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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 $x={\sum }_n{a}_n{e}_n$. Its basis projection isand its basis constant is $K_b={\sup }_n\Vert P_n\Vert <\infty$.

For the sequence space $Y$ in the question, defineConvergence in the definition of $Y$ makes $J$ well defined, and uniqueness of basis coefficients makes it a linear bijection. Moreover,whereasThus $J$ is an isomorphism and, in particular, the displayed supremum really is a complete norm on $Y$.

The coordinate functional of a Schauder basis isIt is bounded because $a_n{e}_n=(P_n-P_{n-1})x$. On finite linear combinations of the $e_n^{\ast }$, the nth partial-sum operator is the restriction of $P_n^{\ast }$, sinceThe operators have norms at most $K_b$, so the standard basis criterion shows that the dual sequence of a Schauder basis $(e_n^{\ast })$ is a basic sequence in $X^{\ast }$. For every $x^{\ast }\in X^{\ast }$ and $x\in X$,which is precisely $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in the weak-star topology.

Suppose now that $X$ is a reflexive Banach space. If $\Vert x^{\ast }-P_n^{\ast }{x}^{\ast }\Vert$ did not tend to zero, approximation by finite basis blocks would give a bounded block sequence $(u_j)$ and an $\epsilon >0$ such that $|x^{\ast }(u_j)|\ge \epsilon$. 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 $|x^{\ast }(u_j)|\ge \epsilon$. Hence $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in norm for every $x^{\ast }$, so the basis is shrinking and $(e_n^{\ast })$ is a basis of $X^{\ast }$.

The converse fails. The standard basis of $c_0$ is shrinking because its dual sequence is the standard basis of ${\ell }^1=(c_0{)}^{\ast }$, but $c_0$ is not reflexive.

Finally assume $(e_n^{\ast })$ is a basis of $X^{\ast }$. Map $x^{\ast \ast }\in X^{\ast \ast }$ toFor $s_n={\sum }_{k=1}^n{a}_k{e}_k$,so $(a_n)\in Z$. Conversely, if $(a_n)\in Z$ and $M={\sup }_n\Vert s_n\Vert$, defineThe limit exists: for $m>n$,because $(e_n^{\ast })$ is a basis. Also $|F(x^{\ast })|\le M\Vert x^{\ast }\Vert$, so $F\in X^{\ast \ast }$ and $F(e_n^{\ast })=a_n$. These two constructions are inverse and satisfyThus $X^{\ast \ast }$ and $Z$ are isomorphic.

Solved by gpt-5.6-sol high.

A Schauder basis is shrinking when its dual sequence is a Schauder basis of the whole dual space. Every Schauder basis of a reflexive Banach space is shrinking.