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!



gap in a Banach spaces ultrapower proof
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.


