I'm working on the following problem, I'm having trouble with the reverse direction. My question is bolded below. Also could someone check my forward direction?:
Let $(X, \mathcal{M}, \mu)$ be a $\sigma$ finite measure space and $\{f_n\},f \in L^P(X)$. Prove that $f_n \rightharpoonup f$ in $L^p(X)$ iff $\|f_n\|_p \leq c$ for all $n$ and $\int_A f_n\, d\mu \rightarrow \int_A f \, d\mu$ for all $A$ with $\mu(A) < \infty$.
For the reverse direction, we can use the characteristic functions in $L^q$ to build arbitrary functions in $L^q$ and use Monotone Convergence on $A$ equals a ball. Then increase the radius of the ball at each step making error $\epsilon/2^n$. However, I'm having trouble seeing how I use the boundedness of the sequence $f_n$)
(For the forward direction, choosing $\chi_{A}\in L^q(X)$ will get the integral condition and the $\|f_n\|_p$ were bounded because the sequence originally lived in $L^p(X)$.)
Best Answer
Here is how the boundedness of the sequence $\{f_n\}$ should enter your argument in the reverse direction:
Let $g\in L^q$. You want to show that $\int f_n g\, d\mu \rightarrow \int f g \,d\mu$. To this end, construct a sequence of simple functions $g_m$ such that $g_m\rightarrow g$ strongly in $L^q$ (I guess this is what you mean by "build"). You can then deduce
$$ \begin{split} \left|\int (f_n-f) g\, d\mu \right| &\leq \left|\int (f_n-f) (g-g_m)\, d\mu \right| + \left|\int (f_n-f) g_m\, d\mu\right|\\ &\leq \|f_n-f\|_p\|g-g_m\|_q + \left|\int (f_n-f) g_m\, d\mu\right| \\ &\leq (\|f_n\|_p+\|f\|_p)\|g-g_m\|_q + \left|\int (f_n-f) g_m\, d\mu\right| \\ &\leq (C+\|f\|_p)\|g-g_m\|_q+ \left|\int (f_n-f) g_m\, d\mu\right| <\epsilon \end{split} $$
for $n$ sufficiently large, provided $\|f_n\|_p$ is bounded by $C$. More precisely, you first choose $m$ sufficiently large so that the first term above is less than, say, $\frac{\epsilon}{2}$, then choose $n$ sufficiently large so that the second term above is less than $\frac{\epsilon}{2}$.
P.S. Your should include your assumptions on $p$. I am assuming $p\in(1,\infty)$.