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.



A formally weaker form of the extendable local reflexivity for Banach spaces
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 $\epsilon N^{1/2}$.


