Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.

Question

Given that $$\int_{\Omega} |f_n -f | \, d \mu \to 0,$$ $\mu(\Omega )=1$ and $ \|f_n\|_{L^2}^2 \leq M$, show $ \|f\|_{L^2}^2 \leq M$.

Attempt:

Note first that $f_n \in L^1(\Omega)$ since $$\|f_n \cdot 1\|_{L^1} \leq M^\frac{1}{2}$$ by C-S. Let $1>\epsilon > 0$. Let $N \in \mathbb N$ be such that, if $n \geq N$, $$\|f_n -f\|_{L^1} < \epsilon.$$ Since $f_n \to f$ in $L^1(\Omega)$, we have that, $$\|f\|_{L^1} \leq \|f_n-f\|_{L^1} + \|f_n\|_{L^1},$$ we see that $|f| \leq |f_n| + \epsilon$ a.e. by (i) the montonicity of the integral and (ii) since $\mu(\Omega) = 1$.

In particular, this is because $$\int_{\Omega} |f| \, d\mu \leq \int_{\Omega} |f_n| + \epsilon \, d\mu.$$

Thus, $$\|f\|_{L^2}^2 \leq \|f_n\|_{L^2}^2 + 2\epsilon\|f_n\|_{L^1} + \epsilon^2$$

Thus, we get $\|f\|_{L^2}^2 \leq M+ \epsilon$, since $\epsilon$ was arbitrary as desired.

Anything wrong with this proof? Any suggestions?

Answer 1

Since $f_n\xrightarrow{L^1(\Omega)} f$, there is a subsequence $(f_{n_k})$ of $(f_n)$ such that $f_{n_k} \rightarrow f$ pointwise a.e. on $\Omega$. By Fatou's lemma,

$$\|f\|_{L^2}^2 \le \varliminf_{k\to \infty} \|f_{n_k}\|_{L^2}^2\le \varliminf_{k\to \infty} M = M.$$