As far as I understand, for linear operators on bra vectors,
Let $H = L_2 (\mathbb{R})$, and $\phi \in H^*$ and $f \in H$ be arbitrary, and let $\alpha : H \to H$ be a linear map.
Then for any $\alpha$, we have an induced dual map $\alpha^*: H^* \to H^*$ such that
$$\langle\phi, \alpha (f) \rangle = \langle\alpha^* (\phi), f\rangle \quad \forall \phi \in H^*, \forall f \in H,$$ where $\langle,\rangle$ is the non-degenerate bilinear map between $H$ and $H^*$.
However, the quote from Dirac's book on Quantum Mechanic:
"ii) The eigenvalues associated with eigenkets are the same as the eigenvalues associated with eigenbras.
iii) The conjugate imaginary (dual map) of any eigenket is an eigenbra belonging to the same eigenvalue, and conversely."
and with the above definition that I gave, I get the following result:
Let $\lambda$ be an eigenvalue associated with $f$ of $\alpha$.Then we have
$$\langle\alpha^* (\phi), f\rangle = \langle\phi, \alpha (f) \rangle = \langle\phi, \lambda f\rangle \Rightarrow \\ \langle\alpha^* (\phi), f\rangle = \langle\phi, \lambda f\rangle = \langle\lambda \phi, f\rangle \Rightarrow \\ \alpha^* (\phi) = \lambda \phi \quad \forall \phi \in H^*$$
Even though, this result does not conflict with the first condition that Dirac gave, it conflicts with the second one because with this result, if $\lambda$ is an eigenvalue for $\psi$ which is the corresponding bra vector of the ket $g$, then g has also have the eigenvalue $\lambda$, and since $\psi$ is arbitrary, $\alpha$ cannot have more than one eigenvalue on $H$.
So my question is that is there any flow in my derivation and the above definition, (I mean there should be), if not, what am I missing ?
Edit:
We consider $H = L_2 (\mathbb{R})$ as a complex vector space.
Edit 2:
The bilinear form $$\langle \psi, f\rangle = (g, f) = \int \bar g f dx,$$ where $\psi$ is the eigenket of $g$, and $(,)$ is the inner product defined on $H$.
Edit 3:
I have realised my mistake, which is after the line $$ \langle\alpha^* (\phi), f\rangle = \langle\phi, \lambda f\rangle = \langle\lambda \phi, f\rangle $$ because f is not arbitrary, it is fixed according to $\lambda$, so we cannot assume the next step, and we are left with
$$<\alpha^* (\phi) - \lambda \phi, f > = 0.$$
But, now how can we derive what Dirac says ?