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I found this on a qualifier exam, and I think it will help me understand $L^p$ spaces better.

Let $f_n$ be a sequence of measurable function on a finite measure space. Suppose that $$\sup_n \int_X |f_n(x)|^2 d\mu < \infty$$ and that $\lim_{n\to \infty}f_n(x) =: f(x)$ exists $\mu$-almost everywhere. Which of the following are true (proving or providing a counterexample):

(1) $\int_X |f(x)|^2 d\mu < \infty$

(2) $ \int_X |f(x)| d\mu < \infty$

(3) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)|^2 d\mu = 0$

(4) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)| d\mu = 0$

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1 Answer 1

up vote 5 down vote accepted
  1. It's true by an application of Fatou's lemma to $\{|f_n|^2\}$, a sequence of non-negative measurable functions.

  2. It's true by Cauchy-Bunyakovsky-Schwarz inequality (and the fact that the measure space is finite).

  3. Take $f_n(x):=\sqrt n \chi_{(0,n^{—1})}(x)$ on $[0,1]$; then $\int_{[0,1]}|f_n|^2dx=1$, and $f_n\to 0$ almost everywhere.

  4. It's true: apply Egoroff theorem and Cauchy-Bunyakovsky-Schwarz inequality. We can assume that $f=0$ (otherwise we consider $g_n:=f_n-f$ instead of $f_n$, which is integrable and $g_n\to 0$ pointwise). Fix $\varepsilon>0$. Then we can find $C$ measurable such that $\sup_{x\in C}|f_n(x)|\to 0$ and $\mu(X\setminus C)\lt \varepsilon$. We have $$\int_X|f_n(x)|d\mu(x)\leqslant \mu(X)\sup_{x\in C}|f_n(x)|+\sqrt{\varepsilon}\sqrt{\int_X |f_n|^2}d\mu,$$ so $$\limsup_{n\to+\infty}\int_X|f_n(x)|d\mu(x)\leqslant \sqrt \varepsilon\sup_{k\in\Bbb N}\sqrt{\int_X |f_k|^2}d\mu.$$

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Can you explain why we may assume $f=0$ in part 4? –  analysis_qual May 1 at 21:57
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See edit. ${}{}{}$ –  Davide Giraudo May 1 at 22:16
    
I guess I'm having trouble seeing why replacing $f_n$ with $f_n - f$ preserves the assumption $\sup_n \| f_n \|_2^2 < \infty$. Is this obvious? –  analysis_qual May 1 at 22:41
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It's not so obvious: we use (1). –  Davide Giraudo May 2 at 5:57

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