Let $\{f_n\}$ $\subset$ $L^1(E)$.
Show that if $f_n$ $\rightarrow$ $0$ in measure and $\sup||f_n||_1 < \infty$ then for every $g \in L^1(E)$:
$\int_E \sqrt{f_n(x) \cdot g(x)} dx \rightarrow 0$
I tried to tackle the question as follows:
Let $g \in L^1(E)$, then we can set C = $\int_E |g(x)|dx < \infty$.
Now, by the Cauchy-Schwarz inequality, we get:
$|\int_E \sqrt{|f_n(x) \cdot g(x)|}dx\ |^2 \leqslant \Bigl( \int_E \bigr (\sqrt{|f_n(x)|}\ \bigr)^2 \Bigl) \cdot \Bigl( \int_E \bigr (\sqrt{|g(x)|}\ \bigr)^2 \Bigl) = $
$=\Bigl( \int_E|f_n(x)|dx \Bigl) \cdot \Bigl( \int_E|g(x)|dx \Bigl)=C \cdot \Bigl( \int_E|f_n(x)|dx \Bigl)$
It is given that $\{f_n\} \rightarrow 0$ in measure and $\sup||f_n||_1 < \infty$. So from what I understand, if I can show that (*) for every $\epsilon>0$ there exists a measurable set $A$ with $\mu(A)<\infty$ such that $\bigl| \int_\text{E\A} f_n\bigl|<\epsilon$, then by Vitali's theorem, $\{f_n\}$ converges in $L_1$ to $0$. That is, $||f_n||_1 \rightarrow 0$
$\implies |\int_E \sqrt{|f_n(x) \cdot g(x)|}dx\ |^2 \leqslant C \cdot \Bigl( \int_E|f_n(x)|dx \Bigl) = C\cdot||f_n||_1 \rightarrow 0.$
As requested.
I'm having trouble showing (*), can anyone help me with this question? Maybe there is a different direction? I will appreciate it very much.