Q:

If $||g_j-g||_{1,\infty}\to0$, then $||g_j||_{1,\infty}\to||g||_{1,\infty}$

I have some problems with my notes: my teacher wrote that if a sequence $\{g_j\}_j\subseteq L^{1,\infty}(\Bbb R^n)$ (which is the weak $L^1$ space, endowed with the quasinorm $||\cdot||_{1,\infty}$) converges here, i.e. $\exists g\in
L^{1,\infty}(\Bbb R^n)$ such that
$||g_j-g||_{1,\infty}\to0$, then $||g_j||_{1,\infty}\to||g||_{1,\infty}$, and this is because given $f,g\in L^{1,\infty}(\Bbb R^n)$ then
$$
||f+g||_{1,\infty}\le(1+\varepsilon)||f||_{1,\infty}+C(\varepsilon)||g||_{1,\infty},\;\;\exists C(\varepsilon)$$
Can someone shade a light on these two facts? Many thanks

A:

The problem here is that we do not have the triangle inequality and its reversed version. However, with the given hint, we have for each positive $\varepsilon$, 
$$\lVert g_j\rVert_{1,\infty}\leqslant (1+\varepsilon)\lVert g\rVert_{1,\infty}+C(\varepsilon)\lVert g-g_j\rVert_{1,\infty}$$
hence 
$$\tag{1}\limsup_{j\to +\infty}\lVert g_j\rVert_{1,\infty}\leqslant (1+\varepsilon)\lVert g\rVert_{1,\infty}.$$
We also have 
$$\lVert g\rVert_{1,\infty}\leqslant (1+\varepsilon)\lVert g_j\rVert_{1,\infty}+C(\varepsilon)\lVert g-g_j\rVert_{1,\infty}$$
hence 
$$\lVert g\rVert_{1,\infty}\leqslant (1+\varepsilon)\liminf_{j\to +\infty}\lVert g_j\rVert_{1,\infty}\tag{2}.$$
Combining (1) and (2), we derive that for each positive $\varepsilon$, 
$$\frac 1{1+\varepsilon}\lVert g\rVert_{1,\infty}\leqslant\liminf_{j\to +\infty}\lVert g_j\rVert_{1,\infty}\leqslant \limsup_{j\to +\infty}\lVert g_j\rVert_{1,\infty}\leqslant (1+\varepsilon)\lVert g\rVert_{1,\infty},$$
which gives the wanted result. 

