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I am having trouble with the following exercise:

Using $||a+b|-|a|-|b|| \leq 2 |b|$, prove that given a sequence $(f_n)_n \subset L^1(\Omega)$ with

  • $f_n(x) \to f(x)$ a.e.

  • $(f_n)$ is bounded in $L^1$, i.e. $\|f_n\|_1 \leq M$ for some $M\in \mathbb{R}$,

it follows that $f\in L^1$ and that $$\lim_{n\to\infty} \int > \{|f_n|-|f_n-f|\} = \int |f|.$$

(Set $a = f_n-f$ and $b = f$ in the inequality above.)

So far I managed to do the following: The inequality above is actually more specifically $|a|+|b| - |a+b| \leq 2\min\{a,b\}$ (because $|a+b|\leq |a|+|b|$). Then we set $a$ and $b$ as suggested and get

$|f_n-f| + |f| - |f_n| \leq 2\min\{|f_n-f|, |f|\}$,

hence in particular

$|f_n-f| + |f| - |f_n| \leq 2|f|$, which means that

$|f_n-f| - |f_n| \leq |f|$.

This looks pretty much like the term in the integral above, but it's also totally useless because the left hand side has the wrong sign (because $|f_n-f| \to 0$ pointwise) and it seems to state only that "this negative number is less or equal than some positive number". What I can salvage from that is the inequality (multiply by $(-1)$)

$-|f| \leq |f_n| -|f_n-f|$

and combining this with the triangle inequality ($|f_n| \leq |f_n-f| + |f|$), we get also an upper bound, hence

$-|f| \leq |f_n| -|f_n-f| \leq |f|$

But now I don't know what to do next.

Tera
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Philipp Wacker
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1 Answers1

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First, by Fatou's lemma, $\int |f| = \int \liminf\limits_{n \rightarrow \infty} |f_n| \leq \liminf\limits_{n\rightarrow \infty} |f_n| \leq M$. Thus, $f \in L^1(\Omega)$.

We have by the inequality that $||f_n| - |f_n-f| - |f|| \leq 2 |f|$. Then $$-2 |f| \leq |f_n| - |f_n-f| - |f| \leq 2 |f|,$$ so $$-|f| \leq |f_n| - |f_n - f|\leq 3 |f|,$$ so $$ ||f_n| - |f_n - f|| \leq 3 |f|.$$

Since we've already shown that $f \in L^1(\Omega)$, we now have that $(|f_n| - |f_n - f|)$ satisfies the hypotheses of the dominated convergence theorem. Thus, $$\lim\limits_{n\rightarrow \infty} \int (|f_n| - |f_n - f|) = \int \lim\limits_{n\rightarrow \infty} (|f_n| - |f_n - f|) = \int |f| $$

thanderhop
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