5
$\begingroup$

a homework question from measure and integraiton theory course.

Suppose $f_n \in L_1(\mathbb R^d)$ for each $n\in\mathbb N$, $f_n\geq 0$ , $f_n\to f$ a.e and $\int f_n\to\int f<\infty$.
Prove that $\int|f_n-f| \to 0$

(Hint: ($f_n - f)_-\leq f$. Use dominated convergence theorem )

I am thinking that since $|f_n -f | \to 0$ a.e and $|f_n-f|=(f_n - f)_+ + (fn-f)_-$ . If I can show $(f_n-f)_+ ≤ f$ and $(f_n-f)_-≤f$ . Then since $f_n$ and $f$ are integrable , by DCT, $\int|f_n-f| \to \int0 =0$ . I think my approach might be wrong ..

$\endgroup$
1
$\begingroup$

Thomas E.'s approach is fine, but you can get this with a different use of the triangle inequality somewhat easier using the inequality $0 \leq |a - b| + |a| - |b| \leq 2|a|$. The proof of this is straightforward: We have $|a-b|+|a|-|b|\leq2|a|$ since $|a-b|\leq|a|+|b|$. Further, $||a|-|b||\leq|a-b|$ so that if $|a|\leq|b|$ then $0\leq|a-b|+|a|-|b|$. Similarly if $|b|\leq|a|$ then we have $0 \leq |a-b| + |a|-|b|$.

Apply the dominated convergence theorem to $0 \leq |f_n - f| + f - f_n \leq 2f$ (since in your case everything is positive) and the result follows.

$\endgroup$
3
$\begingroup$

Hint:

Going to positive and negative parts is not really necessary. Here's another type of approach that is also quite straight-forward.

Denote $g_{n}=|f|+|f_{n}|-|f-f_{n}|$ for all $n$, which are non-negative (by triangle-inequality) and measurable. Apply Fatou's Lemma and use the fact that $\liminf(-a_{n})=-\limsup(a_{n})$.

If I calculated them correctly, then what you should get is $$\liminf_{n\to\infty}\int g_{n}=\|f\|-\limsup_{n\to\infty} \|f-f_{n}\|$$ and $$\int \liminf_{n\to\infty} g_{n} =\|f\|,$$

where $\|\cdot\|$ is the $L^{1}$-norm. Do you see how this implies your result?

$\endgroup$
  • 2
    $\begingroup$ You can use \| for norm, as it has a correct spacing. In large quantities it even improves readability by quite a lot. $\endgroup$ – Asaf Karagila May 25 '12 at 15:23
  • $\begingroup$ Sure, didn't know that. Thanks. $\endgroup$ – T. Eskin May 25 '12 at 15:24
  • $\begingroup$ I have a small question but I haven't found an answer on the internet for it, so I hope you can please help. I was wondering if $u_n$ converges to $u$ in $L^2$, then does $u_nv$ converge to $uv$ if $v$ is a measurable function? $\endgroup$ – Fareed AF Sep 12 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.