Generalization of Lp; $L_{\phi}(\mu)=\{f: \exists M>0 \int \phi(\frac{f(x)}{M})d\mu < +\infty\}$ Let $(\Omega, \Sigma, \mu)$ a measure space and $\phi(t)$ defined on $[0,+\infty]$ a continuous, increasing, and convex function with $\phi(0)=0$. 
We can define the space of the measurable functions defined a.e.
$$L_{\phi}(\mu)=\left\{f: \exists M>0 \int \phi(\frac{f(x)}{M})d\mu < +\infty\right\}$$
and a function
$$||f||_{\phi}=\inf\left\{M>0:\int \phi(\frac{f(x)}{M})d\mu \leq 1\right\}.$$
Prove $L_{\phi}(\mu)$ is a vector space, $||\cdot||_{\phi}$ a norm and $(L_{\phi}(\mu),||\cdot||_{\phi})$ a Banach space.
I have tried to prove $L_{\phi}(\mu)$ is a vector space but I can't demonstrate it's closed under sums and scalar multiplication. I've tried to use the convexity of $\phi$ without success. 
I have no idea what to do.
 A: $\newcommand{\norm}[1]{\left\|{#1}\right\|}\newcommand{\set}[1]{\left\{{#1}\right\}}\newcommand{\of}[1]{\left({#1}\right)}\newcommand{\abs}[1]{\left|{#1}\right|}$
First of all, note that the definition should be 
$$\norm{f}_\phi := \inf\set{M > 0 : \int\phi\of{\frac{\abs{f(x)}}{M}} \,d\mu \le 1 },$$
otherwise the definition doesn't make sense, since the functions $f:\Omega\to [0,\infty]$ don't form a vector space in the first place.
Now with this definition, we can take $f$ to have values in $\newcommand{\RR}{\mathbb{R}}\RR$ or $\newcommand{\CC}{\mathbb{C}}\CC$ so that we can make sense of things like $-f$. 
First we show that $\norm{f}_\phi < \infty$ and $\norm{g}_\phi < \infty$ implies that $\norm{f+g}_\phi < \infty$. In fact, we prove the triangle inequality. First fix some $\epsilon > 0$.
Now, let $F=\norm{f}_\phi+\epsilon$, $G=\norm{g}_\phi+\epsilon$, $M=F+G= \norm{f}_\phi+ \norm{g}_\phi+2\epsilon$. 
Then since $\phi$ is increasing and convex, we have
$$ \phi\of{\frac{\abs{f+g}}{M}}
\le \phi\of{\frac{\abs{f}+\abs{g}}{M}}
\le \phi\of{
\frac{F}{M}\cdot \frac{\abs{f}}{F}+
\frac{G}{M}\cdot \frac{\abs{g}}{G}
}
\le
\frac{F}{M}
\phi\of{
\frac{\abs{f}}{F}}
+
\frac{G}{M}\phi\of{\frac{\abs{g}}{G}
}
.
$$
Then because $\phi$ is increasing, and $F > \norm{f}_\phi$,
$$\int \phi\of{\frac{\abs{f}}{F}}\,d\mu \le 1,$$ 
and similarly since $G > \norm{g}_\phi$,
$$\int \phi\of{\frac{\abs{g}}{G}}\,d\mu \le 1$$ 
as well.
Thus 
$$\int \phi\of{\frac{\abs{f+g}}{M}}\,d\mu 
\le\frac{F}{M}\int \phi\of{\frac{\abs{f}}{F}}\,d\mu
+\frac{G}{M}\int \phi\of{\frac{\abs{g}}{G}}\,d\mu
\le \frac{F+G}{M}=1. 
$$ 
Hence since $\epsilon$ was arbitrary, we have $\norm{f+g}_\phi\le \norm{f}_\phi + \norm{g}_\phi,$
and in particular, if $\norm{f}_\phi<\infty$, and $\norm{g}_\phi < \infty$, then we have $\norm{f+g}_\phi < \infty$. 
Now to show that $\norm{\alpha f}_\phi = \abs{\alpha}\norm{f}_\phi$, when
$\norm{f}_\phi < \infty$, 
which will show that $L_\phi(\mu)$ is closed under scalar multiplication, we do almost the same thing.
First note that the case $\alpha=0$ is obvious, so we may assume $\alpha\ne 0$. Then let $\epsilon > 0$, $F=\norm{f}_\phi+\epsilon$,
then
$$\int\phi\of{\frac{\abs{\alpha f}}{\abs{\alpha}F}} = \int\phi\of{\frac{\abs{f}}{F}}\le 1,$$
so $\norm{\alpha f}_\phi \le \abs{\alpha}F$ for all $\epsilon>0$, so
$\norm{\alpha f}_\phi \le \abs{\alpha}\norm{f}_\phi$.
Applying this inequality to $1/\alpha$ and $\alpha f$,
we also have $\norm{f}_\phi \le \norm{\alpha f}_\phi/\abs{\alpha}$. 
Hence $\norm{\alpha f}_\phi =\abs{\alpha}\norm{f}_\phi$.
