Let $X$ be a normed linear space and let $N$ be a weak*-closed subspace of $X^*$. How to show that there exists a normed linear space $Y$ and $T\in B(X,Y)$ such that $T^*(Y^*)=N$?
I feel that if $T$ is defined in such a way that $(\text{ker }T)^{\perp}=N$, then $N=T^*(Y^*)$. But how to define such a $T$?
Let $N_0\subset X$ be the subspace $N_0=\{x\in X:\ \varphi(x)=0,\ \forall\varphi\in N\}$. Define $Y=X/N_0$, and let $T:X\to Y$ be the quotient map. Then $\ker T=N_0$. So $$\tag1 \overline{\operatorname{ran}} T^*=(\ker T)^\perp=N_0^\perp=\{\psi\in X^*:\ \psi(x)=0,\ \forall x\in N_0\}. $$ We have that $N\subset N_0^\perp$. If the inclusion were proper, we would have nonzero $\psi\in X^*\setminus N$ such that $\psi(x)=0$ for all $x\in N_0$. The fact that $N$ is weak$^*$ closed allows us to separate in the predual (because the dual of $X^*$ with weak$^*$ topology is $X$); that is, there exists $z\in X$ with $\psi(z)=1$ and $z|_N=0$, that is $\varphi(z)=0$ for all $\varphi\in N$. But then $z\in N_0$, and we get a contradiction. So $N=N_0^\perp$.
Finally, note that $T$, being a quotient map, has closed range (since its onto). Then $T^*$ also has closed range. Thus $(1)$ becomes $$\tag2 \operatorname{ran}T^*=N. $$