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 quantity measuring weak non-compactness
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 $c\_0^{\*\*} = \ell\_\infty$.


