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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!

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    $\begingroup$ You ask some interesting questions, Dongyang, but rarely give any indication of why you are interested in the question or what information you have about it. You would probably get more "action" if you would take the time to give this information. $\endgroup$ Jan 4, 2016 at 20:22
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    $\begingroup$ Thanks for your helpful suggestions, Bill. I'll do it later. $\endgroup$ Jan 4, 2016 at 21:37

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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 $c_0^{**} = \ell_\infty$.

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  • $\begingroup$ if we take $X=c_{0},Y=l_{\infty}$ and $A$ be the summing basis of $c_{0}$, then the weak* closure of $A$ is $A\cup \{e_{0}\},$ where $e_{0}=(1,1,1,...)$. So $wk_{Y}(A)=0$ and $wk_{X}(A)=d(e_{0},c_{0})>0$.So it seems that the right inequality is not true. $\endgroup$ Jan 5, 2016 at 16:55
  • $\begingroup$ Wrong, Dongyang. You are confused by the two ways of embedding $X$ into $X^{(4)}$ I mentioned in my answer. Just think about $\ell_\infty$ as being a space that contains $c_0$ rather than as being $c_0^{**}$ and you'll see. In general, if $X$ is a closed subspace of $Y$, then the intersection of the weak$^*$ closure of $X$ in $Y^{**}$ with $Y$ is just $X$ itself. $\endgroup$ Jan 5, 2016 at 17:22
  • $\begingroup$ since the weak* clousre of $A$ in $c_{0}^{**}$ is equal to the weak* closure of $A$ in $l_{\infty}^{**}$, the weak* closure of $A$ in $l_{\infty}^{**}$ is also equal to $ A\cup \{e_{0}\}$. Thus $wk_{Y}(A)=0$. I am not sure that $wk_{Y}(A)=0$. If this is false, could you tell me what is $wk_{Y}(A)$? $\endgroup$ Jan 5, 2016 at 17:41
  • $\begingroup$ Certainly not! The weak$^*$ topology on $Y^{**}$, relativized to $Y$, is the weak topology on $Y$. Apply this to $Y=\ell_\infty$. That $\ell_\infty$ is a dual space is irrelevant. $\endgroup$ Jan 5, 2016 at 18:51
  • $\begingroup$ let $i:c_{0}\rightarrow l_{\infty}$ be the inclusion map and $\epsilon>0$, my question: $d(e_{0},c_{0})>(2-\epsilon)d(i^{**}e_{0},l_{\infty})$? where $e_{0}=(1,1,1,...)$. $\endgroup$ Jan 6, 2016 at 6:37

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