The map $A \to A'$ is a linear isometry from $B(X,Y) \to B(Y',X')$. Let $X$ and $Y$ be normed linear spaces. Show that the map $A \to A'$ is a linear isometry from $B(X,Y) \to B(Y',X')$.
$A'$ is the dual of $A$.
Need some hints for the problem.
 A: Recall that the norm on the dual of a normed vector space $X$ is given (for $L\in X^\star$) by:
$$ \Vert L \Vert_{X^\star} = \sup_{\Vert x \Vert_X \leq 1} \vert L(x) \vert.$$
Let
$$H: L(X,Y) \rightarrow L(Y^\star, X^\star), A \mapsto \left(A^\star: B \mapsto B\circ A\right).$$
One easily checks that this is a linear map. Using the definition of the operator norm and the norm on the dual of $X$, we get for $A\in L(X,Y)$
$$ \Vert H(A) \Vert_{L(Y^\star, X^\star)}  
= \sup_{\Vert B \Vert_{Y^\star}\leq 1, \ \Vert x \Vert_X \leq 1} \vert (B \circ A)(x) \vert \leq \Vert A \Vert_{L(X,Y)} .$$
Now use the Hahn-Banach Theorem zu construct a suitable sequence $(B_n)_{n\geq 1} \subseteq Y^\star$ to prove
$$ \Vert H (A) \Vert_{L(Y^\star, X^\star)} \geq \Vert A \Vert_{L(X,Y)}$$
and conclude that $H$ is an isometry.
A: Let $T:B(X,Y) \rightarrow B(Y',X')$ such that $T(A)=A'$
Hint
You can prove that the norm of the operator $A$ in $B(X,Y)$ equals  the norm of $A'$ in $B(Y',X')$.
For this you can use the Hahn-Banach Theorem.
