 Solution

The open mapping theorem says that a bounded surjective linear map $T:E\to F$ between Banach spaces is open. Here is a proof using only Baire and completeness. Surjectivity gives $F={\bigcup }_{n\ge 1}T(n{\overline{B}}_E)$, hence also a cover by their closures. Baire makes one such closure contain a ball $B(y_0,r)$. Taking differences gives $B(0,r)\subseteq \overline{T(2n{\overline{B}}_E)}$. Scaling yields some $\delta >0$ withIf $\Vert y\Vert <\delta /2$, recursively use scaled versions of this inclusion to choose $x_k$ with $\Vert x_k\Vert \le 2^{-k}$ and residual $\Vert y-T{\sum }_{j=1}^k{x}_j\Vert <\delta 2^{-k-1}$. The sum converges in $E$ to an element of norm at most one, whose image is $y$. Thus $B(0,\delta /2)\subseteq T({\overline{B}}_E)$, and scaling gives $B(0,\delta /4)\subseteq T(\frac{1}{2}{\overline{B}}_E)\subseteq T(B_E)$. Translating and scaling proves openness on every open set. This proof does not use the closed graph or inverse mapping theorem.

Apply it to the evaluation map $E:X\to {\ell }^{\infty }$, $Ef=(f(z_n){)}_n$. It is bounded with norm at most one and onto by hypothesis. Hence its image of the unit ball contains a ball about zero, yielding a constant $c$ and, for every $\alpha$, an interpolant $g$ withFor completeness, scale a ball of radius $\eta$ inside $E(B_X)$ by $2\Vert \alpha {\Vert }_{\infty }/\eta$ to obtain $c=2/\eta$; zero data use $g=0$.

For each $n$, interpolate the data which are one at index $n$ and zero elsewhere, obtaining $g_n$ of norm at most $c$. The supplied factorization property gives $g_n=k_n{h}_n$ with $\Vert k_n{\Vert }_{\infty }=\Vert g_n{\Vert }_{\infty }\le c$. Evaluation at $z_n$ gives $1=|k_n(z_n)h_n(z_n)|\le c|h_n(z_n)|$, soThe three enumerated properties of $h_n$ are supplied hypotheses for this argument, rather than additional requests to prove them.