$f_{n}$ converges in $L^{1}$ and $\|g_{n}\|_{L^{1}(\mathbb R)}\leq \|f_{n}\|_{L^{1}(\mathbb R)}\implies {g_{n}}$ converges in $L^{1}(\mathbb R)$?

Question

Let $f_{n}, g_{n}\in L^{1}(\mathbb R)$ (Lebesgue space). Suppose there exist $f\in L^{1}(\mathbb R)$ such that $\|f_{n}-f\|_{L^{1}(\mathbb R)}\to 0$ as $n\to \infty;$ that is, the sequence $\{f_{n}\}$ converges in $L^{1}(\mathbb R)$. Also, assume that, $\|g_{n}\|_{L^{1}(\mathbb R)}\leq \|f_{n}\|_{L^{1}(\mathbb R)}$, for all $n\in \mathbb N.$

My Question: What can we say about $g_{n}$; can we expect $g_{n}$ converges in $L^{1}(\mathbb R) $ ? Or, we can produce a counter example ?

Thanks,

Answer 1

There is no reason to expect that the $g_n$ converge, the analogous implication does also not hold in $\mathbb{R}$. For an - admittedly boring - explicit example, take $f_n = \chi_{[0,2]}$ for all $n$ (where $\chi_A$ is the characteristic function of the measurable set $A\subset \mathbb{R}$), and $g_n = \chi_{[n,n+1]}$. Evidently the constant sequence $f_n$ converges, but $\lVert g_n - g_m\rVert = 2$ for $m\neq n$.