4
$\begingroup$

Let $\{f_n\}$ be a sequence of Lebesgue integrable functions such that

(i) $\int_0^1|f_n(x)|^2 dx < 100$

(ii) $ f_n \to 0$ almost everywhere

We must show that

$$ \lim_{n\to\infty}\int_0^1 |f_n(x)| \, dx = 0$$

I have a solution using Egoroff and Schartz Inequality. Is that necessary? Any other ideas ? Also I prove it by myself without that. I will edit the post later.

$\endgroup$
  • 2
    $\begingroup$ You can also prove this using the fact that uniform $L^2$-boundedness implies uniform integrability. See for instance the Vitali Convergence Theorem. $\endgroup$ – Shalop Aug 19 '17 at 5:39
5
$\begingroup$

Hint: By the Cauchy-Schwarz inequality,

$$\begin{align*} \int_0^1 |f_n(x)| \, dx &= \int_0^1 1_{\{|f_n(x)| \leq \epsilon\}} |f_n(x)| \, dx + \int_0^1 1_{\{|f_n(x)|>\epsilon\}} |f_n(x)| \, dx \\ &\leq \epsilon + \sqrt{\int_0^1 1_{\{|f_n(x)|>\epsilon\}} \, dx} \cdot \underbrace{\sqrt{\int_0^1 f_n(x)^2 \, dx}}_{\leq 10}. \end{align*}$$

Now let first $n \to \infty$ and then $\epsilon \to 0$.

$\endgroup$
  • $\begingroup$ Yeah that's the same that I have. I mention that in the post. By the way is less or equal to 100 not 10. Thank you because you give more details. Why you want $\epsilon \to 0$ ? When I prove limits I just use the absolute value and prove it is less than epsilon and I never take limit to infinity and epsilon to 0. I guess is the same but I'm not used to. $\endgroup$ – Richard Clare Aug 19 '17 at 17:25
  • $\begingroup$ @RichardClare No, you didn't mention it in your post; you mentioned Egorov and Schwar(t)z ... there is no Egorov in my answer. Regarding your question: Yes, it's essentially the same... $\endgroup$ – saz Aug 19 '17 at 18:34
  • $\begingroup$ My bad, I have TWO solutions, one using Egoroff and the other using the Schartz. I Like this way of reasoning better. Thank you anyway. I will mark this as the correct answer. I notice that is 10 because you take the square root. $\endgroup$ – Richard Clare Aug 19 '17 at 18:47

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.