A formally weaker form of the extendable local reflexivity for Banach spaces

Question

Rosenthal introduce the notion of the extendable local reflexivity for Banach spaces as follows: Let $X$ be a Banach space and let $\lambda\geq 1$. $X$ is said to be $\lambda$-extendably locally reflexive ($\lambda$-ELR) if for every finite-dimensional subspaces $E \subset X^{**}$ and $F \subset X^{*}$, and for every $\varepsilon > 0$, there exists a bounded linear operator $T:X^{**}\rightarrow X^{**}$ with $\|T\|\leq \lambda+\varepsilon$ such that $T(E)\subset X$ and $(Tx^{**})(x^*)=x^{**}(x^*)$ for every $x^{**} \in E$ and $x^* \in F$. A Banach space $X$ is said to be ELR if $X$ is $\lambda$-ELR for some $\lambda \geq 1$. Rosenthal, Johnson and Oikhberg proved that a Banach space $X$ has the BAP and is ELR if and only if $X^{*}$ has the BAP. Now if we replace the "$(Tx^{**})(x^*)=x^{**}(x^*)(x^{**} \in E, x^* \in F)$" in the definition by "$|(Tx^{**})(x^*)-x^{**}(x^*)| \leq \varepsilon \|x^{**}\|\|x^*\|(x^{**} \in E, x^* \in F)$", we have a formally weaker form of the ELR. Moreover, we can prove that this formally weaker form of the ELR also permits to lift the BAP from a Banach space to its dual space. My question is: Is this formally weaker form of the ELR equivalent to the ELR? The answer,at least, is yes whenever the Banach space $X$ has the BAP.

Answer 1

The answer is yes by a small perturbation argument. For a small $\epsilon$, enlarge $F$ so that $F$ $1+\epsilon$-norms $E$. Get $T$ from the condition. Choose any Auerbach basis $u_k$ for $E$ and take biorthogonal functions $x_k^*$ to $u_k$ with $x_k^*$ in $F$. So $\|u_k\|=1$ and by the norming condition you can have $\|x_k^*\| \le 1 + \epsilon$. There are $x_k$ in $X$ biorthogonal to $x_k^*$ with $\|x_k\| \le (1 +\epsilon)^2$ (for example, by usual local reflexivity, or just by Helly's theorem). Define $S:X^{**} \to X$ by $$ Sx^{**} = \sum_k \langle x^{**},x_k^*\rangle (1-\langle x_k^*, Tu_k \rangle ) x_k. $$

Then $S$ has small norm if $\epsilon$ times the dimension of $E$ is small. Replace $T$ with $T+S$.

EDIT, 7/20/14. I wrote down the wrong expression for $S$. Rather than giving the most natural correct $S$, I'll explain in a more conceptual manner how to get $S$. Having gotten $u_k$, $x_k^*$, and $x_k$ as above, you define $S$ on the span of $u_k$ so that for each $k$,

$$ Su_k = (1-\langle x_k^*, Tu_k \rangle ) x_k - \sum_{j\not= k} \langle x_j^*, Tu_k \rangle x_j $$

so that for all $k$ and $j$ you have $\langle (T+S)u_k, x_j^* \rangle = \langle u_k, x_j^* \rangle$. The coefficients of each $x_j$ in the expression for $S$ are of order $\epsilon$, so, since $u_k$ is Auerbach and $x_k$ is almost Auerbach, The norm of $S$ on $E$ is of order at most $\epsilon N$, where $N$ is the dimension of $E$. So $S$ can be extended to an operator from $X^{**}$ to $SE$ having norm of order at most $\epsilon N^{1/2}$.