Suppose $ \pi: E \rightarrow E $ and $ \pi^{*}: E^{*} \rightarrow E^{*} $ are dual mappings. Assume that $ \pi $ is a projection operator in $ E $. Prove that $ \pi^{*} $ is a projection operator in $ E^{*} $.
Definition: suppose that $E,E^*$ and $F,F^*$ are two pair of dual spaces and $\varphi : E \rightarrow F $, $\varphi^*: E^* \leftarrow F^*$ are linear mappings. The mappings $\varphi $ and $\varphi^*$ are called dual if $\langle y^* , \varphi (x) \rangle = \langle \varphi^*(y^*) , x \rangle$, $y^* \in F^*$, $x \in E $. (Recall that $\langle x,y \rangle$ is called the scalar product of $x $ and $y$).
Definition: A linear transfomation $\varphi :E \rightarrow E $ is called a projection operator in $E $, if $\varphi^2 = \varphi $
This exercise seemed interesting to me and I want to find a proof. I must show that $ {\pi^ *}^2 = \pi^*$. I have that
\begin{equation}
\begin{split}
\langle y^*, \pi(x) \rangle &= \langle y^*, \pi^2(x) \rangle &= \langle \phi^*(y^*), x \rangle
\end {split}
\end{equation}
$y^{*} \in E^{*}, x \in E$, by definition, but could not obtain the conclusion. Could you give me a suggestion? Please.
Perhaps the fact should be used that if $\varphi $ is a projection operator in $E $, then $E = \ker\varphi \oplus \rm im\varphi$
Take any $x \in E$ and any $y^* \in E^*$ we have
$$\begin{aligned} \langle \left(\left(\pi^*\right)^2 - \pi^*\right)(y^*), x \rangle & = \langle (\pi^* \circ \pi^*)(y^*), x \rangle-\langle \pi^*(y^*), x \rangle\\ &=\langle \pi^*(y^*), \pi(x) \rangle-\langle y^*, \pi(x) \rangle\\ &=\langle y^*, (\pi \circ \pi)(x) \rangle-\langle y^*, \pi(x) \rangle\\ &=\langle y^*, (\pi \circ \pi)(x) - \pi(x)\rangle\\ &=0 \end{aligned}$$
Because $\pi$ is a projection operator. This allows to conclude that $\pi^*$ is itself also a projection operator.