Complemented set in X*** Let X be a normed space, then prove that the image of $J:X^* \rightarrow X^{***}$ is a complemented subset of $X^{***}$
My attempt:
I know that the image of the canonical embedding will be closed in $X^{***}$ but I can't see how to show it's a complemented subspace, though of proving that the codimension of $Im(J)$ has finite dimension.
 A: I'll call $J=J_{X^{*}}:X^{*}\to X^{***}$, which is given by:
$$ J_{X^{*}}(f)(\varphi)=\varphi(f), $$
for all $f\in X^{*}$ and all $\varphi \in X^{**}$.
Since, as you said, the image $J_{X^{*}}(X^{*})$ is closed in $X^{***}$, the ideia is to find
$$P:X^{***}\to X^{***}$$ linear and bounded such that $P^{2}=P$ and $P(X^{***})=J_{X^{*}}(X^{*})$.
Given $J_{X}:X\to X^{**}$ the canonical evaluation map
$$J_{X}(x)(f)=f(x),$$
for all $x\in X$ and all $f\in X^{*}$, you can consider its dual transform, $(J_{X})^{*}:X^{***}\to X^{*}$, defined in this question. Then for this case, this means,
$$ (J_{X})^{*}(h)(x)=h(J_{X}(x)),$$
for all $h\in X^{***}$ and all $x\in X$.
Now, choose $P=J_{X^{*}} \circ (J_{X})^{*}:X^{***}\to X^{***}$, which is linear and bounded. (Why?)
Then, $P^{2}=J_{X^{*}} \circ (J_{X})^{*} \circ J_{X^{*}} \circ (J_{X})^{*}= J_{X^{*}} \circ Id_{X^{*}} \circ (J_{X})^{*}= J_{X^{*}} \circ (J_{X})^{*}=P$. Here I used that $(J_{X})^{*} \circ J_{X^{*}}= Id_{X^{*}}$, which you can prove using the definitions, please check it.
Also, $P(X^{***})= J_{X^{*}}[(J_{X})^{*}(X^{***})]= J_{X^{*}}(X^{*})$. Here, I used that $[(J_{X})^{*}(X^{***})]=X^{*}$. It is also straightforward to prove, using $J_{X^{*}}= Id_{X^{*}}$.
And now we are done. If you need more details let me know.
