L^p Spaces and Convergence of Functions

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This is a functional analysis qualifying exam problem that I can't figure out. Any assistance would be appreciated since I have to take a similar qual soon. I was able to make some limited progress in the p=2 case using Holders inequality.

Suppose f_n, f\in L^p where 1\le p <\infty and that f_n \rightarrow f a.e. Show that \|f_n-f\|_p \rightarrow 0 iff \|f_n\|_p \rightarrow \|f\|_p.
 
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Hi tornado28! :smile:

First reduce everything to the case p=1. That is, assume that you know the result for p=1, show that it also holds for p>1.

In reducing this case, you will need the inequality:

For x,y\geq 0, then |x-y|^p\leq |x^p-y^p|. Try to prove this.
 
tornado28 said:
This is a functional analysis qualifying exam problem that I can't figure out. Any assistance would be appreciated since I have to take a similar qual soon. I was able to make some limited progress in the p=2 case using Holders inequality.

Suppose f_n, f\in L^p where 1\le p <\infty and that f_n \rightarrow f a.e. Show that \|f_n-f\|_p \rightarrow 0 iff \|f_n\|_p \rightarrow \|f\|_p.
This is in fact a difficult question. I had the same when I took my analysis qual a few years ago. Here is how you prove it. First, notice that \|f_n\|_p =\| f_n-f+f \|_p \leq \| f_n-f \|_p + \| f\|_p This implies that \| f_n\|_p-\| f\|_p \leq \| f_n-f \|_p. Thus if lim(\| f_n-f \|_p)=0 then lim(\| f_n\|_p - \| f \|_p)=0
Next, to show the second part, use the fact that 2^p( |f_n|^p+|f|^p-|f_n - f|^p) \geq 0 and lim( 2^p( |f_n|^p+|f|^p-|f_n - f|^p))=2^{p+1}|f|^p and apply Fatou's lemma. I will let you finish the rest.
Vignon Oussa
 
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Ok, I understand why \|f_n-f\|_p \rightarrow 0 \Rightarrow \|f_n\|_p \rightarrow \|f\|_p, but I still don't have a solution for the other direction. In your solution, Vig, it is not necessarily true that 2^p( |f_n|^p+|f|^p-|f_n - f|^p) \geq 0 since we could, for instance, have f_n=1, f = -1, and p=2.

I think I could do it using Egorov's Theorem in the case that the underlying space is sigma compact, but since both of you seem to think that it can be done in general I wonder if I could have another hint.

I thought I had a proof using the inequality you suggested Micro, but I'm running into the same problem. It's not necessarily the case that |f_n - f|^p \le \left| |f_n|^p - |f|^p \right| when f_n and f have opposite signs.
 
tornado28 said:
Ok, I understand why \|f_n-f\|_p \rightarrow 0 \Rightarrow \|f_n\|_p \rightarrow \|f\|_p, but I still don't have a solution for the other direction. In your solution, Vig, it is not necessarily true that 2^p( |f_n|^p+|f|^p-|f_n - f|^p) \geq 0 since we could, for instance, have f_n=1, f = -1, and p=2.

I think I could do it using Egorov's Theorem in the case that the underlying space is sigma compact, but since both of you seem to think that it can be done in general I wonder if I could have another hint.

I thought I had a proof using the inequality you suggested Micro, but I'm running into the same problem. It's not necessarily the case that |f_n - f|^p \le \left| |f_n|^p - |f|^p \right| when f_n and f have opposite signs.

Ok, let's do this in the way I know works (I was trying to find an easy way out). Firstly, establish that

\lim_{n\rightarrow +\infty}{2^p(|f_n|^p+|f|^p)-|f_n-f|^p}=2^{p+1}|f|^p

Now, apply Fatou's lemma to calculate

\int{2^{p+1}|f|^pd\mu}

(note: to be able to apply Fatou's lemma, you'll need to know that 2^p(|f_n|^p+|f|^p)-|f_n-f|^p\geq 0. To show that this is the case, apply that the function \Phi(x)=x^p is convex)
 
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