If $\pi$ and $\pi^*$ are dual projection operators in $E, E^*$, show that $Im ( \pi^*) = (ker\pi)^\perp$.

Question

Let $E,E^*$ be dual spaces($E^*$ doesn't have to be $L(E)$).Suppose $\pi: E \to E$ and $\pi^*: E^* \to E^*$ are dual mappings.Assume that $\pi$ is a projection operator in E. Prove that $\pi^*$ is a projection operator in $E^*$ and that
$$Im ( \pi^*) = (ker\pi)^\perp$$ and $$Im(\pi) = (ker \pi^*)^\perp$$

I have shown that $\pi^*$ is also a projection operator, but I have now idea how to show the following relations.

I would appreciate any help.

Note: The sign $A^\perp$ means the orthogonal space of A, i.e $$(ker\pi)^\perp = \{ x^* \in E^* | <x^*, x> = 0 \quad \forall x \in ker\pi \}$$

Note: We are not assuming neither $E$ nor $E^*$ is finite.

Answer 1

$\newcommand{\im}{\operatorname{Im}}$ $\newcommand{\ang}[1]{\langle #1\rangle}$ First, for a general linear operator $u:E\to E$,

$$(\im u)^\perp=\ker u^*;(\im u^*)^\perp=\ker u.$$

In fact, $$\begin{align*} (\im u)^\perp &=\{x^*\in E^*|\ang{x^*,ux} = 0,\forall x\in E\}\\ &=\{x^*\in E^*|\ang{u^*x^*,x}=0,\forall x\in E\}\\ &=\{x^*\in E^*|u^*x^*=0\}=\ker u^*. \end{align*} $$

Second, for a projection $\pi$,

$$(\im\pi)^{\perp\perp}=\im \pi.$$

Indeed, $E=\pi E\oplus (1-\pi)E,$

so $$\forall x\in (\pi E)^{\perp\perp},\quad x=y+z,$$ where $y\in \pi E$ and $z\in (1-\pi) E,$ $$0=(1-\pi)x=(1-\pi)y+(1-\pi)z=0+z=z$$ since $1-\pi\in (\pi E)^\perp$. Thus $x=y\in \pi E$, another direction is just by definition.

Therefore, $\im \pi=(\im\pi)^{\perp\perp}=(\ker u^*)^\perp.$
And the other is similary.

Answer 2

Let $y^* \in Im ( \pi^*)$. Then $y^*=\pi^*(x^*)$ for some $x^* \in E^*$. Then we have for $x \in ker \pi$:

$y^*(x)=x^*( \pi(x))=0$, hence $y^* \in (ker\pi)^\perp$, therefore

$Im ( \pi^*) \subseteq (ker\pi)^\perp$.

The rest is your turn !