gap in a Banach spaces ultrapower proof

Question

This is an adaptation of a Heinrich proof, but I'm missing a key ingredient.

Conjecture. Suppose $(x_n)_{n=1}^\infty$ is a Schauder basis for a Banach space $X$ whose canonical isometric copy in $X^{**}$ is complemented. Then for any free ultrafilter $\mathcal{U}$ on $\mathbb{N}$, the canonical copy of $X$ in $X^\mathcal{U}$ is complemented in $X^\mathcal{U}$.

Proof idea. Denote by $q:X\to X^{**}$ the canonical isometric embedding. Set $X_N=\text{span}(x_n)_{n=1}^N$, and note that there exists $C\in[1,\infty)$ such that for each $N\in\mathbb{N}$ there is an operator $\widehat{P}_N:X^{**}\to X$ which is a $C$-bounded linear projection onto $X_N$ and for which $\widehat{P}_Nq$ acts as the identity on $X_N$. Let's define the linear map $J:\text{span}(qx_n)_{n=1}^\infty\to X^\mathcal{U}$ by the rule $$ Ju=(\widehat{P}_Nu)_\mathcal{U}. $$ Note that if $x\in\text{span}(x_n)_{n=1}^\infty$ then there is $k\in\mathbb{N}$ and $x\in X_N$ for $N>k$, whence $$ Jqx=(\widehat{P}_1qx,\cdots,\widehat{P}_kqx,x,x,x,\cdots)_\mathcal{U}=x^\mathcal{U}. $$ It follows that $Jq$ and hence also $J$ are continuous. Now we can extend $J$ to $qX$ via continuity, so that $Jq$ is the canonical embedding $r:X\to X^\mathcal{U}$. Due to the fact that $qX$ is complemented in $X^{**}$, we can extend $J$ again to a continuous linear operator $J:X^{**}\to X^\mathcal{U}$ with range $rX$.

Next we define the linear map $V:X^\mathcal{U}\to X^{**}$ via the rule $$ V(y_n)_\mathcal{U}=\underset{\mathcal{U}}{\text{weak*-lim}}\,qy_n $$ which exists by the weak*-compactness of $B_{X^{**}}$ together with the fact that if $K$ is a compact Hausdorff space then for each $(k_n)_{n=1}^\infty\in K^\mathbb{N}$ the (unique) limit $\lim_\mathcal{U}k_n$ exists in $K$. We now have $$ \left\langle VJqx,f\right\rangle =\left\langle Vx^\mathcal{U},f\right\rangle =\langle qx,f\rangle $$ for $x\in X$ and $f\in X^*$. This means $VJ$ is the identity on $qX$ and hence that $JV$ is an idempotent with range $rX$.

GAP. We need to show that $JV$ is continuous, from which the conjecture will follow. Of course $J$ is continuous, and so if $V$ was too then that would be sufficient. But due to the clumsiness of my weak-star intuitions, I can't seem to show it. Or, alternatively, maybe there's a theorem about linear idempotents being continuous under certain conditions.

Any ideas would be much appreciated. Thanks!

Answer 1

Theorem. Suppose $T: X \to Y^*$ is a bounded linear operator and $\mathcal{U}$ is a free ultrafilter on $\Bbb{N}$. Then $T$ extends to an operator $S:X^\mathcal{U} \to Y^*$ with $\|S\| = \|T\|$.

Proof: Define $V:\ell_\infty(X)$ to $Y^*$ by letting $V(x_n)_n$ be the weak$^*$ limit along $\mathcal{U}$ of $Tx_n$. Identify $X$ with its diagonal in $\ell_\infty(X)$. Then $V$ extends $T$ and has the same norm as $T$. If $\|x_n\| \to 0$ along $\mathcal{U}$, then $V(x_n) =0$, so $V$ induces an operator from $X^\mathcal{U}$ into $Y^*$.

Your conjecture follows. No approximation condition on $X$ is needed.