Hello!
Like Mr Yuan suggested, call the first one 'dual' and write $f^\*$$f^\ast$ and the second one adjoint and write $f^\dagger$. Then a fairy simple calculation shows, that $f^\*$$f^\ast$ and $f^\dagger$ are closely related to each other:
Let $i: X \to X^\*$$i: X \to X^\ast$ and $j: Y \to Y^\*$$j: Y \to Y^\ast$ be the operators coming from the Riesz's represantation theorem. Then for any $y' \in Y^\*$$y' \in Y^\ast$ and $x \in X$ there holds:
$\langle j^{-1}\cdot y', f \cdot x\rangle = \langle f^\dagger \cdot j^{-1} \cdot y, x\rangle$.
On the right hand side we have: $\langle f^\dagger \cdot j^{-1} \cdot y', x\rangle = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$,
while on the right hand side there is: $\langle j^{-1} \cdot y', f \cdot x \rangle = y' \cdot f\cdot x = f^\* \cdot y' \cdot x$$\langle j^{-1} \cdot y', f \cdot x \rangle = y' \cdot f\cdot x = f^\ast \cdot y' \cdot x$
That for we get: $f^\* \cdot y' \cdot x = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$$f^\ast \cdot y' \cdot x = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$. Since this holds for all $x \in X$, there must be $f^\* \cdot y' = i \cdot f^\dagger \cdot j^{-1} y'$$f^\ast \cdot y' = i \cdot f^\dagger \cdot j^{-1} y'$ for all $y' \in Y^\*$$y' \in Y^\ast$ and we can conclude, that
$f^\* = i\cdot f^\dagger \cdot j^{-1}$$f^\ast = i\cdot f^\dagger \cdot j^{-1}$.
If you don't destinguish between $X$ and $X^\*$$X^\ast$ and $Y$ and $Y^\*$$Y^\ast$ respectively, then $f^\* = f^\dagger$$f^\ast = f^\dagger$.
Kind regards Konstantin
