A quantity measuring weak non-compactness Let $A$ be a bounded subset of a Banach space $X$. Set: $wk_{X}(A)=\inf\{\epsilon>0:\overline{A}^{w^{*}}\subset X+\epsilon B_{X^{**}}\}$, where $\overline{A}^{w^{*}}$ denotes the $weak^{*}$ closure of $A$ in $X^{**}$. Since $A$ also can be considered to be a bounded subset of $X^{**}$, my question is:$wk_{X^{**}}(A)\leq wk_{X}(A)$? Thank you!
 A: Suppose that $A$ is a bounded subset of $X$ and $X$ is a subspace of $Y$.  Then
\begin{equation}
wk_Y(A) \le wk_X(A) \le 2 wk_Y(A),\ \ \ (\#)
\end{equation}
and $2$ is the best constant in the right inequality. The choice $Y:=X^{**}$ gives the inequality you want. 
First, notice that $wk_X(A)$ is the supremum of the distance from $F$ to $X$ as $F$ ranges over the weak$^*$ closure of $A$ in $X^{**}$.  
The left inequality follows from the duality theory  that is taught in a beginning course in functional analysis.  You can identify $X^{**}$ with $X^{\perp\perp} \subset Y^{**}$. Under this identification, the  weak$^*$ topology on $X^{**}$ is the relativization of the weak$^*$ topology on $Y^{**}$ to $X^{\perp\perp}$. So under this identification, the  $ \text{weak}^*$ closure of $A$ in $X^{**} = X^{\perp\perp}$ is equal to the weak$^*$ closure of $A$ in $Y^{**}$! This makes the left inequality completely obvious. Observe that the proof for a general $Y$ becomes confusing if you specialize to  $Y=X^{**}$ because in that case you have two distinct copies of $X^{**}$ in $X^{(4)}$--itself and $X^{\perp\perp}$. 
To prove the right inequality in (#), take any $F$ in $X^{\perp\perp}\subset Y^{**}$ and any $y\in Y$ and set $d:= \|F-y\|$. We use a duality argument to estimate the distance from $y$ to $X$.  This distance is (arbitrarily close to) $\langle y^*, y \rangle $ for some norm one $y^* \in X^\perp \subset Y^*$. But since $F\in X^{\perp\perp}$, we have $\langle y^*, y \rangle = \langle y^*, y - F\rangle \le d$. So the distance from $F$ to $X$ is at most twice the distance from $F$ to $Y$, which gives the right inequality in (#).
To see that $2$ is the best constant in the right side of (#), let $A$ be the summing basis in $c_0$, set $X:=c_0$ and let $Y$ be $c$ or $c_0^{**} = \ell_\infty$.
