I have an Operator $T: L^1(\mathbb{R}^d) \rightarrow w-L^1(\mathbb{R}^d)$, where $w-L^1$ denotes the weak $L^1$ space endowed with the quasi-norm: $$ \|f\|_{w-L^1}=\sup_{\alpha>0} \ \alpha \cdot \lambda(\{x\in\mathbb{R}^d: \ |f(x)|\ge\alpha\}) $$ where $\lambda$ denotes the Lebesgue measure.
If I know that T is bounded on a dense subset (set of all simple functions), why can't T be extended to a bounded operator on whole $L^1$?
As far as I know, the problem ist due to the relaxed triangle inequality in the Quasi-Banach space, i.e. $\|f+g\|_{w-L^1}\le C\cdot (\|f\|_{w-L^1}+\|g\|_{w-L^1})$. But I can't identify the problem in imitating the proof for Banach spaces as $w-L^1(\mathbb{R}^d)$ is complete and boundedness of an operator is equivalent to continuity also in Quasi-Banach spaces.
EDIT: I found the post "Is quasinorm always continuous?" and probably the weak $L^1$ quasi-norm is not continuous and therefore the proof fails?