The set ${L}^{\Phi}$ is absolutely convex.That is, if $f,g\in {L}^{\Phi}$ and $\alpha,\beta$ are scalars such that $|\alpha|+|\beta|\leq 1$, then $\alpha f +\beta g\in {L}^{\Phi}$. Also $h\in {L}^{\Phi} \mbox{ with } |f|\leq |h|$ and f measurable implies $f\in {L}^{\Phi}.$
$\textbf{Proof:}$ Let $f,g\in {L}^{\Phi}$.Then by the monotonicity and convexity of $\Phi,$ for $0<\gamma =|\alpha|+|\beta|\leq 1,$
$$\Phi(|\alpha f+\beta g|)\leq \Phi(|\alpha||f|+|\beta||g|)$$
$$\leq \gamma\Phi\left(\frac{|\alpha|}{\gamma}|f|+\frac{|\beta|}{\gamma}|g| \right) \quad(1)$$
$$\leq |\alpha|\Phi(|f|)+|\beta|\Phi(|g|)$$
integrating both side over $\Omega,$
$$\int_{\Omega}\Phi(|\alpha f+\beta g|)d\mu\leq\int_{\Omega} |\alpha|\Phi(|f|)+|\beta|\Phi(|g|)d\mu$$
My question is How we got $(1),\ $ I mean how $\gamma$ came out?