I had asked a question pertaining to Orlicz norms here. However, in the book I was reading, it said (and I paraphrase) "It's not difficult to show it is a norm on the space of integrable random variables and for which $\|X\|_{\psi}$ is finite". So I decided to prove this.
To recap, if $\psi$ is a monotone nondecreasing, convex function with $\psi(0) = 0$, the Orlicz norm of an integrable random variable X wrt $\psi$ is given by
$$\|X\|_{\psi} = \inf \left\{u>0: \mathbb{E}\left[\psi\left(\frac{|X|}{u}\right) \right] \leq 1\right\}$$
Here is what I got so far:
1) $\|aX\|_{\psi} = |a|\|X\|_{\psi} \quad \forall a \in \mathbb{R}$
$$ \|aX\|_{\psi} = \inf \left\{u>0: \mathbb{E}\left[\psi\left(\frac{|aX|}{u}\right) \right] \leq 1\right\}$$
Replace $u$ with $|a|v$ $$ = \inf \left\{|a|v>0: \mathbb{E}\left[\psi\left(\frac{|X|}{v}\right) \right] \leq 1\right\}$$ $$ = |a|\inf \left\{v>0: \mathbb{E}\left[\psi\left(\frac{|X|}{v}\right) \right] \leq 1\right\}$$ $$ = |a|\|X\|_{\psi}$$
2) $\|X\|_{\psi} = 0 \iff X = 0 \quad a.e$
($\Leftarrow$) This directly follows from substitution. You will however have to split the expectation into two parts, one where $X=0$ and the other where it isn't. But I got it nonetheless.
Now for ($\Rightarrow$).
If $\|X\|_{\psi} = 0$, then $\forall u > 0$, $$ \mathbb{E}\left[\psi\left(\frac{|X|}{u}\right) \right] \leq 1$$
Now suppose, the implication were false, then let $A = \left\{\omega \subset \Omega : X(\omega) \neq 0\right\}$. Now by Jensen's Inequality, $$\psi \left[\mathbb{E}\left(\frac{|X|1_A}{u}\right) \right] \leq \mathbb{E}\left[\psi\left(\frac{|X1_A|}{u}\right) \right]$$
Now unless I assume that $\psi$ is unbounded, I don't know how to proceed. Incidentally, in the book, the $\psi$ they use are all unbounded. If it is unbounded, then for a sufficiently small $u$, I can "cross" 1, which will be a desired contradiction. So I need help for the bounded case.
3) Triangle inequality
I tried several convexity tricks, but I didn't get this. Essentially, we have to show $$\|X+Y\|_{\psi} \leq \|X\|_{\psi} +\|Y\|_{\psi}$$ Perhaps there is a trick I haven't tried... Anyway I would appreciate any hints, tips and answers for the parts I didn't get. Thanks for your time.
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$\begingroup$For triangle inequality, take $X,Y$ for which $\lVert\cdot\rVert_{\psi}$ is finite, and for a fixed $\varepsilon$, consider $u$ and $v$ such that $u-\lVert X\rVert_\psi<\varepsilon$, $v-\lVert Y\rVert_\psi<\varepsilon$, and$$\max\left\{E\left[\psi\left(\frac{|X|}u\right)\right],E\left[\psi\left(\frac{|Y|}v\right)\right]\right\}\leqslant 1.$$Then\begin{align} \psi\left(\frac{|X+Y|}{u+v}\right)&\leqslant\psi\left(\frac{|X|+|Y|}{u+v}\right)&\phi\mbox{ is non-decreasing}\\ &=\psi\left(\frac u{u+v}\frac{|X|}u+\frac v{u+v}\frac{|Y|}v\right)&\\ &\leqslant \frac u{u+v}\psi\left(\frac{|X|}u\right)+\frac v{u+v}\psi\left(\frac{|Y|}v\right)&\mbox{ by convexity}. \end{align}This is what we do when we have the definition of the Orlicz norm.
Once can check that actually, the infimum is attained (taking a sequence $(u_n,n\geqslant 1)$ approaching the infinimum, and using Fatou's lemma). This will avoid the reasoning with $\varepsilon$.
$\endgroup$ 2 $\begingroup$Actually $\phi(x)$ can not be bounded, as long as it is not zero function. For convex functions, there is a super additive property if $\phi(0)\leq0$, which is $\phi(x)+\phi(y)\leq\phi(x+y)$, you can see Wiki page for this simple property. Now there exists $a\in R$ such that $\phi(a)>0$, then $\phi(na)>n\phi(a)$.
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