Suppose we have a norm defined as below
$$N_{\Phi}(f)=\inf\left\{k>0:\frac{1}{k}f\in B_{\Phi}\right\}=\inf\left\{k>0: \int_{\Omega}\Phi\left({\frac{f}{k}}\right)\leq 1\right\}.$$
where, $\Phi:\mathbb{R}\to\mathbb{R^+}$ is a convex function such that $\Phi(0)=0,\Phi(-x)=\Phi(x),\lim_{x\to\infty}\Phi(x)=\infty$ and $f:\Omega\to\mathbb{R}$ is a measurable function on an abstract measure space $(\Omega,\Sigma,\mu),$ such that $\int_{\Omega}\Phi(\alpha f)d\mu<\infty$ for some $\alpha>0.$
My question is: how can we say that the above norm is well defined?
By convexity, nonnegativity and $\Phi(0)=0$ the function $\Phi(t)$ is increasing for $t\ge 0,$ as for $0<\lambda <1$ we get $$\Phi(\lambda t)=\Phi(\lambda t+(1-\lambda)0)\le \lambda \Phi(t)+(1-\lambda)\Phi(0)\le \Phi(t)$$ As the function $\Phi$ is even we get $\Phi(t)=\Phi(|t|).$ The Lebesgue dominated convergence implies $$\lim_{\lambda\to 0^+}\int\limits_{\Omega}\Phi(\lambda f(x))\,d\mu(x)=\lim_{\lambda\to 0^+}\int\limits_{\Omega}\Phi(\lambda |f(x)|)\,d\mu(x)=0$$ Therefore the set $$\left \{ k\in \mathbb{R}^+ \,:\,\int\limits_{\Omega}\Phi(k^{-1}|f(x)|)\,d\mu(x)\le 1\right \}$$ is nonempty. Let $A_\delta=\{x\in\Omega\,:\, |f(x)|\ge \delta>0\}.$ If $f\neq 0,$ there exists $\delta>0$ such that $\mu(A_\delta)>0.$ Then $$ \int\limits_{\Omega} \Phi(k^{-1}|f(x)|)\,d\mu(x)\ge \int\limits_{A_\delta} \Phi(k^{-1}|f(x)|)\,d\mu(x)\ge \mu(A_\delta)\Phi(k^{-1}\delta)$$ Therefore $$\lim_{k\to 0^+}\int\limits_{\Omega} \Phi(k^{-1}|f(x)|)\,d\mu(x)=\infty$$ Hence $$\inf\left \{ k\in \mathbb{R}^+ \,:\,\int\limits_{\Omega}\Phi(k^{-1}|f(x)|)\,d\mu(x)\le 1\right \}>0$$ This shows that $N_\Phi(f)>0$ for $f\neq 0.$ Next for $\lambda\neq 0$ we get $$N_\Phi(\lambda f)=\inf\left \{ k\in \mathbb{R}^+ \,:\,\int\limits_{\Omega}\Phi(k^{-1}|\lambda|\,|f(x)|)\,d\mu(x)\le 1\right \}\\ =|\lambda| \inf\left \{ k|\lambda|^{-1}\in \mathbb{R}^+ \,:\,\int\limits_{\Omega}\Phi(k^{-1}|\lambda|\,|f(x)|)\,d\mu(x)\le 1\right \}=|\lambda|\,N_\Phi(f)$$ Let $f$ and $g$ satisfy $$\int\limits_\Omega \Phi(k^{-1}|f(x)|)d\mu(x)\le 1\ \int\limits_\Omega \Phi(l^{-1}|g(x)|)d\mu(x)\le 1$$ for some $k,l>0.$ Then $$\Phi((k+l)^{-1}|f(x)+g(x)|)\le \Phi((k+l)^{-1}(|f(x)|+|g(x)|))\\ \le {k\over k+l}\Phi(k^{-1}|f(x)|)+{l\over k+l}\Phi(l^{-1}|g(x)|)$$ Hence $$\int\limits_\Omega \Phi((k+l)^{-1}|f(x)+g(x)|)d\mu(x)\\ \le {k\over k+l}\int\limits_\Omega \Phi(k^{-1}|f(x)|)d\mu(x)+{l\over k+l}\int\limits\Phi(l^{-1}|g(x)|)d\mu(x)$$ Assume $$\int\limits_\Omega \Phi(k^{-1}|f(x)|)d\mu(x)\le 1,\ \int\limits_\Omega \Phi(l^{-1}|g(x)|)d\mu(x)\le 1$$ then $$\int\limits_\Omega \Phi((k+l)^{-1}|f(x)+g(x)|)d\mu(x)\le 1$$ Hence $N_\Phi(f+g)\le k+l,$ which implies $$N_\Phi(f+g)\le N_\Phi(f)+N_\Phi(g)$$