No, this is not true. The following argument is about spaces of finite measure, but it can easily be adapted to your question about $\mathbb R$.
Let $(X, m)$ be a space with $m(X) < \infty$. Take $g_i = g = 1$, and $f=0$. You are then asking the following:
if $f_i \in L^2 (X) \subseteq L^1 (X)$, and if $f_i \to 0$ in $L^1 (X)$, does it follow that $f_i \to 0$ in $L^2 (X)$?
No, it does not: take $X = [0,1] \subset \mathbb R$, and $f_i = i \; 1_{[0, \frac 1 {i^2}]}$ (with $1_A$ the indicator function of the subset $A$). Then $f_i \to 0$ in $L^1$ but $\| f_i \| _{L^2} = 1 \not\to 0$ in $L^2$.