Dual space and conditions for weak convergence in Orlicz Space not having $\Delta_{2}$ property

Question

I am interested in conditions for weak convergence on Orlicz spaces where the corresponding Young function, $\Phi:[0,\infty) \rightarrow [0,\infty)$, does not have the $\Delta_{2}$ condition, i.e. does not satisfy the following inequality: $$ \displaystyle \sup_{t \in [0,\infty)} \dfrac{\Phi(2t)}{\Phi(t)} < \infty. $$ I have been attempting to apply the following theorem for Kosaku Yosida's book, Functional Analysis (MR617913, Zbl 0435.46002, see Chapter 5, Section 1 Theorem 3 on page 121):

Theorem 3. A sequence in a normed linear space $X$ converges weakly to an element $x_{\infty} \in X$ iff the following two conditions are satisfied:

1. $\sup_{n} \|x_{n}\| < \infty$, and
2. $\lim_{n \rightarrow \infty} f(x_{n}) = f(x_{\infty})$ for every $f$ from any strongly dense subset $D^{*}$ of $X^{*}$ (where $X^{*}$ denotes the dual of $X$).

So suppose I have a Orlicz Space $L^{\Phi}(\mu)$ where $(\mathcal{X}, \mu)$ is a finite measure space and $\Phi$ is a Young function without the $\Delta_{2}$ property. In particular, I am interested in the Young function $\Phi(t) = e^{|t|} - 1$. Let $\|x_{n}\|$ be a sequence in $L^{\Phi}(\mu)$ satisfying condition i.): $\sup_{n} \|x_{n}\| < \infty$.

To finish showing weak convergence, I would then need to show that condition ii.) holds. Theorem 4 and its proof (see pages 121-122) give an approach on how to do this on the space $L^{1}(\mathcal{X},\mu)$. We define an appropriate set function as a limit: $\displaystyle \psi(B) = \lim_{n \rightarrow \infty} \int_{B} x_{n}(s) \mu(ds)$, use the Lebesgue-Nikodym theorem to show there exists $x_{\infty} \in L^{1}(\mathcal{X},\mu)$ such that $$ \lim_{n \rightarrow \infty} \int_{B} x_{n}(s) \mu(ds) = \int_{B} x_{\infty}(s) \mu(ds) \text{ for all } B, $$ and it follows that this equation holds for any simple function $g(s) = \sum_{j=1}^{k} \alpha_{j} \chi_{B_{j}}(s)$ ($\chi_{B_{j}}$ denoting the characteristic function): $$ \lim_{n \rightarrow \infty} \int_{B} x_{n}(s) g(s) \mu(ds) = \int_{B} x_{\infty}(s) g(s) \mu(ds). $$ Since simple functions are dense in $L^{\infty}(\mu) = (L^{1}(\mu))^{*}$ the second condition of Theorem 3 is satisfied.

It is this last step that seems to breakdown for Orlicz spaces without the $\Delta_{2}$ condition. The main hurdle is that, since $\Phi$ is not $\Delta_{2}$, its dual is not straightforwards. For $\Phi(t) = e^{|t|} - 1$, looking at Corollary 12 on page 124 of Rao & Ren's "Theory of Orlicz Spaces", elements of the dual are of the form $$ y^{*}(f) = \int f(s)h(s) \mu(ds) + \int f(s) \nu_{1} $$ where $h \in L^{\Psi}(\mu)$, the complementary Orlicz space, and $\nu_{1} \in B_{\Psi}$, where $B_{\Psi}$ is the set of finitely additive scalar set functions vanishing on $\mu$-null sets and with support contained in some $f \in L^{\Phi}(\mu) - \mathcal{M}^{\Phi}$. Note $\mathcal{M}^{\Phi}$ is the closed span of all step functions from $L^{\Phi}(\mu)$ and in the case of $\Phi(t) = e^{|t|}-1$, $\mathcal{M}^{\Phi} = M^{\Phi} = \left \{ f \in L^{\Phi}(\mu): \int \Phi(\alpha f) \mu < \infty \text{ for all } \alpha > 0 \right \}$ (see Rao & Ren, Chapter 3.4, Proposition 3, page 75).

Questions/observations:

1. Are there known results for weak convergence involving non-$\Delta_{2}$ Orlilcz spaces? Ideally results that get around the more difficult dual spaces present under non-$\Delta_{2}$ conditions.
2. Are there references (besides Rao & Ren) that deal with the dual spaces of non-$\Delta_{2}$ Orlicz spaces?
3. Are there references detailing the role of the $\nu_{1}$ term: $\int f(s) \nu_{1} $ in the decomposition of an element of $(L^{\Phi}(\mu))^{*}$?
4. I believe (but have not proved or found a reference) that the dual of $\Phi(t) = e^{|t|}-1$ is $\Psi(t) = t\log^{+}(t)$, where $\log^{+}(t) = \max(0,\log(t))$. Since this is a $\Delta_{2}$ function, simple functions are dense in $L^{\Psi}(\mu)$ (see "Stopping Times and Directed Processes" page 47 Corollary 2.1.18 or "Theory of Orlicz Spaces by Rao & Ren, page 77 Corollary 5). Is this somehow enough to prove weak convergence even with the presence of the $\int f(s) \nu_{1}$ term?

Answer 1

Giving a partial answer to my own question in the event someone else has a similar question. Further comments/corrections welcome.

Within the 2 statements of Question 1, I will refer to each part as:

1.) Question 1.1: Are there known results for weak convergence involving non-$\Delta_{2}$ Orlicz spaces?

2.) Question 1.2: results that get around the more difficult dual spaces present under non-$\Delta_{2}$ conditions.

$\textbf{Question 1.1:}$ "Stopping Times and Directed Processes" by Edgar & Sucheston, Complement 2.1.25(4), page 49 gives a counter example for weak sequential completeness in the case of $\Phi \not\in \Delta_{2}$ on a finite measure space (the space $[0,1]$ with Lebesgue measure). Note that the cited counterexample uses a Young Function of $\Phi_{2}(t) = e^{t} - t - 1$, as opposed to $\Phi_{1}(t) = e^{|t|} - 1$ referenced in the question above. However, since $\mu$ is a finite measure and both $\Phi_{1}$ and $\Phi_{2}$ dominated each other at infinity: $$ \lim_{t \rightarrow +\infty} \dfrac{\Phi_{1}(t)}{\Phi_{2}(t)} < \infty $$ and $$ \lim_{t \rightarrow +\infty} \dfrac{\Phi_{2}(t)}{\Phi_{1}(t)} < \infty $$ then the Orlicz spaces induced by the $\Phi_{1}$ and $\Phi_{2}$ are in fact equal (see Definition 2.2 and Theorem 2.3(2) on page 52 of "Stopping Times"). Note that it can be shown $L^{\Phi_{1}}(\mu)$ and $L^{\Phi_{2}}(\mu)$ are equal as sets with equivalent norms (see "Foundations of Symmetric Spaces of Measurable Functions" by Rubshtein, Grabarnik, Muratov, & Pashkova, Chapter 16.3, Theorem 16.3.2, page 212). I am still not entirely certain how the given counterexample shows the result (I can see boundedness in norm but not the weak convergence since I am still unclear on elements of the dual). I will likely ask about this as a follow up question.

$\textbf{Question 1.2:}$ My hope had been that under more generous conditions, namely $(\mathcal{X},\mu)$ being a finite measure space, that $(L^{e^{|t|}-1}(\mu))^{*} = L^{\Psi}(\mu)$, allowing us to avoid the term $\displaystyle \int f(s) \nu_{1}$ or space $B_{\Psi}$ referenced above. This is $\textbf{NOT}$ true.

$\textbf{Short Answer for Question 1.2: }$ For a quick reference, see Rao & Ren's "Theory of Orlicz Spaces" Chapter 4.1, Corollary 12, page 113, which states that for a finite measure space, $L^{\Phi}(\mu)$ is reflexive if and only if $\Phi \in \Delta_{2} \cap \nabla_{2}$. $\Phi_{1}(t) = e^{|t|}-1$ is not $\Delta_{2}$.

$\textbf{Longer Answer for Question 1.2: }$For completeness, I will also include a longer explanation that shows directly where the $\Delta_{2}$ condition shows up.

Letting $(\Phi, \Psi)$ be complementary Young functions, Theorem 10 of Rao & Ren's book (page 112) states $L^{\Phi}(\mu)$ is reflexive if and only if $L^{\Phi}(\mu) = M^{\Phi}$ and $L^{\Psi}(\mu) = M^{\Psi}$, or equivalently both $L^{\Phi}(\mu)$ and $L^{\Psi}(\mu)$ have absolutely continuous norms.

My focus was on the second condition involving absolutely continuous norms. An element $f \in L^{\Phi}(\mu)$ is said to have an absolutely continuous norm if $N_{\Phi}(f \chi_{A_{n}}) \rightarrow 0$ for each sequence of measurable sets $A_{n}$ decreasing to the nullset. The space $L^{\Phi}(\mu)$ is said to have absolutely continuous norm if every $f \in L^{\Phi}(\mu)$ has this property. Note that $\displaystyle N_{\Phi}(f) = \inf_{k > 0} \left( \int \Phi \left(\frac{f}{k} \right) \ d\mu \leq 1 \right)$ is the Gauge or Luxembourg norm and $\chi_{A_{n}}$ is the characteristic function of $A_{n}$. This is given as Definition 13 in Chapter 3.4, page 84 of Rao & Ren's "Theory of Orlicz Spaces" book.

Theorem 14 of the same chapter gives equivalent characterizations of absolute continuity of the norm, namely that $L^{\Phi}(\mu)$ has absolutely continuous norm at $f$ if and only if for each measureable $f_{n}$ such that $f_{n} \rightarrow f \text{ a.e.}$, and $|f_{n}| \leq |f| \text{ a.e.}$, we have $N_{\Phi}(f_{n} - f) \rightarrow 0$.

In trying to prove this equivalent characterization of absolutely continuous norm, we can see where the $\Delta_{2}$ condition is required.

$\textbf{Assumptions:}$ Suppose $(\mathcal{X},\mu)$ is a finite measure space, $\Phi$ is a continuous Young function, $f_{n} \rightarrow f \text{ a.e.}$, and $|f_{n}| \leq |f| \text{ a.e.}$. Letting $k > 0$ be arbitrary, we attempt to show $\displaystyle \int \Phi(k(f_{n}-f))\mu \rightarrow 0$ since this is equivalent to $f_{n}$ converging to $f$ in $L^{\Phi}(\mu)$ (see Proposition 2.1.10(5) in "Stopping Time").

Since $f_{n} \rightarrow f \text{ a.e.}$, then $kf_{n} \rightarrow kf \text{ a.e.}$. Further, since $\Phi$ is continuous, $\Phi(kf_{n}) \rightarrow \Phi(kf) \text{ a.e.}$ (see Folland, Exercise 2.37(a)).

Since $\mu$ is a finite measure, $\Phi(kf_{n}) \rightarrow \Phi(kf) \text{ a.e.}$ implies $\Phi(kf_{n})$ converges in $\mu-$measure to $\Phi(kf)$ ("Real Analysis for Graduate Students" by Bass, Proposition 10.2(1)). Additionally, $\Phi(kf_{n}) \leq \Phi(kf)$ since $|kf_{n}| \leq |kf| \text{ a.e.}$ and $\Phi$ is a Young function (non-decreasing). Therefore, by the measure convergence version of the Dominated Convergence Theorem, we have $$ \lim \int_{\mathcal{X}} \Phi(kf_{n}) \ \mu(dx) = \int_{\mathcal{X}} \Phi(kf) \ \mu(dx) $$.

Together, we now have:

1.) $kf_{n} \rightarrow kf \text{ a.e.}$

2.) $\displaystyle \int \Phi(kf_{n}) \mu \rightarrow \int \Phi(kf) \mu$

If this were an $L^{p}$ space with $\Phi(t) = |t|^{p}$ this would be enough to conclude $f_{n}$ converges to $f$ in $L^{p}$, using the inequality $$ 2^{p-1}(|x| + |y|^{p}) - |x-y|^{p} \geq 0. $$ The corresponding inequality we would need for our Young function would be $$ \Phi(x) + \Phi(y) - \Phi(x-y) \geq 0. $$

However, for $\Phi \notin \Delta_{2}$, this inequality cannot hold. Considering the special case when $x = y$, the needed inequality resembles the $\Delta_{2}$ condition: $$ \Phi(x) + \Phi(y) - \Phi(x-y) = 2\Phi(x) - \Phi(2x). $$ For $\Phi \notin \Delta_{2}$ there is no coefficient we could place in front of $\Phi(x)$ such that the needed inequality would hold. Thus, the inequality needed to replicate the typical $L^{p}$ argument does not hold.

$\textbf{Question 4:}$ See "Application of Orlicz Spaces" also by Rao & Ren, Chapter 1.3, Example 5. In this example it is shown the complement to $\Phi(t) = e^{|t|}-1$ is $\Psi_{2}(t) = $

\begin{cases} 0, &\quad |t| \leq 1 \\ |t| \log{|t|} - |t| + 1, &\quad |t| > 1. \end{cases}

If we compare this with $\Psi_{1}(t) = |t|\log^{+}|t|$ in the manner outlined in the answer to Question 1.1: $$ \lim_{t \rightarrow \infty} \dfrac{t \log{t} - t + 1}{t \log{t}} = \lim_{t \rightarrow \infty} \dfrac{\log{t}}{ \log{t}+1} = \lim_{t \rightarrow \infty} \dfrac{1/t}{ 1/t} = 1$$ and similarly $$ \lim_{t \rightarrow \infty} \dfrac{t \log{t}}{t \log{t} - t + 1} = 1. $$ Therefore, $L^{\Psi_{1}}(\mu) = L^{\Psi_{2}}(\mu)$.

By Corollary 9 of Chapter 4.2 in "Theory of Orlicz Spaces" we can express $x^{*} \in (L^{\Phi}(\mu))^{*}$ as $$ x^{*}(f) = \int fg d\mu + z^{*}(f) $$ where $g \in L^{\Psi_{1}} = L^{\Psi_{2}}$ to connect the dual of $e^{|t|}-1$ to $|t|\log^{+}|t|$. For the sake of completion, $z^{*}(f) \in (M^{\Phi})^{\perp}$.