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Consider $f_n \to f$ a.e., $f_n \ge 0$, $f_n \in L^2(d\mu), f \in L^2(d\mu)$ and $\int |f_n|^2 \, d\mu \mathop{\longrightarrow}\limits_{n \to \infty} \int |f|^2 \, d\mu$.

Find an elementary argument to show that $f_n \to f$ in $L^2(d\mu)$.

We need to show that:

\begin{align*} \lim\limits_{n \to \infty} \lVert f_n - f \rVert_2 &= 0 \\ \lim\limits_{n \to \infty} \left(\lVert f_n - f \rVert_2\right)^2 &= 0 \\ \end{align*}

Starting with:

\begin{align*} \lim\limits_{n \to \infty} \left(\lVert f_n - f \rVert_2\right)^2 &= \lim\limits_{n \to \infty} \int (f_n - f)^2 \, d\mu \\ &= \lim\limits_{n \to \infty} \int (f_n^2 + f^2 - 2 f f_n) \, d\mu \\ &= \lim\limits_{n \to \infty} \int f_n^2 + \int f^2 - 2 \lim\limits_{n \to \infty} \int f f_n \, d\mu \\ \end{align*}

Since $f_n \to f$ in $L^2(d\mu)$, we can simplify:

\begin{align*} &= 2 \int f^2 - 2 \lim\limits_{n \to \infty} \int f f_n \, d\mu \\ \end{align*}

If $f_n \nearrow f$ we could use the monotone convergence theorem to conclude this proof. But we don't have that. I'm not sure how to conclude this.

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clay
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1 Answers1

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Indeed you can't use Beppo Levi, but you can use Fatou. Specifically, $$-2\lim_{n\to\infty} \int f_nf\,d\mu\le-2\int f^2\,d\mu$$ and thus $\lim_{n\to\infty} \lVert f_n-f\rVert^2\le 0$.

Added: Techincally, you chose for yourself the bad notation, and so now I have to rewrite everything.

\begin{align}\limsup_{n\to\infty} \lVert f-f_n\rVert^2&=\limsup_{n\to\infty}\int f_n^2+f^2-2ff_n\,d\mu=\\&=\lim_{n\to\infty} \int f_n^2+f^2\,d\mu+\limsup_{n\to\infty}\int -2ff_n\,d\mu=\\&=2\lVert f\rVert^2-2\liminf_{n\to\infty}\int f_nf\,d\mu\le\\ \text{(Fatou) }&\le 2\lVert f\rVert^2-2\int \liminf_{n\to\infty} f_nf\,d\mu=0\end{align}