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: I want to generate a subspace of $L^{\Phi}$ and look for its property. My idea is to put some conditions on $\Phi$ to get the subspace. I was thinking of putting some condition on $\Phi$ such that it will be non-zero in some interval $(a,b)$ or $[a,b]$ and zero otherwise. But think if I do this then this will contradict the property $(3)$ that $\Phi$ is satisfying. Now suppose I modified (3) such that $\Phi(0)=0$ then if I define $\Phi$ as I said above will it still be convex? I think then it will not be convex anymore(Not sure). Can someone clarify this also if someone can help to find some condition on $\Phi$ such that we get subspace,that will be great help.I am trying to understand Orlicz space deeply.