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This may be basic, and if it is I apologize, but I have found no references to it in literature. I would appreciate a reference at least if I am wrong. I have supplied background for those interested, but you are more than welcome to skip it - I have indicated where it begins and ends.

BACKGROUND STARTS HERE

I am currently dealing with certain questions regarding, in very broad terms, the behavior of duals and preduals of Banach spaces, especially trying to establish simple ways to test for the existence of a predual. One class of Banach spaces for which there is a handy iff statement on when they are dual spaces is the family of spaces of the form $C(X)$, $X$ a compact Hausdorff spaces: a predual exists if and only if $X$ is hyperstonean, i.e. extremely disconnected (the closure of any open set is open) and having sufficiently many normal measures (in a sense which is better left described elsewhere, e.g. vol. I of Takesaki's Theory of Operator Algebras).

Cumbersome as this description is, the proof is more cumbersome still, and indeed, it doesn't seem right for a discussion of hyperstonean spaces to be needed to prove that $C([0,1])$ has no predual* (this seems like a silly objection, but bear with me if you have so far). Fortunately, it is not needed; in fact, it can proved that for a compact metric space $K$, $C(K)$ never has a predual, by combining the two facts that 1) $c_0$ embeds isometrically into any such space and 2) $c_0$ does not embed isometrically into any separable space which has a predual.**

*Note that the standard proof based on Krein-Milman only works in the real case; here we are dealing with complex-valued continuous functions.

**Not that this fact is so trivial - in fact, I have not found a proof of it which doesn't go through things like points of weak-to-norm continuity and the like.

All good and well; however, these methods outlive their usefulness quickly, as they immediately break down for even spaces of the form $C^k([a,b])$. However, we know quite well that the latter space can be identified with the direct sum $\mathbb{C}^k \oplus C([a,b])$ - as such, due to this and other examples, it is useful to know whether the presence of a non-dual factor in a direct sum necessarily makes the direct sum itself non-dual.

BACKGROUND ENDS HERE

Question: Let $X$ and $Y$ be Banach spaces such that the direct sum $X \oplus Y$ has a predual $V$. Is it true that $V$ must be of the form $V_X \oplus V_Y$, where $V_X$ and $V_Y$ are preduals of $X$ and $Y$, respectively? In other words: is $X \oplus Y$ a dual space iff $X$ and $Y$ are dual spaces?

I actually have something of an idea for a proof. Suppose $V$ is a predual for $X \oplus Y$. Define subspaces of $V$:

$ V_X = \left \{v \in V | y(v) = 0 \forall y \in Y \right \} $

$ V_Y = \left \{v \in V | x(v) = 0 \forall x \in X \right \} $

These are our candidates for the decomposition, and they at least have the virute of intersecting trivially. It is only left to show that any element of $V$ is the sum of elements from these subspaces; but $V$ naturally embeds into the direct sum $X^* \oplus Y^*$, so we already have a decomposition for $v \in V$ in that space:

$v = v_x + v_y, v_x \in X^*, v_y \in Y^*$

Thus, all that remains to show is that both $v_x$ and $v_y$ are elements of $V$, which is where I am currently stuck.

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up vote 5 down vote accepted

The answer is no as $X$ need not have a predual. Indeed, take a space $X$ which is compelemnted in its bidual but not isomorphic to a dual space (e.g. $X=L_1$) and consider $X^{**} = X \oplus Y$. You can produce even more weird examples: $\ell_1$ (which has lots of projections) has preduals which are hereditarily indecomposable (so only finite/cofinite-dimensional subspaces are complemented).

It seems that you are interested in von Neumann algebras, so it need not hold even when your sum is isomorphic to $\ell_\infty(\Gamma)$ as there exist injective $C(K)$-spaces which are not isomorphic to dual spaces (Rosenthal). Take such a space and embed it into $\ell_\infty(\Gamma)$ for $\Gamma$ big enough.

Moral: there is absolutely no reason for which the projections onto $X$ and $Y$ should be weak*-to-weak* continuous in general.

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I was actually not thinking about Von-Neumann algebras in this context, but either way this was very helpful - thanks! –  Super-Measurable Analyst Jun 12 at 23:04
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