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Proof that the space $L^1$ is complete in its metric

Completeness of $L^1$Convolution with a special approximation to the identity functionProblem with proof that positive infinite series are commuativeProof completeness of $L^p$Showing a metric space is completeProving weak compactness of $L^p$ when $p neq 1$ or $p neq infty$Proof that $L^{p}$ is complete in Folland's Real AnalysisHow can I prove this using a Combinatorial Proof?A problem in the proof of the completeness of $L^1$ space in stein <real analysis>The vector space $L^1$ is complete in its metric

2

$begingroup$

I don't understand a particular detail of the proof on page 70 of Stein and Shakarchi's Princeton Lectures in Analysis III.

We consider a Cauchy sequence ${f_n}$ in $L^1$ and choose a subsequence ${f_{n_k}}$ such that $||f_{n_k+1} - f_{n_k}||_{L^1} leq frac{1}{2^k}$.

Then, we consider a function $f := f_{n_1} + sum_{k=1}^{infty}(f_{n_k+1}-f_{n_k})$ and observe that $|f| leq |f_{n_1}| + sum_{k=1}^{infty} |f_{n_k+1} - f_{n_k}| =: g$. Everything makes sense so far.

The next steps are where I become uncertain about the specific details.

The proof then observes that

$$int|f_{n_1}| + sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq int|f_{n_1}| + sum_{k=1}^{infty}2^{-k} < infty$$

Here is the following questions I have:

1) How is $sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}|leq sum_{k=1}^{infty}2^{-k}$?

I know that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k} implies sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq sum_{k=1}^{infty} frac{1}{2^k} int dmu$, but that begs the question, what do we do with the $int dmu$? In fact, often times the book is so loose with the integration symbol at times that it's not even always clear what space we're integrating over. For instance, if we're integrating over $mathbb{R^d}$, then $int dmu = infty$, so I assume this can't possibly be the case. What am I missing here?

real-analysis proof-explanation complete-spaces

share|cite|improve this question

asked Mar 12 at 21:44

user516079user516079

388310

$endgroup$

add a comment | 

2

$begingroup$

I don't understand a particular detail of the proof on page 70 of Stein and Shakarchi's Princeton Lectures in Analysis III.

We consider a Cauchy sequence ${f_n}$ in $L^1$ and choose a subsequence ${f_{n_k}}$ such that $||f_{n_k+1} - f_{n_k}||_{L^1} leq frac{1}{2^k}$.

Then, we consider a function $f := f_{n_1} + sum_{k=1}^{infty}(f_{n_k+1}-f_{n_k})$ and observe that $|f| leq |f_{n_1}| + sum_{k=1}^{infty} |f_{n_k+1} - f_{n_k}| =: g$. Everything makes sense so far.

The next steps are where I become uncertain about the specific details.

The proof then observes that

$$int|f_{n_1}| + sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq int|f_{n_1}| + sum_{k=1}^{infty}2^{-k} < infty$$

Here is the following questions I have:

1) How is $sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}|leq sum_{k=1}^{infty}2^{-k}$?

I know that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k} implies sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq sum_{k=1}^{infty} frac{1}{2^k} int dmu$, but that begs the question, what do we do with the $int dmu$? In fact, often times the book is so loose with the integration symbol at times that it's not even always clear what space we're integrating over. For instance, if we're integrating over $mathbb{R^d}$, then $int dmu = infty$, so I assume this can't possibly be the case. What am I missing here?

real-analysis proof-explanation complete-spaces

share|cite|improve this question

asked Mar 12 at 21:44

user516079user516079

388310

$endgroup$

add a comment | 

$begingroup$

I don't understand a particular detail of the proof on page 70 of Stein and Shakarchi's Princeton Lectures in Analysis III.

We consider a Cauchy sequence ${f_n}$ in $L^1$ and choose a subsequence ${f_{n_k}}$ such that $||f_{n_k+1} - f_{n_k}||_{L^1} leq frac{1}{2^k}$.

Then, we consider a function $f := f_{n_1} + sum_{k=1}^{infty}(f_{n_k+1}-f_{n_k})$ and observe that $|f| leq |f_{n_1}| + sum_{k=1}^{infty} |f_{n_k+1} - f_{n_k}| =: g$. Everything makes sense so far.

The next steps are where I become uncertain about the specific details.

The proof then observes that

$$int|f_{n_1}| + sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq int|f_{n_1}| + sum_{k=1}^{infty}2^{-k} < infty$$

Here is the following questions I have:

1) How is $sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}|leq sum_{k=1}^{infty}2^{-k}$?

I know that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k} implies sum_{k=1}^{infty} int|f_{n_k+1}-f_{n_k}| leq sum_{k=1}^{infty} frac{1}{2^k} int dmu$, but that begs the question, what do we do with the $int dmu$? In fact, often times the book is so loose with the integration symbol at times that it's not even always clear what space we're integrating over. For instance, if we're integrating over $mathbb{R^d}$, then $int dmu = infty$, so I assume this can't possibly be the case. What am I missing here?

real-analysis proof-explanation complete-spaces

share|cite|improve this question

asked Mar 12 at 21:44

user516079user516079

388310

$endgroup$

share|cite|improve this question

asked Mar 12 at 21:44

user516079user516079

388310

share|cite|improve this question

asked Mar 12 at 21:44

user516079user516079

388310

asked Mar 12 at 21:44

user516079user516079

388310

asked Mar 12 at 21:44

user516079user516079

388310

1 Answer
1

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oldest

votes

1

$begingroup$

Your hypothesis is not that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}$, but that
$$
|f_{n_k+1} - f_{n_k}|_1 leq frac{1}{2^k}.
$$
That's exactly
$$
int |f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}.
$$

share|cite|improve this answer

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!! Thank you so much - I knew my confusion was likely due to a simple misunderstanding. I can't accept this answer until after 12 minutes have passed, but I will be sure to do so!
  $endgroup$
  – user516079
  Mar 12 at 21:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't worry. And, eventually, you'll get used to the "loose" notation for integrals.
  $endgroup$
  – Martin Argerami
  Mar 12 at 21:56
  

  
  
  
  

add a comment | 

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1

$begingroup$

Your hypothesis is not that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}$, but that
$$
|f_{n_k+1} - f_{n_k}|_1 leq frac{1}{2^k}.
$$
That's exactly
$$
int |f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}.
$$

share|cite|improve this answer

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!! Thank you so much - I knew my confusion was likely due to a simple misunderstanding. I can't accept this answer until after 12 minutes have passed, but I will be sure to do so!
  $endgroup$
  – user516079
  Mar 12 at 21:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't worry. And, eventually, you'll get used to the "loose" notation for integrals.
  $endgroup$
  – Martin Argerami
  Mar 12 at 21:56
  

  
  
  
  

add a comment | 

1

$begingroup$

Your hypothesis is not that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}$, but that
$$
|f_{n_k+1} - f_{n_k}|_1 leq frac{1}{2^k}.
$$
That's exactly
$$
int |f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}.
$$

share|cite|improve this answer

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh!! Thank you so much - I knew my confusion was likely due to a simple misunderstanding. I can't accept this answer until after 12 minutes have passed, but I will be sure to do so!
  $endgroup$
  – user516079
  Mar 12 at 21:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't worry. And, eventually, you'll get used to the "loose" notation for integrals.
  $endgroup$
  – Martin Argerami
  Mar 12 at 21:56
  

  
  
  
  

add a comment | 

$begingroup$

Your hypothesis is not that $|f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}$, but that
$$
|f_{n_k+1} - f_{n_k}|_1 leq frac{1}{2^k}.
$$
That's exactly
$$
int |f_{n_k+1} - f_{n_k}| leq frac{1}{2^k}.
$$

share|cite|improve this answer

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

share|cite|improve this answer

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184

answered Mar 12 at 21:46

Martin ArgeramiMartin Argerami

128k1184184