Show that $A^{\dagger^{\dagger}} = A $

Question

How do we show that $A^{\dagger^{\dagger}} = A $ without assuming $A$ to be a explicit matrix. That is, given a linear operator $A$, let us define $A^\dagger$ to be a unique operator such that $\langle A^\dagger u \mid v \rangle = \langle u\mid A v \rangle $ (standard inner product in Hilbert space) for any vectors $u$ and $v$. How can we prove that using this definition alone.

Answer 1

Per the definition, $A^{\dagger\dagger}$ is the unique linear operator such that $$\langle A^{\dagger\dagger}u\mid v\rangle = \langle u\mid A^\dagger v\rangle = \overline{\langle A^\dagger v\mid u\rangle} = \overline{\langle v\mid Au\rangle} = \overline{\overline{\langle Au\mid v\rangle}} = \langle Au\mid v\rangle$$ for all vectors $u$ and $v$, so...