[Math] Prove $g_n f_n \rightarrow gf$ in $L_p([0,1])$. Where $f_n$ converges and $(g_n)_{n=1}^\infty $ is a sequence of bounded measurable functions

lebesgue-integralreal-analysis

Suppose that $ f_n \rightarrow f $ in $ L_p([0,1])$ with respect to the $|| . ||_p$ norm. $\{g_n\}_{n=1}^{\infty} $is a sequence of measurable functions such that $|g_n| \leq M$ for each $n$ and $g_n \rightarrow g$.

Work done so far:

$g_n \rightarrow g$ a.e. implies that $g_n^p \rightarrow g^p$ a.e.

Next since $|g_n|^p \leq M^p \in \mathbb{R} $ by Lebesgue Dominated Convergence Theorem we get
$\int_{[0,1]} g_n^p \rightarrow \int_{[0,1]} g^p. $

Then,
$\int_{[0,1]} |g_n f_n – gf|^p \leq \int_{[0,1]} |g_nf_n – g_nf|^p + \int_{[0,1]} |g_n f – gf|^p $ (by Minkowski's)

$= \int_{[0,1] }|g_n|^p|f_n-f|^p + \int_{[0,1]} |f|^p |g_n – g|^p$

So at this point I will put $ (\int_{[0,1]} |g_n f_n – gf|^p)^{1/p} = || g_n f_n – gf||$

And I can almost show that the last two terms go to 0, but am having trouble dealing with $|g_n|^p$ in the first term and $|f|^p$ in the second term.

Best Answer

For the first term, use the uniform boundedness of $(g_n)$ to write $$\Bigr(\int |g_n|^p|f_n-f|^p\Bigl)^{1/p}\le M\Bigl(\int |f_n-f|^p\Bigl)^{1/p}$$ and use the fact that $(f_n)$ converges to $f$ in $L_p$.

For the second term, perform some epsilonics using the following facts:

  1. Since $f\in L_p$ and since $(g_n)$ is uniformly bounded, for any $\epsilon>0$, there is a $\delta>0$ so that for any set $A$ of measure less than $\delta$, we have $\int_A |f|^p |g_n-g|^p <\epsilon$.
  2. For any $\delta>0$, there is a set $B$ of measure less than $\delta$ so that $(g_n)$ converges uniformly to $g$ on $B^c$. Note for such a $B$, on $B^c$ the quantity $|g_n-g|^p$ can be made uniformly small provided $n$ is sufficiently large.
  3. One may write

$$ \Bigl(\int |f|^p |g_n-g|^p\Bigr)^{1/p} =\Bigl(\int_B |f|^p |g_n-g|^p \,+\,\int_{B^c} |f|^p|g_n-g|^p\Bigr)^{1/p}. $$

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