We have defined the $L^p$ space for $p\geq1$.One thing I know is that for $0<p<1$ the norm will not satisfy the triangle inequality.
$L^p(A)$={$f \colon A\to R$ (measurable) such that $\int|f|^pd\nu<\infty$}
Now I want to define the Orlicz space,which is generalization of $L^p$ spaces,defined as follows,
$L^\phi(A)$={$f \colon A\to R$ measurable such that $\int\phi(kf)d\nu<\infty$ for some $k>0$},where $\phi$ is a convex function.
Question(1): Is there any point I am missing that we cannot define $L^p$ when $0<p<1$ so is there any connection between this and why we are taking convex function to define Orlicz space?
Question(2): What if I take a function that is not a convex function but we can go back to $L^p$ space from $L^\phi$ space by defining $\phi$ other than a convex function, is that possible?
An answer to both your questions:
An Orlicz function, is by definition, a convex function. An Orlicz space, therefore, is always a Banach space. In particular, it has a non-trivial dual. This means that there are always non-zero bounded linear functionals on an Orlicz space. $L^p$ for $0<p<1$ is not a Banach space. In fact, there are no non-zero bounded linear functionals on these spaces. As a result, there is no way of "getting" $L^p$ with $0<p<1$ from a proper Orlicz space.