Let's define $A:=\{f\in L^1(\Bbb R^n)\cap L^2(\Bbb R^n)\;:\;f\;\mbox{has compact support}\}$. So $A$ is dense in $L^1(\Bbb R^n)$.
Given then $f\in L^1(\Bbb R^n)$; by density there exists $\{f_j\}_j\subset A$ such that $f_j$ converges to $f$ in $L^1$.
Let now $T$ be a linear operator of the strong type $(2,2)$ (i.e. if $h\in L^2$ then $Tf\in L^2$ too and $||Th||_2\le C||h||_2$), such that if $h\in A$ then $Th\in L^{1,\infty}(\Bbb R^n)$ and $||Th||_{1,\infty}\le C||h||_1$.
Moreover $T$ has an off-diagonal kernel $K(x,y)$, i.e. $K\in L_{\mbox{loc}}^2(\Bbb R^n\times\Bbb R^n\setminus\{\mbox{diagonal}\})$ such that for every $f\in L^2(\Bbb R^n)$ with compact support we have $Tf(x)=\int_{\Bbb R^n}K(x,y)f(y)\,dy$ when $x\notin\operatorname{supp}(f)$.
Here $L^{p,\infty}(\Bbb R^n)$ denotes the weak $L^p$ space and $||\cdot||_{p,\infty}$ is its quasinorm; we know that $(L^{1,\infty}(\Bbb R^n),||\cdot||_{p,\infty})$ is complete.
I must prove that $\{Tf_j\}_j$ converges in $L^{1,\infty}(\Bbb R^n)$ to $Tf$.
I know such a sequence converges because it is a Cauchy sequence, in fact $$ ||Tf_j-Tf_k||_{1,\infty}=||T(f_j-f_k)||_{1,\infty} \stackrel{\mbox{hyp.}}{\le}C||f_j-f_k||_{1}\to0\;\;\mbox{as}\;\;j,k\to+\infty $$ thus, being $L^{1,\infty}(\Bbb R^n)$ complete (as already recalled) the sequence converges, i.e. there exists some $g\in L^{1,\infty}(\Bbb R^n)$ such that $||Tf_j-g||_{1,\infty}$ tends to $0$ as $j$ goes to infinity. How can I prove that $g=Tf$?
I tried different triangle inequalities but with no useful results (in $L^{1,\infty}$ triangle inequality holds up to a positive multiplicative constant, i.e. $||h_1+h_2||_{1,+\infty}\le C(||h_1||_{1,+\infty}+||h_2||_{1,+\infty})$).
Can someone help me please? Thanks