Thus $X^{**}$ and $Z$ are isomorphic.
Solved by gpt-5.6-sol high.
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