Question about subspace of an Orlicz space

Question

Let me define it first
Definition 1: A mapping $\Phi:\mathbb{R^+}\to\mathbb{R^+}$ is said to be an N function if
(1) $\Phi$ is continuous on $\mathbb{R^+}$
(2) $\Phi$ is convex
(3) $\Phi(x)=0$ iff $x=0$
(4) $\lim_{x\to 0}\frac{\Phi(x)}{x}=0$
(5) $\lim_{x\to \infty}\frac{\Phi(x)}{x}=\infty$

Orlicz space: Let $(\Omega,F,\mu)$ be arbitrary measure space.Then the orlicz space is defined as $$\tag{1} L^{\Phi}(\mu)=\Big\{f:\Omega \to \mathbb{R} \hspace{0.2cm} \text{such that} \int_{\Omega}\Phi(|\alpha f|)\,\mathrm d\mu<\infty \mbox{ for some } \alpha>0\Big\} $$ What we have is that $L^{\Phi}$ is a normed linear space with the norm $$ \|f\|_{\Phi} = \inf\Big\{k>0:\int_{\Omega}\Phi(\tfrac{f}{k})\,\mathrm d\mu\leq 1\Big\} $$

My Question: From the definition of the Orlicz space we can see that if I take $\Phi(x)=|x|^p$ for $p>1$ then it is nothing but $L^p$ space.In $L^p$ we know that we can define the subspace of it. In the same way, I want to define the subspace of $L^{\Phi}$ space.For the subspace, we know that we have to find the subset of $L^{\Phi}$ which should be vector space with the norm obtained by restricting the norm on $L^{\Phi}$ space.But the problem is I know the logic but I am finding it difficult to find the subspace of it. Can someone please help me with this or give some hints that will be a great help. Thanks

Edit: Let me define $\Delta_2$ condition for an $\Phi$.$\Phi\in\Delta_2$ if $\Phi(2x)\leq K\Phi(x)$ $\forall x\geq x_0\geq 0$ and for some absolute value $K>0$ so if I define (1) now that is by putting this one extra condition on $\Phi$ will it be a subspace of $L^{\Phi}$? One thing is clear this new collection of $\Delta_2$-N functions is a subset of N functions.Can someone help me out to verify that what I am thinking is correct or not?

Answer 1

It is possible to define $L^{\Phi}(\mathcal F,\mu)$ as the space of the $\mathcal F$-measurable functions $f$ for which $\lVert f\rVert_\Phi$ is finite.

If $\mathcal G$ is a sub-$\sigma$-algebra of $\mathcal F$, then defining $L^{\Phi}(\mathcal G,\mu)$ in the same way and keeping the norm $\lVert \cdot \rVert_\Phi$.

We can also consider subspaces of $L^{\Phi}(\mathcal F,\mu)$ by considering the vector space generated by a finite number of $L^{\Phi}(\mathcal F,\mu)$.

I afraid that I do not understand what $\Delta_2$ condition will bring in this context.

The comment made by mihaild here will certainly help you to solve the conflict of definition.

Answer 2

I think you are confusing the space $L_\Phi(\mu)$ with its properties. If $\Phi$ satisfies the properties (1)-(3) listed above (such functions are called Young functions), then $L_\Phi$ (after identifying functions that differ in sets of measure $0$) equipped with $\|\;\|_\Phi$ as defined in the OP, defines a complete normed space, and $$Q_\Phi\Big(\frac{|f|}{\|f\|_\Phi}\Big):=\int_X\Phi\Big(\tfrac{|f|}{\|f\|_\Phi}\Big)\,d\mu \leq1$$ for all $f\in L_\Phi\setminus\{0\}$.

• If the underlaying measure space $(X,\mathscr{F},\mu)$ is finite, then it is not difficult to prove that $L_\infty(\mu)\subset L_\Phi(\mu)\subset L_1(\mu)$ and there are constants $\alpha, \,\beta>0$ such that $$\alpha\|f\|_1\leq\|f\|_\Phi\leq \beta\|f\|_\infty$$ This implies that the inclusion maps $I_1:L_\infty(\mu)\rightarrow L_\Phi(\mu)$ and $I_2:L_\Phi(\mu)\rightarrow L_1(\mu)$ are bounded. This also yields an interesting subspace of $L_\Phi(\mu)$, namely $E:=\overline{L_\infty(\mu)}^{L_\Phi(\mu)}$, the closure of $L_\infty(\mu)$ in $(L_\Phi(\mu),\|\;\|_\Phi)$.

• If in addition, $\Phi$ satisfies properties (1)-(5) listed in the OP (such functions are called weak Young functions), $\mu$ is non-atomic, and $\Phi$ the additional property $\Delta_2$ in the OP, then $L_\Phi(\mu)$ satisfies the additional property:
  (a) If $(f_n, f:n\in\mathbb{N})\subset L_\Phi(\mu)$ is a sequence such that $Q_\Phi(f-f_n)<\infty$ for all $n$ large enough, $\lim_nQ_\Phi(f-f_n)=0$ iff $\lim_n\|f-f_n\|_\Phi=0$ and vice versa, if (a) holds, then $\Phi$ satisfies $\Delta_2$.
  In such case, $L_\infty(\mu)$ is dense in $L_\Phi(\mu)$.

Under additional conditions of $\Phi$, duality between $L_\Phi(\mu)$ spaces holds in a similar was as in $L_p(\mu)$ spaces.

• If $\Phi$ is a Young function and its dual (Fenchel-Legendre dual) $\Phi^*(y)=\sup_x(xy -\Phi(x))$ is also a Young function then:
  (1) If $f\in L_{\Phi}(\mu)$ and $g\in L_{\Phi^*}(\mu)$ then $fg\in L_1(\mu)$ and $$\int_X|fg|\,d\mu \leq 2\|f\|_\Phi\|g\|_{\Phi^*}$$ and $\Big(\overline{L_\infty(\mu)}^{L_\Phi(\mu)}\Big)^*=L_{\Phi^*}(\mu)$. In particular, when $\Phi$ satisfies condition $\Delta_2$, then $\Big(L_\Phi(\mu)\big)^*=L_{\Phi^*}(\mu)$.

See Barry Simon's Convex Analysis book, pp. 33-40 for details and examples.

Now, depending on $\Phi$, you may encounter different interesting subspaces. Here is one example from Stein, E. and Shakarchi, R. Functional Analysis, Princeton, pp. 41

• For $1\leq p_0<p_1<\infty$, the space $L_{p_0}(\mu)+L_{p1}(\mu)$ is a Banach space with norm $\|f\|:=\inf\{\|f_0\|_{p_0}+\|f_1\|_{p_1}: f=f_0+f_1\}$. It turns out that $L_0+L_1$ can also be considered as an Orlicz space with $$\Phi(x)=\mathbb{1}_{[0,1)}x^{p_0} + \mathbb{1}_{[1,\infty)}(x)x^{p_1}$$ and that there are constants $a, b>0$ such that $$a\|f\|\leq\|f\|_\Phi\leq b\|f\|$$ In particular, any $L_r(\mu)$ with $p_0\leq r\leq p_1$ is a subspace of $L_{\Phi}(\mu)$. (Here $\mu$ not need be finite).