Passing complemented subspaces to duals

Question

Sorry for this rather basic question from Banach space theory.

Suppose I have a complemented subspace $E$ of a Banach space $X$.

So let's write $i:E\to X$ to be the inclusion map.

Then I have a quotient map $i^{*}:X^{*}\to E^{*}$ with kernel $E^{\perp}$.

Now that implies $E^{*} \cong X^{*}/E^{\perp}$, but is it necessarily true that $X^{*} \cong E^{*}\oplus E^{\perp}$?

Answer 1

You also have a continuous projection $p \colon X \to E$, with $p\circ i = \operatorname{id}_E$. Looking at the dual map $p^\ast \colon E^\ast \to X^\ast$, we see that $p^\ast$ is a continuous injection of $E^\ast$ into $X^\ast$, which by the closed range theorem has closed range. Since $i^\ast \circ p^\ast = \operatorname{id}_{E^\ast}$, it follows that $p^\ast(E^\ast) \cap E^\perp = \{0\}$. And $X^\ast = p^\ast(E^\ast) + E^\perp$, so indeed $X^\ast \cong p^\ast(E^\ast) \oplus E^\perp$ by the open mapping theorem. Since $p^\ast(E^\ast) \cong E^\ast$, we hence have indeed $X^\ast \cong E^\ast \oplus E^\perp$.

From a different angle, if we have an isomorphism $E \oplus F \cong X$, then taking duals, we get an isomorphism $X^\ast \cong E^\ast \times F^\ast$. Under this isomorphism, $F^\ast$ is identified with $E^\perp$.