The Hermitian conjugate of an operator $A:H_1\rightarrow H_2$ between Hilbert spaces is the linear map $A^\dagger:H_2\rightarrow H_1$ such that $$ (A^\dagger h_1,h_2)_{H_2} = (h_1,A h_2)_{H_1} $$
I am now trying to write this in bra-ket notation ($H_1=H_2=$space of states): $$ (A^\dagger \vert \psi\rangle,\vert \phi \rangle) = (\vert \psi\rangle,A \vert \phi \rangle) \\ ?=\langle\psi\vert A \vert\phi\rangle $$ I do not know how to write the LHS. I expected to find something like $$ \langle \psi\vert A^\dagger\vert\phi\rangle=\langle\phi\vert A\vert\psi\rangle^* $$ Which is the definition given in physics books.
Mathematically if $\lvert \phi \rangle \in H$ is your ket, then the bra $\langle \phi \rvert$ is an element of $H^*$, the dual of $H$ defined by $$ \langle \phi \rvert:H \to \mathbb{C}: \lvert \psi \rangle \mapsto \langle \phi \rvert \psi \rangle= (\lvert \phi \rangle, \lvert \psi \rangle). $$
Then we find $$ \langle \psi \rvert A^\dagger \lvert \phi \rangle = (\lvert \psi \rangle, A^\dagger \lvert \phi \rangle) = (A\lvert \psi \rangle, \lvert \phi \rangle) = (\lvert \phi \rangle, A\lvert \psi \rangle)^* = \langle \phi \rvert A \lvert \psi \rangle^*. $$