real analysis - Convergence in $L^1$ and uniformly bounded $L^4$ norm implies $L^2$ convergence

Suppose that $f_n$ and $f$ are $L^1$ functions in a measure space $(X, \mu)$ with $\mu(X) < \infty$ such that $f_n \to f$ in $L^1$. Further, suppose that $$C = \sup_n \int |f_n|^4 < \infty.$$ That is, the $L^4$ norms of the $f_n$ are "uniformly bounded." I am told that this implies $f_n, f \in L^2$, and $f_n \to f$ in $L^2$, but I can't prove this last part.

I can prove $\int |f|^4 \leq C$. (Apply Fatou's lemma to a subsequence of $f_n$ which converges pointwise to $f$.) I can also prove that $f_n$ and $f$ are both $L^2$ (All $L^2$ functions with finite support are automatically $L_1$, and $f_n^2$ and $f^2$ are $L^2$.) I don't see how to deal with $||f_n - f||_2$. I don't see how the fourth powers are supposed to come into the picture.

From $\{f_n\}$ is uniformly integrable if and only if $\sup_n \int |f_n|\,d\mu < \infty$ and $\{f_n\}$ is uniformly absolutely continuous? I can see that my sequence is uniformly integrable, and I've vaguely heard of the Vitali convergence theorem, but I suspect that there is an easier way to prove this.

Answer

I don't think we need $\mu(X)<\infty.$

Lemma: Let $g:X\to [0,\infty)$ be measurable. Then
$\int_X g^2\,d\mu \le \int_X g\,d\mu +\int_X g^4\,d\mu.$

Proof: Let $A=\{x: g(x)\le 1\},$ $B =\{x: g(x)> 1\}.$ Then

$$\int_X g^2\,d\mu = \int_A g^2\,d\mu + \int_B g^2\,d\mu \le \int_A g\,d\mu + \int_B g^4\,d\mu\le \int_X g\,d\mu + \int_X g^4\,d\mu.$$

In our problem we have $f_n\to f$ in $L^1,$ which implies $\int_X |f_n|\,d\mu$ is uniformly bounded. We also know $\int_X |f_n|^4\,d\mu$ is uniformly bounded. By the lemma, $\int_X |f_n|^2\,d\mu$ is uniformly bounded.

As pointed out in the OP, the a.e. pointwise convergence of some subsequence $f_{n_k}$ to $f$ then shows $\int_X |f|^p\,d\mu<\infty$ for $p=1,2,4$ (Fatou's Lemma).

We thus have, by Cauchy-Schwartz,

$$\tag 1 \int_X |f_n-f|^2\,d\mu = \int_X |f_n-f|^{1/2}\cdot |f_n-f|^{3/2}\,d\mu$$ $$ \le \left ( \int_X |f_n-f|\,d\mu \right )^{1/2}\left ( \int_X |f_n-f|^3\,d\mu \right )^{1/2}.$$

Now

$$\int_X |f_n-f|^3\,d\mu = \int_X |f_n-f|\cdot |f_n-f|^2\,d\mu$$ $$ \le \left ( \int_X |f_n-f|^2\,d\mu \right )^{1/2}\left ( \int_X |f_n-f|^4\,d\mu \right )^{1/2}.$$

Insert the last inequality into $(1)$ and do some canceling to get

$$\left (\int_X |f_n-f|^2\,d\mu \right )^{3/4}\le \left ( \int_X |f_n-f|\,d\mu \right )^{1/2}\left ( \int_X |f_n-f|^4\,d\mu \right )^{1/4}.$$

On the right we have terms $\to 0$ times terms that are uniformly bounded. It follows that $\int_X |f_n-f|^2\,d\mu\to 0$ as desired.