On Riesz Representation and containment of the $L^p$ spaces when $E$ has finite measure

Question

Let $E \subseteq \mathbb R$ have finite measure, and let $p_1,p_2 \in (1,\infty)$ with $p_1 < p_2$. We have $L^{p_2}(E) \subseteq L^{p_1}(E)$; let $i: L^{p_2}(E) \hookrightarrow L^{p_1}(E)$ denote the inclusion map. The dual map $i^*: L^{p_1}(E)^* \to L^{p_2}(E)^*$ is given by $$i^*(\phi) = \phi \circ i = \phi|_{L^{p_2}(E)}$$ By the Hahn-Banach Theorem, every bounded linear functional on a linear subspace of a normed linear space extends to a bounded linear functional on the entire space; therefore $i^*$ is a surjective map. Now let $q_1,q_2$ denote the respective conjugates of $p_1,p_2$, defined by $\frac1{p_i} + \frac1{q_i} = 1$. We have $q_2 < q_1$, and so $L^{q_1}(E) \subseteq L^{q_2}(E)$; let $j$ denote the inclusion map. Let $R_1:L^{q_1}(E) \cong L^{p_1}(E)^*$ denote the isomorphism afforded by the Riesz Representation Theorem; namely $R_1(g) = \left(L^{p_1}(E) \ni f \mapsto \int_E fg\right)$, and let $R_2$ be defined similarly. We have the following diagram:

$\require{AMScd}$ \begin{CD} L^{p_1}(E)^* @>i^*>> L^{p_2}(E)^*\\ @| @| \\ L^{q_1}(E) @>>j> L^{q_2}(E) \end{CD}

where the double vertical bars indicate the respective isomorphisms.

Q: Does this diagram commmute?

On the one hand, direct verification seems to suuggest that this is indeed the case: let $g \in L^{q_1}(E)$, then $(f \mapsto \int_E fg)$ is the linear functional $R_1(g)$ on $L^{p_1}(E)$; restricting this to $L^{p_2}(E)$ gives a linear functional on $L^{p_2}(E)$ which is given by integration against an $L^{q_2}(E)$ function (since $g \in L^{q_2}(E)$ also), and by Riesz Representation there is only one $g' \in L^{q_2}(E)$ which accomplishes this, namely $g' = g$ itself. Therefore, one would conclude that the diagram indeed commutes.

On the other hand, $R_2^{-1} \circ i^* \circ R_1$ is a composition of surjective maps and thus surjective, while the containment $L^{q_1}(E) \subseteq L^{q_2}(E)$ is strict; therefore it seems that the diagram can't possibly commute!

What is going on here?

Answer 1

$i^{\ast}$ is not surjective. Let $(X,\Vert\cdot\Vert_{X})$ be a normed space. Hahn-Banach theorem says that if $Y$ is a subspace of $X$ and $T:Y\rightarrow Z$ is linear and continuous with respect to the norm $\Vert\cdot\Vert_{X}$, then you can extend $T$ to a linear continuous function $T_{1}:X\rightarrow Z$. In this case given $T\in L^{p_{2}}(E)^{\ast}$, we have that $T:L^{p_{2}% }(E)\rightarrow\mathbb{R}$ is linear but it is continuous with respect to the norm in $\Vert\cdot\Vert_{L^{p_{2}}(E)}$ but not with respect to the norm in $L^{p_{1}}(E)$. So even if $L^{p_{2}}(E)\subset L^{p_{1}}(E)$ we cannot use Hahn-Banach to extend $T$ to an element of $L^{p_{1}}(E)^{\ast}$. More concretely, since $q_{2}

Answer 2

I've figured it out. The problem is a misapplication of the Hahn-Banach Theorem. This is actually really subtle. The problem is that we have a subspace with a different norm than the ambient space, namely the $||\cdot||_{p_2}$ norm on $L^{p_2}(E)$, compared with the $||\cdot||_{p_1}$ norm on $L^{p_1}(E)$. Any functional $T \in L^{p_2}(E)^*$ is bounded with respect to the $p_2$-norm, but it may not be bounded with respect to the $p_1$-norm on $L^{p_2}(E)$. For Hahn-Banach to apply we need our functional to be bounded with respect to the norm induced by the ambient space. This kind of functional may not be extended to be a functional on the entirety of $L^{p_1}(E)$, and so the surjectivity argument fails. If it were not for this subtlety, and for example if we could suppose that every $||\cdot||_{p_2}$-bounded functional was also $||\cdot||_{p_1}$-bounded, then Hahn-Banach would indeed apply and we could say that every functional in $L^{p_2}(E)$ could be extended to a functional in $L^{p_1}(E)$, and thus that every functional in $L^{p_2}(E)^*$ was the restriction of a functional in $L^{p_1}(E)$, implying surjectivity of $i^*$.