Under given conditions whether $limlimits_{nto infty}...












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$begingroup$



Let ${f_n}_{n=1}^{infty}$ be a sequence of continuous real-valued functions defined on $mathbb R$ which converges pointwise to a continuous real-valued function $f$. Which of the following statements are true?



(i) If $0leq f_n leq f$ for all $nin mathbb N$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$



(ii) If $|f_n(t)|leq |sin t|$ for all $tin mathbb R$ and for all $nin mathbb N,$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$




(i) If $int_{-infty}^{infty} f<infty$, then we can use DOMINATED CONVERGENCE theorem and can say that the statement is true. But if $int_{-infty}^{infty}f=infty$ then what can we say about the statement?



(ii) I was not able to do this one.



Note: At the answer-key it's given that (i) is true but (ii) is false.










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$endgroup$

















    3












    $begingroup$



    Let ${f_n}_{n=1}^{infty}$ be a sequence of continuous real-valued functions defined on $mathbb R$ which converges pointwise to a continuous real-valued function $f$. Which of the following statements are true?



    (i) If $0leq f_n leq f$ for all $nin mathbb N$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$



    (ii) If $|f_n(t)|leq |sin t|$ for all $tin mathbb R$ and for all $nin mathbb N,$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$




    (i) If $int_{-infty}^{infty} f<infty$, then we can use DOMINATED CONVERGENCE theorem and can say that the statement is true. But if $int_{-infty}^{infty}f=infty$ then what can we say about the statement?



    (ii) I was not able to do this one.



    Note: At the answer-key it's given that (i) is true but (ii) is false.










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$



      Let ${f_n}_{n=1}^{infty}$ be a sequence of continuous real-valued functions defined on $mathbb R$ which converges pointwise to a continuous real-valued function $f$. Which of the following statements are true?



      (i) If $0leq f_n leq f$ for all $nin mathbb N$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$



      (ii) If $|f_n(t)|leq |sin t|$ for all $tin mathbb R$ and for all $nin mathbb N,$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$




      (i) If $int_{-infty}^{infty} f<infty$, then we can use DOMINATED CONVERGENCE theorem and can say that the statement is true. But if $int_{-infty}^{infty}f=infty$ then what can we say about the statement?



      (ii) I was not able to do this one.



      Note: At the answer-key it's given that (i) is true but (ii) is false.










      share|cite|improve this question











      $endgroup$





      Let ${f_n}_{n=1}^{infty}$ be a sequence of continuous real-valued functions defined on $mathbb R$ which converges pointwise to a continuous real-valued function $f$. Which of the following statements are true?



      (i) If $0leq f_n leq f$ for all $nin mathbb N$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$



      (ii) If $|f_n(t)|leq |sin t|$ for all $tin mathbb R$ and for all $nin mathbb N,$ then $displaystyle lim_{nto infty} int_{-infty}^{infty}f_n(t)dt=int_{-infty}^{infty}f(t)dt.$




      (i) If $int_{-infty}^{infty} f<infty$, then we can use DOMINATED CONVERGENCE theorem and can say that the statement is true. But if $int_{-infty}^{infty}f=infty$ then what can we say about the statement?



      (ii) I was not able to do this one.



      Note: At the answer-key it's given that (i) is true but (ii) is false.







      integration sequence-of-function






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      edited Jan 10 at 16:19









      Did

      248k23224463




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      asked Jan 10 at 16:10









      nurun neshanurun nesha

      1,0362623




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          (1) Suppose $0 le f_n le f$ and $int_{-infty}^infty f(t); dt = infty$. Given $N > 0$, there is $M$ such that $int_{-M}^M f(t); dt > N$. By dominated convergence $int_{-M}^M f_n(t); dt to int_{-M}^M f(t); dt$, so $int_{-infty}^infty f_n(t); dt ge int_{-M}^M f_n(t); dt > N$ for sufficiently large $n$.



          (2) Try $f_n(t) = sin(t)$ for $n pi < t < (n+1)pi$, $0$ otherwise.






          share|cite|improve this answer









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          • 3




            $begingroup$
            $f$ is stated to be a continuous real-valued function.
            $endgroup$
            – Robert Israel
            Jan 10 at 16:25













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          $begingroup$

          (1) Suppose $0 le f_n le f$ and $int_{-infty}^infty f(t); dt = infty$. Given $N > 0$, there is $M$ such that $int_{-M}^M f(t); dt > N$. By dominated convergence $int_{-M}^M f_n(t); dt to int_{-M}^M f(t); dt$, so $int_{-infty}^infty f_n(t); dt ge int_{-M}^M f_n(t); dt > N$ for sufficiently large $n$.



          (2) Try $f_n(t) = sin(t)$ for $n pi < t < (n+1)pi$, $0$ otherwise.






          share|cite|improve this answer









          $endgroup$









          • 3




            $begingroup$
            $f$ is stated to be a continuous real-valued function.
            $endgroup$
            – Robert Israel
            Jan 10 at 16:25


















          2












          $begingroup$

          (1) Suppose $0 le f_n le f$ and $int_{-infty}^infty f(t); dt = infty$. Given $N > 0$, there is $M$ such that $int_{-M}^M f(t); dt > N$. By dominated convergence $int_{-M}^M f_n(t); dt to int_{-M}^M f(t); dt$, so $int_{-infty}^infty f_n(t); dt ge int_{-M}^M f_n(t); dt > N$ for sufficiently large $n$.



          (2) Try $f_n(t) = sin(t)$ for $n pi < t < (n+1)pi$, $0$ otherwise.






          share|cite|improve this answer









          $endgroup$









          • 3




            $begingroup$
            $f$ is stated to be a continuous real-valued function.
            $endgroup$
            – Robert Israel
            Jan 10 at 16:25
















          2












          2








          2





          $begingroup$

          (1) Suppose $0 le f_n le f$ and $int_{-infty}^infty f(t); dt = infty$. Given $N > 0$, there is $M$ such that $int_{-M}^M f(t); dt > N$. By dominated convergence $int_{-M}^M f_n(t); dt to int_{-M}^M f(t); dt$, so $int_{-infty}^infty f_n(t); dt ge int_{-M}^M f_n(t); dt > N$ for sufficiently large $n$.



          (2) Try $f_n(t) = sin(t)$ for $n pi < t < (n+1)pi$, $0$ otherwise.






          share|cite|improve this answer









          $endgroup$



          (1) Suppose $0 le f_n le f$ and $int_{-infty}^infty f(t); dt = infty$. Given $N > 0$, there is $M$ such that $int_{-M}^M f(t); dt > N$. By dominated convergence $int_{-M}^M f_n(t); dt to int_{-M}^M f(t); dt$, so $int_{-infty}^infty f_n(t); dt ge int_{-M}^M f_n(t); dt > N$ for sufficiently large $n$.



          (2) Try $f_n(t) = sin(t)$ for $n pi < t < (n+1)pi$, $0$ otherwise.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 16:20









          Robert IsraelRobert Israel

          326k23215469




          326k23215469








          • 3




            $begingroup$
            $f$ is stated to be a continuous real-valued function.
            $endgroup$
            – Robert Israel
            Jan 10 at 16:25
















          • 3




            $begingroup$
            $f$ is stated to be a continuous real-valued function.
            $endgroup$
            – Robert Israel
            Jan 10 at 16:25










          3




          3




          $begingroup$
          $f$ is stated to be a continuous real-valued function.
          $endgroup$
          – Robert Israel
          Jan 10 at 16:25






          $begingroup$
          $f$ is stated to be a continuous real-valued function.
          $endgroup$
          – Robert Israel
          Jan 10 at 16:25




















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