Put-$X=\left\{f|f:\mathbb{R}^n\to \mathbb R,\mbox{Lebesgue, mesurable ,bounded with compact support}, f\in L^1\right\}$ ($X$ is a norm space as a sub-space of $L^1$).
Define $T(f):X\to \mathbb{R}$ $$T(f)=\int_{\mathbb{R}^n}x_1\cdot f(x) dx.$$
Update $T$ isn't continuous, in what conditions/changes on $X$ (or the norm type I use) it will be?
Note: The real assertation I need is that- Given $\{f_n\},\{g_n\}\in X$ such that $||f_n-g_n||_{L^1}\to 0$, does it holds that $|T(f_n)-T(g_n)|\to 0$ ? In what conditions/changes it will be?
Let $g_N=0$ and $f_N$ the indicator of $\left[N,N+a_N\right)\times \left(0,1\right)\times\dots\times \left(0,1\right)$ where $\left(a_N\right)_{N\geqslant 1}$ is a sequence which converges to $0$. Then the $\mathbb L^1$-norm of $f_N$ is $a_N$ and goes to zero as $N$ goes to infinity, but $$T \left(f_N \right)= \int_{N}^{N+a_N}x_1\mathrm dx_1=\frac{\left(N+a_N\right)^2-N^2 }2=Na_N+\frac{a_N^2}2.$$ Choose $a_N$ such that $Na_N\geqslant 1$ for any $N$.
Of course, endowing $X$ with the norm $\left\lVert \cdot\right\rVert$ defined by $\left\lVert f\right\rVert:=\int_{\mathbb R^n}\left\lvert x_1\right\rvert \left\lvert f\left(x\right)\right\rvert\mathrm dx$ make $T\colon \left(X,\left\lVert \cdot\right\rVert\right)\to\mathbb R$ continuous.