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

Question

Let $X$ be a Banach space and $Y$ be a closed subspace of $X$. For $1<p<\infty$ 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.

Answer 1

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.