How do I prove that this space is a Banach space? We define the following space 
$$
L^{\Phi}(\Omega)=\left\{u:\Omega\rightarrow \mathbb{R}~\text{measurable};~\int_{\Omega}\Phi\left(\frac{u}{\lambda}\right) dx<+\infty, ~\text{for any}~\lambda>0\right\}
$$
where $\Phi:\mathbb{R}\to\mathbb{R}^+$ satisfies the following conditions: 
$(i)$ $\Phi$ is a continuous and a convex function;
$(ii)$ $\Phi(t)=0$ if, and only if $t=0$;
$(iii)$ $\displaystyle\frac{\Phi(t)}{t}\overset{t\rightarrow0}{\longrightarrow}0$ and $\displaystyle\frac{\Phi(t)}{t}\overset{t\rightarrow+\infty}{\longrightarrow}+\infty$;
$(iv)$ $\Phi$ is even (that is $\Phi(t)=\Phi(-t)$)
Endowed with the norm:
$$
||u||_{\Phi}=\inf
\left\{\lambda>0; \int_{\Omega}\Phi\left(\frac{u}{\lambda}\right)dx\leq 1\right\}
$$
I start the proof by : Let $(f_n)\subset L^{\Phi}(\Omega)$ be a Cauchy sequence that is $||f_n-f_m||_{\Phi}\to0$ when $m,n\to\infty$
How do we find $f\in L^{\Phi}(\Omega)$ such that $||f_n-f||_{\Phi}\to0$ ?
Thank you 
 A: Step 0
I think we could start by proving that 
${L}^{{\Phi}} \left({\Omega}\right) \subset  {L}^{1}_{\text{loc}} \left({\Omega}\right)$. Indeed, if $B$ is some bounded
measurable subset of ${\Omega}$, then for $u \in  {L}^{{\Phi}} \left({\Omega}\right)$ and for all
${\lambda} > {\left\|u\right\|}_{{L}^{{\Phi}} \left({\Omega}\right)}$ one has by Jensen's inequality
$$\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \int_{{\Omega}}^{}{\Phi} \left(\frac{u}{{\lambda}}\right) d x  \leqslant  1&\Rightarrow &\displaystyle  \int_{B}^{}{\Phi} \left(\frac{u}{{\lambda}}\right) d x  \leqslant  1\\
&\Rightarrow &\displaystyle  \frac{1}{\left|B\right|} \int_{B}^{}{\Phi} \left(\frac{\left|u\right|}{{\lambda}}\right) d x  \leqslant  \frac{1}{\left|B\right|}\\
&\Rightarrow &\displaystyle  {\Phi} \left(\frac{1}{\left|B\right|} \int_{B}^{}\frac{\left|u\right|}{{\lambda}} d x\right)  \leqslant  \frac{1}{\left|B\right|}\\
&\Rightarrow &\displaystyle  {\Phi} \left(\frac{{\left\|u\right\|}_{{L}^{1} \left(B\right)}}{{\lambda} \left|B\right|}\right)  \leqslant  \frac{1}{\left|B\right|}\\
&\Rightarrow &{\left\|u\right\|}_{{L}^{1} \left(B\right)}  \leqslant  \left|B\right| {{\Phi}}^{{-1}} \left(\frac{1}{\left|B\right|}\right) {\lambda}
\end{array}$$
Taking the infimum over ${\lambda}$ yields
$${\left\|u\right\|}_{{L}^{1} \left(B\right)}  \leqslant  \left|B\right| {{\Phi}}^{{-1}} \left(\frac{1}{\left|B\right|}\right) {\left\|u\right\|}_{{L}^{{\Phi}} \left({\Omega}\right)}$$
It follows from this that a Cauchy sequence ${f}_{n}$ in ${L}^{{\Phi}} \left({\Omega}\right)$ is also a Cauchy sequence in ${L}^{1} \left(B\right)$. Hence, there is some function $f \in  {L}^{1}_{\text{loc}} \left({\Omega}\right)$
such that ${f}_{n} \rightarrow  f$ in ${L}^{1}_{\text{loc}} \left({\Omega}\right)$. It remains to prove that $f \in  {L}^{{\Phi}} \left({\Omega}\right)$ and that ${f}_{n} \rightarrow  f$ in ${L}^{{\Phi}} \left({\Omega}\right)$.
Edit: sketch of the end of the proof
Step 1
We only need to prove the convergence of a subsequence of the ${f}_{n}$ because if a subsequence of a Cauchy sequence
converges, the whole sequence converges. As ${f}_{n} \rightarrow  f$ in ${L}^{1}_{\text{loc}} \left({\Omega}\right)$, we can suppose, upon extracting
a subsequence, that ${f}_{n} \rightarrow  f$ pointwise almost everywhere.
Step 2
As ${f}_{n}$ is a Cauchy sequence, there is a sequence ${c}_{n} \rightarrow  0$ of numbers such that
${\left\|{f}_{n}-{f}_{m}\right\|}_{{L}^{{\Phi}} \left({\Omega}\right)}  \leqslant  {c}_{n}$ for all $m  \geqslant  n$. For any ${\lambda}  >  {c}_{n}$, we
may apply Fatou's lemma
$$\int_{}^{}{\Phi} \left(\frac{{f}_{n}-f}{{\lambda}}\right) d x = \int_{}^{}\liminf\limits _{m \rightarrow  \infty } {\Phi} \left(\frac{{f}_{n}-{f}_{m}}{{\lambda}}\right) d x  \leqslant  \liminf\limits _{m \rightarrow  \infty } \int_{}^{}{\Phi} \left(\frac{{f}_{n}-{f}_{m}}{{\lambda}}\right) d x  \leqslant  1$$
Step 3
For any ${\lambda} > 0$ the convexity of ${\Phi}$ imply that
$$\int_{}^{}{\Phi} \left(\frac{f}{{\lambda}}\right) d x = \int_{}^{}{\Phi} \left(\frac{1}{2} \frac{{f}_{n}}{{\lambda}}+\frac{1}{2} \frac{f-{f}_{n}}{{\lambda}}\right) d x  \leqslant  \frac{1}{2} \int_{}^{}{\Phi} \left(\frac{{f}_{n}}{{\lambda}}\right) d x+\frac{1}{2} \int_{}^{}{\Phi} \left(\frac{f-{f}_{n}}{{\lambda}}\right) d x$$
By choosing $n$ large enough, one has ${\lambda}  >  {c}_{n}$ and it follows that $\boxed{\int_{}^{}{\Phi} \left(\frac{f}{{\lambda}}\right) d x  <  \infty }$, hence $f \in  {L}^{{\Phi}} \left({\Omega}\right)$
Step 4
The steps 2 and 3 now imply that for all $n$,
$${\left\|{f}_{n}-f\right\|}_{{L}^{{\Phi}} \left({\Omega}\right)}  \leqslant  {c}_{n} \rightarrow  0$$
QED
