# $L_p(I,Y)^\perp=L_q(I,Y^\perp)$?

Let $$X$$ be a Banach space and $$Y$$ be a closed subspace of $$X$$. For $$1 consider the $$p$$-th power Bochner Integrable functions which takes values in $$X$$ and defined on the unit interval $$[0,1]$$, denoted by $$L_p(I,X)$$. It is clear that $$L_p(I,Y)$$ is a closed subspace of $$L_p(I,X)$$. It is well-known that if $$X^*$$ has RNP then $$L_p(I,X)^*=L_q(I,X^*)$$. My question is whether $$L_p(I,Y)^\perp=L_q(I,Y^\perp)$$?

My observations are the following:

Let $$X$$ be an Asplund space then for any $$Y\subseteq X$$, is also Asplund. Now since $$(X/Y)^* = Y^\perp$$, $$Y^\perp$$ has the Radon Nikodym Property and hence $$L_p(I,X/Y)^* = L_q(I,Y^\bot)$$. On the other hand $$(L_p(I,X)/L_p(I,Y))^* = (L_p(I,Y))^\perp (\subseteq L_q(I, X^*))$$.

Now suppose further, $$X$$ is reflexive, hence all the spaces involved will be reflexive.

So if the entities in the LHS quantities are the same, by uniqueness of preduels of reflexive spaces, we get:$$L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$$.

In order to prove the desired result it is enough to prove: $$L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$$. It is still unknown to me whether $$L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$$ is true even if $$X$$ is reflexive.

• One always has $L_q(I, Y^\bot) \subset L_p(I, Y)^\bot$. If $X$ is reflexive and separable, you might use a Hahn-Banach argument to show equality. Commented Jan 26, 2020 at 20:44

Let's consider the following situation: $$E$$ and $$F$$ are Banach spaces, $$D\subset E$$ is a dense subspace, $$Q: E \to F$$ is a continuous linear operator, and its restriction $$Q_0$$ to $$D$$ is a quotient map onto its range $$R$$, i.e., $$Q_0$$ takes the open unit ball of $$D$$ to the open unit ball of $$R$$. Suppose $$R$$ is dense in $$F$$.
If the kernel of $$Q_0$$ is dense in the kernel of $$Q$$, then $$x+\ker Q_0 \mapsto x+\ker Q$$ is an isometry. Hence the canonical isometry $$D/\ker Q_0 \cong R$$ extends to an isometry $$E/\ker Q\cong F$$.
This applies in the notation of the question to $$E=L_p(X)$$, $$F=L_p(X/Y)$$, $$Q(f)=q\circ f$$ with $$q:X\to X/Y$$ the canonical map, and $$D=$$ the space of $$X$$-valued step functions.