Proving $int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$












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I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.



Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.






calculus integration







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  • $begingroup$
    Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
    $endgroup$
    – Dzoooks
    Nov 30 '18 at 4:43
















1












$begingroup$


I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.



Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
    $endgroup$
    – Dzoooks
    Nov 30 '18 at 4:43














1












1








1





$begingroup$


I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.



Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.






calculus integration







share|cite|improve this question











$endgroup$




I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.



Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.






calculus integration




calculus integration






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edited Nov 30 '18 at 6:24









user1551

72.6k566127




72.6k566127










asked Nov 30 '18 at 4:39









dmsj djsldmsj djsl

35517




35517












  • $begingroup$
    Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
    $endgroup$
    – Dzoooks
    Nov 30 '18 at 4:43


















  • $begingroup$
    Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
    $endgroup$
    – Dzoooks
    Nov 30 '18 at 4:43
















$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43




$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43










3 Answers
3






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2












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You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.






share|cite|improve this answer









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    2












    $begingroup$

    Hint: The integral of a positive function is positive.






    share|cite|improve this answer









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    • 2




      $begingroup$
      I am curious why there is a downvote?
      $endgroup$
      – Tianlalu
      Nov 30 '18 at 4:57



















    1












    $begingroup$

    Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,



    $int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,



    a contradiction.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.






          share|cite|improve this answer









          $endgroup$



          You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 3:04









          YiFanYiFan

          3,4841425




          3,4841425























              2












              $begingroup$

              Hint: The integral of a positive function is positive.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                I am curious why there is a downvote?
                $endgroup$
                – Tianlalu
                Nov 30 '18 at 4:57
















              2












              $begingroup$

              Hint: The integral of a positive function is positive.






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                I am curious why there is a downvote?
                $endgroup$
                – Tianlalu
                Nov 30 '18 at 4:57














              2












              2








              2





              $begingroup$

              Hint: The integral of a positive function is positive.






              share|cite|improve this answer









              $endgroup$



              Hint: The integral of a positive function is positive.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 30 '18 at 4:41









              Jimmy R.Jimmy R.

              33.1k42157




              33.1k42157








              • 2




                $begingroup$
                I am curious why there is a downvote?
                $endgroup$
                – Tianlalu
                Nov 30 '18 at 4:57














              • 2




                $begingroup$
                I am curious why there is a downvote?
                $endgroup$
                – Tianlalu
                Nov 30 '18 at 4:57








              2




              2




              $begingroup$
              I am curious why there is a downvote?
              $endgroup$
              – Tianlalu
              Nov 30 '18 at 4:57




              $begingroup$
              I am curious why there is a downvote?
              $endgroup$
              – Tianlalu
              Nov 30 '18 at 4:57











              1












              $begingroup$

              Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,



              $int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,



              a contradiction.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,



                $int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,



                a contradiction.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,



                  $int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,



                  a contradiction.






                  share|cite|improve this answer











                  $endgroup$



                  Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,



                  $int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,



                  a contradiction.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 2 '18 at 22:52

























                  answered Nov 30 '18 at 4:53









                  Mustafa SaidMustafa Said

                  2,9611913




                  2,9611913






























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