The Adjoint of the adjoint of an operator I came across a relation in a book stating that the adjoint of the adjoint of an operator, is the operator back itself. So for instance, if an operator $\hat O$ has an adjoint $\hat O^\dagger$, how is it possible to show that $$(\hat O^\dagger)^\dagger = O$$ I tried computing this with the definition of the adjoint, $$\int dx \,\, \overline{f(t)}\,\,(\hat O^\dagger g(t)) \equiv \int dx \,\,(\overline{\hat O\,f(t)})\,\, g(t)$$
But got stuck and couldn't proceed. Is this the correct way to compute this?
 A: Given a vector space over ${\mathbb C}$ with a inner product $\langle,\rangle$ the adjoint $O^\dagger$ of an operator $O$ is defined by the requirement
$$
\langle u,O v\rangle=\langle O^\dagger u, v\rangle~. 
$$
Now, take a complex conjugate of both sides. Using $\langle u,v\rangle^* = \langle v,u\rangle$, we find
$$
\langle O v , u\rangle=\langle v , O^\dagger u\rangle~. 
$$
However, by definition
$$
\langle v , O^\dagger u\rangle = \langle (O^\dagger)^\dagger v , u\rangle~. 
$$
Thus,
$$
O =  (O^\dagger)^\dagger ~. 
$$
A: We can define the adjoint of an operator $A$ as being $A^\dagger$ such that,
$$\langle A f_1,f_2\rangle = \langle f_1,A^\dagger f_2 \rangle$$
with $\langle \cdot,\cdot\rangle$ being the inner product on the appropriate Hilbert space. Now, if we have $(A^\dagger)^\dagger$ then this is the adjoint of the operator $A^\dagger$ and thus,
$$\langle A^\dagger f_1,f_2 \rangle = \langle f_1,(A^\dagger)^\dagger f_2\rangle$$
is its defining property. Over the field of reals, we have that the inner product is symmetric, and so we can write,
$$\langle A^\dagger f_1,f_2 \rangle = \langle f_2, A^\dagger f_1\rangle = \langle Af_2,f_1\rangle.$$
This relies solely on the definition of the adjoint of $A$. We can then identify by comparing to the second equation that $A = (A^\dagger)^\dagger$ and the proof over $\mathbb C$ is similar. Note that there is a caveat to this property, namely I believe involutiveness is guaranteed for bounded operators.
