Prove that $ \int_X |f_n|^2d\mu<\infty $ for all $ n\in\mathbb N $ 
Let $ (X,\mathcal M,\mu) $ be a measure space with $ \mu(X)<\infty $. Let $ (f_n)_{n=1}^\infty $ be a sequence of functions in $ L^1(X,\mathcal M,\mu) $, and let $ f\in L^1(X,\mathcal M,\mu) $. Suppose that
  $$ \lim_{n\to\infty}\int_X |f_n-f|d\mu=0 .$$
  Suppose also that $$ C:=\sup\left\{\int_X|f_n|^4d\mu:n\in\mathbb N\right\}<\infty .$$
  Prove that 
  
  
*
  
*$ \int_X |f|^4d\mu\le C $.
  
*$  \int_X |f_n|^2d\mu<\infty $ for all $ n\in\mathbb N $, and $ \int_X|f|^2d\mu<\infty $
  
*$ \lim\limits_{n\to\infty}\int_X|f_n-f|^2d\mu=0 .$

My attempt:


*

*Since $$\lim_{n\to\infty}\int_X |f_n-f|d\mu=0, $$ we know that $ f_n $
converges to $ f $ in measure, i.e., for every $ \epsilon>0, $
$$ \lim_{n\to\infty}\mu(\{x\in X:|f(x)-f_n(x)|\ge\epsilon\})=0. $$ Hence we can find a subsequence $ f_{n_k} $ of $ f_n $ such that $$ f_{n_k}\to f\quad a.e.\ \text{as}\ k\to\infty. $$ Now it suffices to prove that $$ \liminf_{k\to\infty}\int_X |f_{n_k}|^4d\mu\le C=\sup\left\{\int_X|f_n|^4d\mu:n\in\mathbb N\right\} $$
which is obviously true and we have applied the Fatou's lemma: $$ \int_X |f|^4d\mu=\int_X \lim_{k\to\infty}|f_{n_k}|^4d\mu\le\liminf_{k\to\infty}\int_X |f_{n_k}|^4d\mu .$$ And we are done!

*I am stuck on this one...... How to deal with $ |f_n|^2 $? I am thinking of Holder's inequalities, but nothing helps.
 A: $$\int |f_n|^2 d\mu\leq\sqrt{ \int d\mu}\cdot\sqrt{\int |f_n|^4 d\mu}$$
A: Use our favourite AM-GM inequality to get 
$$2|f_n|^2\leq |f_n|^4+1$$ and then integrate both sides to get $$2\int_X|f_n|^2d\mu\leq \int_X|f_n|^4d\mu +\mu(X)\leq C+\mu(X)$$
$$\implies sup\ _n \ \int_X|f_n|^2d\mu\leq \frac{C}{2}+\frac{\mu(X)}{2}<\infty$$
A: Notice that the assumption that $\mu(X) < \infty$ is not actually necessary for this part of the problem.  Specifically, write
$$|f_n| = |f_n|^{1/3}|f_n|^{2/3}$$
Then, by Holder's inequality,
$$\lVert f_n \rVert_{L^2} \leq \lVert f_n^{1/3} \rVert_{L^3} \lVert f_n^{2/3} \rVert_{L^6}$$
(where, for brevity, I've dropped the absolute value bars).  But,
$$\lVert f_n^{1/3}\rVert_{L^3} = \left(\int_X \left(f_n^{1/3}\right)^3\;d\mu\right)^{1/3} = \lVert f_n\rVert_{L^1}^{1/3}$$
and
$$\lVert f_n^{2/3}\rVert_{L^6} = \left(\int_X \left(f_n^{2/3}\right)^6\;d\mu\right)^{1/6} = \lVert f_n\rVert_{L^4}^{2/3}$$
So,
$$\lVert f_n \rVert_{L^2} \leq \lVert f_n\rVert_{L^1}^{1/3}\lVert f_n \rVert_{L^4}^{2/3}$$
The $L^4$ norm is bounded by assumption, and the $L^1$ norm is bounded based on the fact that $f_n$ is convergent in $L^1$.

This same estimate could also be derived more quickly by using Riesz-Thorin interpolation on the identity operator $I: L^p \to L^p$.  In particular, we have
$$\lVert I \rVert_{L^p \to L^p} = 1$$
(can you see why?  Once you write down the definition, this becomes pretty clear.)
Notice that $f_n$ lies in $L^1$ and $L^4$.  Taking $\theta = \frac{1}{3}$, we find
$$\frac{1}{2} = \frac{\theta}{1} + \frac{1-\theta}{4}$$
Thus, by Riesz-Thorin,
$$\lVert f_n \rVert_{L^2} = \lVert If_n \rVert_{L^2} \leq \lVert If_n \rVert_{L^1}^\theta  \lVert If_n \rVert_{L^4}^{1-\theta} = \lVert f_n \rVert_{L^1}^{1/3} \lVert f_n \rVert_{L^4}^{2/3}$$
