The third dual can be "decomposed" with the dual and his annihilator My teacher says that is an easy exercise to see that $X^{***}= X^* \bigoplus J(X)^\bot$. That is, $X^*$ is complemented in $X^{***}$ and $J(X)^\bot$ is the topological complement. 
Here $J$ is the canonical James map $J:X\to X^{**}$.
( I've already read other similar questions on this forum, but they did not help me ) 
 A: Note that we should be careful with technicalities. For the statement to make strict sense we actually want to prove that, for the canonical embedding $J_*: X^* \rightarrow X^{***}$, $J_*(X^*)$ is complemented in $X^{***}$.
Let us consider the map $\Phi:X^{***}\to X^{***}$ defined by $\Phi=J_*\circ J^*$. Here $J^*$ is the adjoint of $J$. It is a simple exercise to prove that $\Phi$ is a projection onto $J_*(X^*)$, that is, $\Phi$ is continuous and $\Phi^2 = \Phi$. We thus have $X^{***}=X^*\oplus_A(1-\Phi)X^{***}$ (Technically here we are identifying $X^*$ with $J_*(X^*)$). Here subscript $A$ denotes the algebraic direct sum, because we do not yet know that this is a topological complement. (If you know that a continuous projection gives rise to a topological complement with this decomposition then you actually do, but you do not need this knowledge for this proof.)
However, if we can show that $J(X)^\perp=(1-\Phi)X^{***}$ we will be done, because we know that the annihilator of a subspace is closed. Now $f\in(1-\Phi)X^{***}$ if and only if $f=g-\Phi g$ for some $g\in X^{***}$. We note that $\Phi g$ is the functional $\Phi_g$ satisfying $\Phi_g x=g(x)$ for all $x\in J(X).$ Thus if $x\in J(X)$ we have $f(x)=g(x)-\Phi_g x=0$. Conversely, if $h\in X^{***}$ is such that $h(x)=0$ for all $x\in J(X)$, we can write $h=h-\Phi h$. Thus $J(X)^\perp=(1-\Phi)X^{***}$ and we are done.
