Let $A: B_1 \rightarrow B_2$ be a mapping, where $B_1$, $B_2$ are two Banach Spaces.
Then the adjoint of $A$, $A^*$, is defined by
$A^*:B_2^* \rightarrow B_1^*$ s.t for all $f \in B_1^*, f:B_1 \rightarrow \mathbb{R}$ is linear.
$\langle Ax{,}y\rangle$=$\langle x{,}A^*y\rangle$
My Question is: why when dealing with Hilbert Spaces $H_1$ and $H_2$, we simply say: IF $A:H_1 \rightarrow H_2$, THEN $A^*:H_2 \rightarrow H_1$, without taking into consideration $H_2^*$ and $H_1^*$ ?
The reason behind this is that every Hilbert space is isometrically isomorphic to its dual by the Riesz representation theorem, stating that any functional $\phi$ on a Hilbert space $H$ is of the form $\phi(\cdot)=\langle\cdot,z\rangle$ for some $z\in H$. Using this one-to-one correspondence between $H^*$ and $H$, one can see that if $A:H_1\to H_2$ is a bounded linear operator, then the "Hilbert space adjoint: $A^*:H_2\to H_1$ which is defined by $\langle Ax,y\rangle=\langle x,A^*y\rangle$ for all $x,y$ is "essentially the same" as the "Banach space adjoint: $A^*:H_2^*\to H_1^*$.
If $\sigma:H_1\xrightarrow{\cong}H_1^*$ and $\rho:H_2\xrightarrow{\cong}H_2^*$ are the isometric isomorphisms, then $\sigma^{-1}\circ A^{*,Ban}\circ\rho=A^{*,Hil}$.
The only essential difference of the two adjoints is that in the Hilbert setting we have that $(\lambda A)^*=\bar{\lambda}A^*$ while on the Banach setting we have that $(\lambda A)^*=\lambda\cdot A^*$.
Related: Hilbert space adjoint vs Banach space adjoint?