TITLE: Proof check about bounded operator
QUESTION [5 upvotes]: Let $X$ and $Y$ be Banach spaces, and fix a bounded linear operator $A \in \mathcal{B}(X, Y)$. Choose $\mu \in Y^{*}$,  and define a functional  $A^{*} \mu: X \rightarrow \mathbf{F}$  by $\left(A^{*} \mu\right)(x)=\mu(A x)$,  for $x \in X$. I want to show that the mapping $A^*\colon \mu \mapsto A^{*} \mu$ is a bounded linear mapping of $Y^{*}$ into $X^*$. Linear part is easy and my thought about bounded part is 
\begin{align}\|A^*\|&=\sup_{\|\mu\|=1}\|A^*\mu\|=\sup_{\|\mu\|=1}(\sup_{\|x\|=1}\|(A^*\mu)(x)\|)\\&=\sup_{\|\mu\|=1}(\sup_{\|x\|=1}\|\mu(Ax)\|)=\sup_{\|\mu\|=1}\|\mu A\|=\|A\|
\end{align}
Is that correct?  Also, is there a way to show $\|A\|=\|A^*\|$?Any help is appreciated.

REPLY [3 votes]: [The last part was added later and answer to that part is in my comments below]. 
The proof is OK if the last equality is replaced by an inequality :$ \|\mu (A) \|\leq \|A\|$ if $\|\mu \| \leq 1$. I don't think equality holds in general.