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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









$endgroup$

















    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









    $endgroup$















      2












      2








      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









      $endgroup$




      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




      real-analysis proof-explanation complete-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 12 at 21:44









      user516079user516079

      388310




      388310






















          1 Answer
          1






          active

          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









          $endgroup$









          • 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











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          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









          $endgroup$









          • 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
















          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









          $endgroup$









          • 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














          1












          1








          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









          $endgroup$



          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












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 12 at 21:46









          Martin ArgeramiMartin Argerami

          128k1184184




          128k1184184








          • 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














          • 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








          1




          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






          $begingroup$
          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




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


















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