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.

  • 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

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 $\epsilon \to 0$.

  • $\begingroup$ 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

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