To prove following is a vector space First let me define $\bar{L^{\Phi}}$ space Let $\Phi:\mathbb{R}\to\mathbb{R^+}$ be a convex function such that
$$\Phi(0)=0,\Phi(-x)=\Phi(x) \quad\text{ and}\quad \lim_{x\to \infty}\Phi(x)=\infty$$
$\bar{L^{\Phi}}$=$\{f:\Omega\to\mathbb{R} \text{ measurable such that } \int_{\Omega}\Phi(|f|)d\mu<\infty \}$
In general it is not a vector space, since it is not closed under scalar multiplication. For example let $\Omega=(0,1)$, $\Phi(t)=e^t$ and take $f(x)=\frac{-1}{2}\log x$ then $2f\notin \bar{L^\Phi}$
It is given in the book that $\bar{L^\Phi}$ is a vector space if $\Phi\in \Delta_2$ globally when $\mu(\Omega)=\infty$ and locally when $\mu(\Omega)<\infty$
Let me define $\Delta_2$: $\Phi$ is said to be $\Delta_2$ if $\Phi(2x)\leq K\Phi(x)$ for some $K>0$ and $x\geq x_0 \geq 0$. But I am not understanding how this $\Delta_2$ condition matters.Like if I take $f\in \bar{L^{\Phi}}$ then I have to prove that $\alpha f\in \bar{L^\Phi}$ for every scalar $\alpha$ but for that we need $\Delta_2$ condition.
$\textbf{Proof:}$  to show that $2f\in \bar{L^{\Phi}}$
For that the proof goes like this and I have some doubts in the proof,/
since $\Phi\in \Delta_2$ so if 
$\textbf{Case-1:}$ if $\mu(\Omega)=\infty$ that condition holds globally so we have $\Phi(2|f|)\leq K\Phi(|f|),K>0$ which implies that $2f\in \bar{L^\Phi}$.
$\textbf{Case-2:}$ if suppose $\mu(\Omega)<\infty$ that condition does not holds globally that is $\Phi(2x)\leq K\Phi(x)$ for $x\geq x_0\geq 0$ for some $K>0$ so if I let
$$f_1=\begin{cases}
               f :\text{ if } |f|\leq x_0\\
               0 :\text{ Otherwise }\\
            \end{cases}$$
set $f_2=f-f_1$ so we have $f=f_1+f_2$ and
$$\Phi(2|f|)=\Phi(2|f_1|)+\Phi(2|f_2|)\leq \Phi(2|f_1|)+K\Phi(|f_2|)......(A)$$ hence,
$$\int_{\Omega}\Phi(2|f|)d\mu\leq \Phi(2x_0)\mu(\Omega)+K\int_{\Omega}\Phi(|f|)d\mu<\infty$$ this gives $2f\in \bar{L^{\Phi}}......(B)$
$\textbf{My Question:}$ I have doubt in case 2, In case 2 we have two conditions that is when $\mu(\Omega)=\infty$ so in this case we directly use the $\Delta_2$ condition since this condition holds globally. And then we have $\mu(\Omega)<\infty$ in this case we cannot directly use the $\Delta_2$ condition since it does not hold globally so we consider a function $f_1=f$ when $|f|\leq x_0$ that is now we have $x\geq x_0\geq |f|\geq 0$.After this we have defined $f_2=f-f_1$ so
$$f_2=\begin{cases}
               0 :\text{ if } |f|\leq x_0\\
               f :\text{ Otherwise }\\
            \end{cases}$$
and we have applied the $\Delta_2$ condition on $f$
I want to know if whatever I have understood about case 2 is this correct or I am missing something ?If I am on right path can someone then explain me the step $(A)$ and $(B)$
 A: Seems like you overthink this. The function $\Phi$ is the generalization of a
norm such as $|x|$ on $\mathbb R$. Since $\Phi$ is convex and even it is easy to show that it must increase on $[0,\infty)$ (and decrease on $(-\infty,0]$).
It also seems to me that the condition $\Delta_2$ has two cases which you did not mention very clearly:

*

*When $\mu(\Omega)<\infty$ then there exists an $x_0\ge 0$ such that $\Phi(2x)\le K\Phi(x)$ for all $x\ge x_0\,.$


*When $\mu(\Omega)=\infty$ then $\Phi(2x)\le K\Phi(x)$ for all $x\ge 0\,.$
Case 1. When $\Phi$ satisfies $\Delta_2\,,$ then
\begin{align}
\int_\Omega\Phi(|2f|)\,d\mu&=\int_{\{|2f|\le x_0\}}\Phi(|2f|)\,d\mu+\int_{\{|2f|> x_0\}}\Phi(|2f|)\,d\mu\\
&\le\int_{\{|2f|\le x_0\}}\Phi(x_0)\,d\mu+K\int_{\{|2f|> x_0\}}\Phi(|f|)\,d\mu\,.
\end{align}
The second integral is finite because $f\in \bar{L}^\Phi$ by assumption. The first integral is bounded from above by
$$
\Phi(x_0)\,\mu\{|2f|\le x_0\}\,,
$$
which is finite because $\mu(\Omega)$ is finite.
Case 2. Is even simpler:
$$
\int_\Omega\Phi(|2f|)\,d\mu\le K\int_\Omega\Phi(|f|)\,d\mu<\infty.
$$
