Does $f^2inmathcal{L}^1(mathbb{R})$ imply $finmathcal{L}^1(mathbb{R})$? [closed]











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Suppose $f^2inmathcal{L}^1(mathbb{R})$.



i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.



Is it true that $finmathcal{L}^1(mathbb{R})$?






real-analysis integration lebesgue-integral lebesgue-measure lp-spaces







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closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41


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    – amWhy
    Nov 21 at 1:18















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Suppose $f^2inmathcal{L}^1(mathbb{R})$.



i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.



Is it true that $finmathcal{L}^1(mathbb{R})$?






real-analysis integration lebesgue-integral lebesgue-measure lp-spaces







share|cite|improve this question













closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom

If this question can be reworded to fit the rules in the help center, please edit the question.













  • What's the motivation for this question? Can you add your own thoughts on the question?
    – amWhy
    Nov 21 at 1:18













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Suppose $f^2inmathcal{L}^1(mathbb{R})$.



i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.



Is it true that $finmathcal{L}^1(mathbb{R})$?






real-analysis integration lebesgue-integral lebesgue-measure lp-spaces







share|cite|improve this question













Suppose $f^2inmathcal{L}^1(mathbb{R})$.



i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.



Is it true that $finmathcal{L}^1(mathbb{R})$?






real-analysis integration lebesgue-integral lebesgue-measure lp-spaces




real-analysis integration lebesgue-integral lebesgue-measure lp-spaces






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Nov 21 at 1:16









usmndj

484




484




closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom

If this question can be reworded to fit the rules in the help center, please edit the question.












  • What's the motivation for this question? Can you add your own thoughts on the question?
    – amWhy
    Nov 21 at 1:18


















  • What's the motivation for this question? Can you add your own thoughts on the question?
    – amWhy
    Nov 21 at 1:18
















What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18




What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18










2 Answers
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Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.



This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$



Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.






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    No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.






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      2 Answers
      2






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      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      up vote
      2
      down vote













      Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.



      This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$



      Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.






      share|cite|improve this answer

























        up vote
        2
        down vote













        Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.



        This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$



        Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.



          This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$



          Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.






          share|cite|improve this answer












          Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.



          This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$



          Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 at 1:31









          user3482749

          2,086414




          2,086414






















              up vote
              1
              down vote













              No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.






              share|cite|improve this answer

























                up vote
                1
                down vote













                No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.






                  share|cite|improve this answer












                  No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 1:30









                  Saucy O'Path

                  5,7341626




                  5,7341626















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