Let $A$ be a bounded operator on a Hilbert space $H$ and $D_A$ its domain. We can define the following functional on $D_A$: $$f_\eta(\xi)=(\eta,A\xi).$$ Then we have that: $$||f_\eta(\xi)||=||(\eta,A\xi)||≤||\eta||⋅||A\xi||<\infty.$$ So $f_\eta$ is a bounded functional and hence it's continuous on its domain. Now we can use Riesz' theorem to say that:$$\exists! f^\eta: (f^\eta,\xi)=f_\eta(\xi)=(\eta,A\xi).$$ We define adjoint of $A$ the operator $A^♱$ such that: $$A^♱(\eta)=f^\eta.$$ My first question is: since $f_\eta$ is continuos for every $\eta$, and hence Riesz' theorem holds for every $\eta$ in $H$, can we say that the domain of $A^♱$ can be $H$ itself?
If this is the case, why can't we say that, if $(A,D_A)$ is symmetric, and hence bounded, we can always restrict $D_A^♱=H$ to $D_A$ so that $A $ is also self-adjoint?
I will add an example of my problem. Let $H=L_2([0,1])$ with $\mu(dx)=dx$. Let's consider: $$\hat{p}=-i\frac{d}{dx}.$$ $\hat{p}$ is symmetric when: $$D_A=\{ \psi \in L_2([0,1]): \psi \,\,\text{absolutely continuous}, \psi'\in L_2([0,1]), \psi(0)=\psi(1)=0 \}.$$ Lecturer told us that this operator is not self adjoint beacause the domain of $A^♱$ is: $$D_A^♱= \{ \psi \in L_2([0,1]): \psi \, \,\text{absolutely continuous}, \psi'\in L_2([0,1])\}$$ But this contraddicts what I thought was true: that $D_A^♱=H$.
The domain of the adjoint is not $H$ in general. It is restricted by the existence of the vector in Riesz's lemma in the following way $$D_{A^\dagger}=\{\psi \in H | \exists \eta \in H :\forall v \in D_A, \langle \psi, Av \rangle= \langle \eta, v \rangle \}$$ That is, not everywhere in $H$ does there exist such an $\eta$.
EDIT:
I had ignored the boundedness of the operator in question. Yes, because the operator $A$ is densely defined by continuity one may prove the domain of the adjoint is the entire space. As for your second question, that's exactly right, a bounded symmetric operator is self-adjoint because it coincides with the restriction of the adjoint to its (the operator's) domain