# Convergence to zero in measure and bounded $L^1$ norm implies convergence of $\sqrt{f*g}$

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.

• Unfortunately you cannot show it, because it's not true in general. However it would be true if you had a sequence of functions which are bounded in $L^2$ and are converging to 0 in measure; so my advice is not to use this technique on $f_n$ but on $\sqrt f_n$ instead (for example, start from the case $g$ uniformly bounded) – Lucio Jan 26 '18 at 23:59
• and why do you think the square roots are defined? – zhw. Jan 27 '18 at 1:13
• zhw that was a question I came across, so I guess it is given that the square roots are defined as well, so f_n and g are integrable so in real analysis course we’ve learned that f_n g integrable as well, there for you can bound the integral of sqrt(fg) with the integral of |fg| which is finite and there for sqrt(fg) is integrable. Thank you Lucio and zhw. – Gil Or Jan 27 '18 at 8:05