I have some problems with this exercise: $\forall\varepsilon>0$ there exists $C(\epsilon)>0$ such that for all $f,g\in L^{1,\infty}(\Bbb R^n)$ we have that $$ ||f+g||_{1,\infty}\le(1+\varepsilon)||f||_{1,\infty}+C(\varepsilon)||g||_{1,\infty}$$
I don't know where to start. It's a variance on the triangular inequality of the weak $L^1$ space, but I can't get out from there. Can someone give me some hint please?
Let $\rho_f(s) := \mu\big(\{x : |f(x)| > s\}\big)$ and let $\|f\|_{1,\infty} := \sup_{s > 0}s\rho_f(s)$. Now let $f,g \in L^{1,\infty}$ and notice that $$\{x : |f(x) + g(x)| > s\} \subset \{x : |f(x)| > s(1 + \epsilon)^{-1}\} \cup \{x : |g(x)| > sC(\epsilon)^{-1}\},$$ where $(1 + \epsilon)^{-1} + C(\epsilon)^{-1} = 1.$ This shows that \begin{align} s\rho_{f+g}(s) \le &\ s\rho_f(s(1 + \epsilon)^{-1}) + s\rho_g(sC(\epsilon)^{-1})\\ = &\ (1 + \epsilon)\color{blue}{s(1+\epsilon)^{-1}}\rho_f(\color{blue}{s(1 + \epsilon)^{-1}}) + C(\epsilon)\color{green}{sC(\epsilon)^{-1}}\rho_g(\color{green}{sC(\epsilon)^{-1}}) \\ \le &\ (1 + \epsilon)\|f\|_{1,\infty} + C(\epsilon)\|g\|_{1,\infty}. \end{align}
Take the supremum over $s$ to conclude.