Proving a distribution limit with a non-integrable part in its integrand without Riemann-Lebesgue's lemmaCalculating $lim_lambda rightarrow infty T_fracsin(lambda x)x$ in $mathcalD'(mathbbR)$principal value as distribution, written as integral over singularityProof that limit goes to zero without Riemann-Lebesgue lemma$ lim_ varepsilon rightarrow 0^+ int_x frac varphi(x) xdx = - int_-infty^infty phi'(x) ln(|x|) dx$Why Dirac's Delta is not an ordinary function?The difference between Riemann integrable function and Lebesgue integrable functionEquations in D' (the space of distributions)Existence of the Dirac δ-function defined as a distribution?Density of tensor product of test functionsProving $f$ is not Riemann integrable but is Lesbesgue with definition involving simple step functionsCalculate $lim_ to +infty int_mathbbRf(x)|sin(lambda x)|dx $

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Proving a distribution limit with a non-integrable part in its integrand without Riemann-Lebesgue's lemma


Calculating $lim_lambda rightarrow infty T_fracsin(lambda x)x$ in $mathcalD'(mathbbR)$principal value as distribution, written as integral over singularityProof that limit goes to zero without Riemann-Lebesgue lemma$ lim_ varepsilon rightarrow 0^+ int_ geq varepsilon frac varphi(x) xdx = - int_-infty^infty phi'(x) ln(|x|) dx$Why Dirac's Delta is not an ordinary function?The difference between Riemann integrable function and Lebesgue integrable functionEquations in D' (the space of distributions)Existence of the Dirac δ-function defined as a distribution?Density of tensor product of test functionsProving $f$ is not Riemann integrable but is Lesbesgue with definition involving simple step functionsCalculate $lim_ int_mathbbRf(x)|sin(lambda x)|dx $













2












$begingroup$


Let $f_t(x) = fracsin txx$ and $varepsilon>0$. Further, consider a test function $varphi in C_0^infty(mathbb R)$ with $operatornamesupp varphi subseteq [-R, R]$ for some $R>0.$



I have proved that $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$$ when considering the limit $tto infty$. I want to prove $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) varphi(x) ; mathrm dx to 0$$ from this.



The problem is that $f_t notto 0$, so dominated convergence doesn't work. Moreover; $f_t$ is not absolutely integrable and, even more so, the integral $displaystyleint_mathbb R f_t(x) ; mathrm dx$ only exists in a generalized Riemann sense (with an undefined Lebesgue integral).



Can I prove this without using Riemann-Lebesgue's lemma (as they do here), if the limit $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$ is known?






real-analysis functional-analysis limits distribution-theory







share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Let $f_t(x) = fracsin txx$ and $varepsilon>0$. Further, consider a test function $varphi in C_0^infty(mathbb R)$ with $operatornamesupp varphi subseteq [-R, R]$ for some $R>0.$



    I have proved that $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$$ when considering the limit $tto infty$. I want to prove $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) varphi(x) ; mathrm dx to 0$$ from this.



    The problem is that $f_t notto 0$, so dominated convergence doesn't work. Moreover; $f_t$ is not absolutely integrable and, even more so, the integral $displaystyleint_mathbb R f_t(x) ; mathrm dx$ only exists in a generalized Riemann sense (with an undefined Lebesgue integral).



    Can I prove this without using Riemann-Lebesgue's lemma (as they do here), if the limit $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$ is known?






    real-analysis functional-analysis limits distribution-theory







    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Let $f_t(x) = fracsin txx$ and $varepsilon>0$. Further, consider a test function $varphi in C_0^infty(mathbb R)$ with $operatornamesupp varphi subseteq [-R, R]$ for some $R>0.$



      I have proved that $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$$ when considering the limit $tto infty$. I want to prove $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) varphi(x) ; mathrm dx to 0$$ from this.



      The problem is that $f_t notto 0$, so dominated convergence doesn't work. Moreover; $f_t$ is not absolutely integrable and, even more so, the integral $displaystyleint_mathbb R f_t(x) ; mathrm dx$ only exists in a generalized Riemann sense (with an undefined Lebesgue integral).



      Can I prove this without using Riemann-Lebesgue's lemma (as they do here), if the limit $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$ is known?






      real-analysis functional-analysis limits distribution-theory







      share|cite|improve this question









      $endgroup$




      Let $f_t(x) = fracsin txx$ and $varepsilon>0$. Further, consider a test function $varphi in C_0^infty(mathbb R)$ with $operatornamesupp varphi subseteq [-R, R]$ for some $R>0.$



      I have proved that $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$$ when considering the limit $tto infty$. I want to prove $$displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) varphi(x) ; mathrm dx to 0$$ from this.



      The problem is that $f_t notto 0$, so dominated convergence doesn't work. Moreover; $f_t$ is not absolutely integrable and, even more so, the integral $displaystyleint_mathbb R f_t(x) ; mathrm dx$ only exists in a generalized Riemann sense (with an undefined Lebesgue integral).



      Can I prove this without using Riemann-Lebesgue's lemma (as they do here), if the limit $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$ is known?






      real-analysis functional-analysis limits distribution-theory




      real-analysis functional-analysis limits distribution-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 10 at 19:25









      Markus KlyverMarkus Klyver

      407414




      407414




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          You can get the result integrating by parts. For $varepsilon<R$
          $$
          int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx =int_-R^-varepsilonf_t(x), varphi(x),mathrm dx+int_
          varepsilon^Rf_t(x,) varphi(x),mathrm dx.
          $$
          Now
          beginalign
          int_
          varepsilon^Rf_t(x), varphi(x),mathrm dx&=int_varepsilon^Rfracvarphi(x)x,sin(t,x),mathrm dx\
          &=-fracvarphi(x)x,fraccos(t,x)tBigr|_varepsilon^R+int_varepsilon^RBigl(fracvarphi(x)xBigr)'fraccos(t,x)t,mathrm dx.
          endalign






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
            $endgroup$
            – Markus Klyver
            2 days ago






          • 1




            $begingroup$
            No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
            $endgroup$
            – Julián Aguirre
            yesterday










          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You can get the result integrating by parts. For $varepsilon<R$
          $$
          int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx =int_-R^-varepsilonf_t(x), varphi(x),mathrm dx+int_
          varepsilon^Rf_t(x,) varphi(x),mathrm dx.
          $$
          Now
          beginalign
          int_
          varepsilon^Rf_t(x), varphi(x),mathrm dx&=int_varepsilon^Rfracvarphi(x)x,sin(t,x),mathrm dx\
          &=-fracvarphi(x)x,fraccos(t,x)tBigr|_varepsilon^R+int_varepsilon^RBigl(fracvarphi(x)xBigr)'fraccos(t,x)t,mathrm dx.
          endalign






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
            $endgroup$
            – Markus Klyver
            2 days ago






          • 1




            $begingroup$
            No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
            $endgroup$
            – Julián Aguirre
            yesterday















          1












          $begingroup$

          You can get the result integrating by parts. For $varepsilon<R$
          $$
          int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx =int_-R^-varepsilonf_t(x), varphi(x),mathrm dx+int_
          varepsilon^Rf_t(x,) varphi(x),mathrm dx.
          $$
          Now
          beginalign
          int_
          varepsilon^Rf_t(x), varphi(x),mathrm dx&=int_varepsilon^Rfracvarphi(x)x,sin(t,x),mathrm dx\
          &=-fracvarphi(x)x,fraccos(t,x)tBigr|_varepsilon^R+int_varepsilon^RBigl(fracvarphi(x)xBigr)'fraccos(t,x)t,mathrm dx.
          endalign






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
            $endgroup$
            – Markus Klyver
            2 days ago






          • 1




            $begingroup$
            No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
            $endgroup$
            – Julián Aguirre
            yesterday













          1












          1








          1





          $begingroup$

          You can get the result integrating by parts. For $varepsilon<R$
          $$
          int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx =int_-R^-varepsilonf_t(x), varphi(x),mathrm dx+int_
          varepsilon^Rf_t(x,) varphi(x),mathrm dx.
          $$
          Now
          beginalign
          int_
          varepsilon^Rf_t(x), varphi(x),mathrm dx&=int_varepsilon^Rfracvarphi(x)x,sin(t,x),mathrm dx\
          &=-fracvarphi(x)x,fraccos(t,x)tBigr|_varepsilon^R+int_varepsilon^RBigl(fracvarphi(x)xBigr)'fraccos(t,x)t,mathrm dx.
          endalign






          share|cite|improve this answer









          $endgroup$



          You can get the result integrating by parts. For $varepsilon<R$
          $$
          int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx =int_-R^-varepsilonf_t(x), varphi(x),mathrm dx+int_
          varepsilon^Rf_t(x,) varphi(x),mathrm dx.
          $$
          Now
          beginalign
          int_
          varepsilon^Rf_t(x), varphi(x),mathrm dx&=int_varepsilon^Rfracvarphi(x)x,sin(t,x),mathrm dx\
          &=-fracvarphi(x)x,fraccos(t,x)tBigr|_varepsilon^R+int_varepsilon^RBigl(fracvarphi(x)xBigr)'fraccos(t,x)t,mathrm dx.
          endalign







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Julián AguirreJulián Aguirre

          69.4k24096




          69.4k24096











          • $begingroup$
            Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
            $endgroup$
            – Markus Klyver
            2 days ago






          • 1




            $begingroup$
            No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
            $endgroup$
            – Julián Aguirre
            yesterday
















          • $begingroup$
            Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
            $endgroup$
            – Markus Klyver
            2 days ago






          • 1




            $begingroup$
            No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
            $endgroup$
            – Julián Aguirre
            yesterday















          $begingroup$
          Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
          $endgroup$
          – Markus Klyver
          2 days ago




          $begingroup$
          Can you generalize this to the case you don't know what $f_t$ is, i.e. just knowing $displaystyle int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x) ; mathrm dx to 0$?
          $endgroup$
          – Markus Klyver
          2 days ago




          1




          1




          $begingroup$
          No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
          $endgroup$
          – Julián Aguirre
          yesterday




          $begingroup$
          No without further conditions. Consider for instance $f_t(x)=tsin xchi_[-pi,pi](x)$. Then $int_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), mathrm dx=0$ for all $t$, but nothing can be said about $lim_ttoinftyint_mathbb Rsetminus[-varepsilon, varepsilon] f_t(x), varphi(x) , mathrm dx$.
          $endgroup$
          – Julián Aguirre
          yesterday

















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