Let $X, Y$ be Banach spaces, then any map $T:X\rightarrow Y^*$ admits a unique weak* to weak* continuous extension $\tilde{T}:X^{**} \rightarrow Y^{*}$ such that $\|\tilde{T}\|=\|T\|$.
I want to take $\tilde{T}=(T^*|Y)^*$, which is a map from$X^{**}$ to $Y^{**}$, but how to show it is a unique weak* to weak* continuous extension such that $\|\tilde{T}\|=\|T\|$.
I will denote by $i_X$ etc. the canonical inclusions into bidual $X\rightarrow X^{**}$.
Let $p\colon Y^{***}\rightarrow Y^*$ be defined as the transpose of $i_{Y}$.
Define $\tilde{T}=p\circ T^{**}\colon X^{**}\rightarrow Y^{***}\rightarrow Y^*$. It is weak$^*$-weak$^*$ continuous because p and $T^{**}$ are weak$^*$-weak$^*$ continuous. Next, for $x\in X,y\in Y$ we have $$ \langle\tilde{T}(i_X(x)),y\rangle=\langle T^{**}(i_X(x)),i_Y(y)\rangle=\langle i_X(x),T^*(i_Y(y))\rangle=\langle T^*(i_Y(y)),x\rangle= \langle Tx,y\rangle $$ which proves $\tilde{T}\circ i_X=id$.
Uniqueness of $\tilde{T}$ follows from the Goldstine theorem.