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If $f_n, f \in L^p, 1\leq p < +\infty$ and $f_n \rightarrow f$ almost everywhere, and $\|f_n\|_p \rightarrow \|f\|_p$, then $f_n\rightarrow f$ as for the norm.

Could you give me some hints how to show it??

What does convergence as for the norm mean??

EDIT:

From Fatou`s lemma we have that $$\int \lim \inf [2^{p-1}(|f_n|^p+|f|_p)-|f_n-f|^p]d\mu \leq \\ \lim \inf \int [2^{p-1}(|f_n|^p+|f|^p)-|f_n-f|^p]d\mu \\ \Rightarrow 2^{p-1}\int \lim \inf (|f_n|^p+|f|^p)d\mu+\int \lim \inf (-|f_n-f|^p)d\mu \leq 2^{p-1}(\lim \inf \int |f_n|^pd\mu +\lim \inf \int |f|^pd\mu )+\lim \inf (-\int |f_n-f|^p d\mu) \\ \Rightarrow 2^{p-1}[\int \lim \inf |f_n |^pd \mu+\int \lim \inf |f|^pd \mu]-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}[\lim \inf \int |f_n|^pd \mu+\lim \inf \int |f|^p d\mu]-\lim \sup \int |f_n-f|^pd \mu \ \ \ \ \ (*) $$

Knowing that $||f_n||_p\rightarrow ||f||_p \Rightarrow \left ( \int |f_n|^p\right )^{1/p}\rightarrow \left ( \int |f|^p\right )^{1p}$ , we have that $\lim \inf |f_n|^p=|f|^p$

Therefore, $$(*)\Rightarrow 2^{p-1}\int (|f|^pd \mu+\int |f|^pd\mu)-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}(\lim \inf \int |f_n|^pd \mu +\int |f|^pd \mu)-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow 2^{p-1}\int |f|^pd \mu-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}\lim \inf ||f_n||^p_p-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow 2^{p-1}||f||^p_p-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1} ||f||^p_p-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow \lim \sup \int |f_n-f|^pd \mu \leq \int \lim\sup |f_n-f|^pd \mu \\ \Rightarrow \lim \sup ||f_n-f||^p_p \leq \int \lim \sup |f_n-f|^pd \mu =0, \text{ since } f_n\rightarrow f \text{ almost everywhere } $$

So, we conclude that $||f_n-f||_p\rightarrow 0$.

Mary Star
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  • Maybe it means that $f_{n}$ converges to $f$ in norm, i.e., $||f_{n} - f||_{p} \to 0$ as $n \to \infty$. – layman Nov 12 '14 at 00:29
  • HINT: $2^p(|f_n|^p+|f|^p)-|f_n - f|^p \geq 0$ a.e. and apply Fatou's Lemma – aram Nov 12 '14 at 00:36
  • @Aram what does it mean that $f_n\rightarrow f$ almost everywhere?? Does it mean that $\mu({|f_n-f|>1/n})=0$ ?? – Mary Star Nov 12 '14 at 09:59
  • @MathIsHardNoItsNot what does it mean that $f_n\rightarrow f$ almost everywhere?? Does it mean that $\mu({|f_n-f|>1/n})=0$ ?? – Mary Star Nov 12 '14 at 10:00
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    @MaryStar Well, first you need to know what pointwise convergence is. We say a sequence of functions $f_{n}(x)$ converges pointwise to $f(x)$ if for each $x \in X$ and $\forall \epsilon > 0$, $\exists N$ such that $n \geq N$ implies $|f_{n}(x) - f(x)| < \epsilon$. Now, convergence almost everywhere means that the above is true for all $x \in X$ except for $x \in M$, where $M$ is some subset of $X$ of measure $0$. – layman Nov 12 '14 at 11:03
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    Whenever you see the words almost everywhere, it means the statement is true for all $x$ in the measure space, except for $x$ in a set of measure $0$. So, for example, if you say a function $g: X \to \mathbb{R}$ is continuous almost everywhere, then you are saying that there exists a set $M \subseteq X$ of measure $0$ ($M$ could be the emptyset) such that if $x \not \in M$, $g$ is continuous at $x$. – layman Nov 12 '14 at 11:05
  • @MathIsHardNoItsNot Ahaa... Ok!! So, we have that $$\mu({|f_n(x)-f(x)|\geq \epsilon })=0$$ right?? Then we have that $||f_n||_p\rightarrow ||f||_p$ that means that $\left (\int |f_n|^p d\mu \right )^{1/p} \rightarrow \left (\int |f_n|^p d\mu \right )^{1/p} $ right?? – Mary Star Nov 12 '14 at 11:38
  • @MaryStar What you need to do is show that $||f_{n} - f||{p} \to 0$ as $n \to \infty$. Why do we have $\mu({ x \mid |f{n} - f| \geq \epsilon }) = 0$? Take a look at the answer posted by mookid for the argument of why $\lim \limits_{n \to \infty} \left ( \int |f_{n} - f|^{p} ,d\mu \right )^{\frac{1}{p}} = 0$. – layman Nov 12 '14 at 13:47
  • @MathIsHardNoItsNot I edited my post... Could you tell me if I have understood it right?? – Mary Star Nov 12 '14 at 15:50

1 Answers1

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The convergence as for the norm means that $$ \| f - f_n\|_p \to 0 $$

Thanks to the Jensen inequality: $$ 2^{p-1} (|f(x)|^p + |f_n(x)|^p) - |f_n(x) - f(x)|^p \ge 0 $$ Now apply the Fatou theorem to get $$ \int \liminf \left[2^{p-1} (|f|^p + |f_n|^p) -|f_n - f|^p \right] \le \liminf\int \left[ 2^{p-1} \int (|f|^p + |f_n|^p) - \int |f_n - f|^p\right] $$ and after simplifications: $$ \limsup \int |f_n - f|^p \le \int \limsup |f_n - f|^p = 0 $$

mookid
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  • I edited post... Could you tell me if I have understood it right?? – Mary Star Nov 12 '14 at 15:51
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    @MaryStar seems good to me. be careful: $|f_n|_p\to |f|_p$ is a result on the integrals, not a pointwise result. – mookid Nov 12 '14 at 19:21
  • So have written it wrong?? Do you mean at the point where I wrote $||f_n||_p\rightarrow ||f||_p \Rightarrow \left ( |f_n|^p\right )^{1/p}\rightarrow \left ( |f|^p\right )^{1p}$? It should be $||f_n||_p\rightarrow ||f||_p \Rightarrow \left ( \int |f_n|^p\right )^{1/p}\rightarrow \left ( \int |f|^p\right )^{1p}$, right?? – Mary Star Nov 12 '14 at 21:54
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    @MaryStar indeed. – mookid Nov 12 '14 at 22:42
  • I changed it now at the initial post... Is it now correct?? – Mary Star Nov 12 '14 at 22:44
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    yes, it is fine now. last thing: "we have that $\liminf|f_n | _p =|f|_p$" because of a.s. convergence, not because of your justification. – mookid Nov 12 '14 at 22:46
  • How does it follows from a.s convergence that $\lim \inf |f_n|^p=|f|^p$ ?? I got stuck right now... – Mary Star Nov 12 '14 at 22:55
  • @MaryStar yes it does! in case of convergence $\lim = \liminf = \limsup$ – mookid Nov 12 '14 at 22:57
  • Doesn't almost everywhere convergence mean $|f_n-f|<\epsilon$ for almost all $x$ ?? How do we get $|f_n|$ and $|f|$?? Why doesn't it follow from $||f_n||_p \rightarrow ||f||_p$ ?? – Mary Star Nov 12 '14 at 23:01
  • ae convergence implies $f_n\to f$ except on a negligible set. So inside the integral, the negligible set cancels and remains $\int \limsup |f_n - f| = 0$ – mookid Nov 12 '14 at 23:04
  • Ok!! But I haven't understood why $\lim \inf |f_n|^p=|f|^p$ follows also from almost everywhere convergence and not from $||f_n||_p \rightarrow ||f||_p$... – Mary Star Nov 12 '14 at 23:08
  • this is continuous embedding. well, i think i have saif enough, it is your turn to put things together. – mookid Nov 12 '14 at 23:28
  • Ok.. I understand why $\lim \inf |f_n|^p=|f|^p$ follows from almost everywhere convergence. Do we get from almost everywhere convergence also that $$\lim \inf ||f_n||_p^p=||f||_p^p$$ ?? – Mary Star Dec 06 '14 at 23:38