2

Here, $\Omega\subseteq\mathbb{R}^n$ is an open, bounded set (as nice as you want). It is well known that $L^1(\Omega)^* = L^\infty(\Omega)$. Moreover, it is also true that if we endow $L^\infty(\Omega)$ with the weak-* topology (let us denote this space $L^\infty_*(\Omega)$), we also have that $$L^\infty_*(\Omega)^* = L^1(\Omega)$$ (see for example here). In this way, these two spaces (one with the strong topology, and one with the weak-* topology) are mutually dual to each other - I think some authors such as Rockaferral refer in general to this situation as being "paired spaces", but I'm not completely sure of that.

Similarly, but more generally, if $X$ is a Banach space, it is also true that $L^1(\Omega;X)^*=L^\infty_*(\Omega;X^*)$, where $L^1(\Omega;X)$ is the (normed) space of Bochner integrable functions, and $L^\infty_*(\Omega;X^*)$ is the (normed) space of weakly-* measurable functions $g:\Omega\to X^*$ such that there exists a $c>0$ such that $|\langle g(\omega),x \rangle| \le c\|x\|_X$ for a.e. $\omega\in\Omega$ and all $x\in X$ (the norm here is given by the smallest $c$). As a reference, I have the Dinculeanu-Foias theorem (Theorem 2.2.12) of Gasinski-Papageorgiou

My question is: is it also true that (as in the standard case) $$ L^\infty_*(\Omega;X^*)^* = L^1(\Omega;X) $$ ?

Any hint would be very much appreciated!

rod
  • 905

1 Answers1

2

I think the star notation is a bit confusing. You should distinguish $L^\infty_* (\Omega):= (L^\infty(\Omega),w^*)$ which is $L^\infty$ with the weak-star topology and $L_*^\infty(\Omega,X^*)$ which is something completely different. I think the commonly used notation for the latter is either $L^\infty(\Omega, X^*_{w^*})$ or $L^\infty_w(\Omega,X^*)$.

If you are asking whether $$L^\infty(\Omega,X^*_{w^*})^*= L^1(\Omega,X)$$ then this is not true as one can see by the one dimensional example of $X=\mathbb R$ since $L^\infty(\Omega,\mathbb R^*_{w^*})= L^\infty(\Omega) $ and $L^\infty(\Omega)^* \neq L^1(\Omega)$.

Since $L^1(\Omega,X)^* = L^\infty(\Omega,X^*_{w^*}) $ you always have that $$ (L_*^\infty(\Omega,X^*_{w^*}))^*=L^1(\Omega,X).$$

Now, when $X^*$ has the RNP then $L^\infty(\Omega,X^*_{w^*})=L^\infty(\Omega,X^*) = L^1(\Omega,X)^*$ and so $$ L_*^\infty(\Omega,X^*)^* = L_*^\infty(\Omega,X^*_{w^*})^*=L^1(\Omega,X) .$$ It could be the case that the last equality is true iff $X^*$ has RNP but I am not sure at the moment.

Evangelopoulos Foivos
  • 4,429
  • Thank you very much for your nice answer! I feel like I am missing something: what is the difference between $L^\infty(\Omega;X^)$ and $L^\infty(\Omega;X^_{w^*})$ in your notation? – rod Sep 11 '24 at 09:53
  • 1
    @rod The former is the classic Bochner space of all strongly measurable $f \colon \Omega \to X^$ that are essentially bounded (i.e. $\inf { M>0 \colon |f(\omega)|_{X^} \le M \text { a.a. } \omega }<\infty$), whereas the latter is the space of all weak-star measurable $f \colon \Omega \to X^*$ such that $\langle f(\omega), x \rangle \le c |x|$ a.a $\omega$. You can find more details in Chapter 4.2 of the book Applied Nonlinear Functional Analysis An Introduction – Evangelopoulos Foivos Sep 11 '24 at 10:11
  • Ok, thanks again. It looks to me that the norms of these two spaces coincide, i.e. if $f\in L^\infty(\Omega;X^)\subseteq L^\infty(\Omega;X^{w^*})$, then $|f|{L^\infty(\Omega;X^)}=|f|_{L^\infty(\Omega;X^{w})}$. If that is the case, it should be $L^\infty(\Omega;X^)={f\in L^\infty(\Omega;X^*{w^*}):f\text{ is strongly measurable}}$. Does that make sense? – rod Sep 11 '24 at 12:12
  • 1
    You have to be careful as in general the null set depends on $x \in X$. In the case $X$ is separable then the null set in the definition of $f \in L^\infty(\Omega,X^_{w^})$ does not depend on $x$ and so you have that the two norms are in fact the same and what you wrote is correct. Example 4.2.31 in the book gives a counterexample in the non separable case. – Evangelopoulos Foivos Sep 11 '24 at 12:55
  • Ok, I read the example. For some reason, it's a picture that I struggle to keep in mind, so maybe my next comment is completely on a wrong track. Let us say that I have a specific $X$ such that if $f(\omega)\in X^*$, then $\langle f(\omega), \nu \rangle =0\ \forall \nu\in X$ implies that $f(\omega)=0$. Do I possibly gain something (such as for example the equality between those two norms in my previous comment)? – rod Sep 12 '24 at 14:18
  • You mean if $f$ is such that $\langle f(\omega), x \rangle =0$ for every $x \in X$ a.e. (again, a.e. depends on $x$) implies that $f =0$ a.e. (so a.e. does not actually depend on $x$) then maybe the two norms are the same? I am not sure I can think of something right now – Evangelopoulos Foivos Sep 12 '24 at 15:00
  • The notation "$f(\omega)$" was redundant as there is no need for a function $f$ - it was just to be consistent with your notation. I meant even something stronger: $X$ is such that for any $x^\in X^$, $\langle x^*,\nu\rangle=0\ \forall \nu\in X$ implies that $x=0$. – rod Sep 12 '24 at 15:04