Let H be a Hilbert space and consider a linear bounded operator $A:H\to H$. The adjoint $A^*\in L(H) $ of A with $\forall x,y\in H:\; \langle Ax,y\rangle_H=\langle x,A^*y\rangle_H$ has the property: $(A^*)^*=A,\; A\in L(H)$.
Now let X be a Banach space, and take $A\in L(X)$. Consider $A^\prime\in L(X^\prime)$, the dual operator of A with $\forall x \in X,\; \forall y^\prime \in X^\prime,\; (A^\prime y^\prime )(x)=y^\prime (Ax)$.
Now, is $(A^\prime)^\prime=A,\; A\in L(X)$ true? I think $X$ has to be reflexive because $X$ reflexive means: $(X^\prime)^\prime$ is isometric-isomorphic to $X$.
For example, the adjoint of the left sift operator $\in L( \ell^1(\mathbb{N}))$ is the right shift operator $\in L( \ell^\infty(\mathbb{N}))$. If I want to know what happens if I try to dualize now the the right shift operator $\in L( \ell^\infty(\mathbb{N}))$, the first problem is $(\ell^\infty(\mathbb{N}))^\prime\neq \ell^1(\mathbb{N})$.
If this is correct, how can I argue more precisely, that $(A^\prime)^\prime=A,\; A\in L(X)$ is true if X is reflexive? Regards.
Suppose $A\in L(X),A'\in L(X')$ such that $A'$ is the dual, i.e. $\forall x,x'\;(A'x')(x)=x'(Ax)$. Then of course
$(Ax)(x')=x'(Ax)=(A'x')(x)=x(A'x')$,
and $A$ satisfies the property that defines the dual operator. Note that it can only work when $X$ is reflexive, since we assume that any element in $X''$ can be written as $x\in X$.