Integral of the product of two functions equals zero












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Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
$$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.






real-analysis calculus







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    0












    $begingroup$


    Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
    $$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.






    real-analysis calculus







    share|cite|improve this question









    $endgroup$















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      0





      $begingroup$


      Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
      $$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.






      real-analysis calculus







      share|cite|improve this question









      $endgroup$




      Let $alpha:[a,b]to mathbb{R}$ a continuous function such that
      $$ int_{a}^{b}alpha(x)h(x) ; dx=0$$ for any continuous function $h:[a,b]to mathbb{R}$ such that $h(a)=h(b)=0$. How can I prove that $alpha(x)=0$ for all $xin [a,b]$. Any help will be very appreciated.






      real-analysis calculus




      real-analysis calculus






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      asked Dec 19 '18 at 23:39









      Diego VargasDiego Vargas

      1227




      1227






















          3 Answers
          3






          active

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          3












          $begingroup$

          HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
            $endgroup$
            – Diego Vargas
            Dec 19 '18 at 23:46






          • 1




            $begingroup$
            Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
            $endgroup$
            – John Hughes
            Dec 19 '18 at 23:48



















          2












          $begingroup$

          Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.






            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









              3












              $begingroup$

              HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
                $endgroup$
                – Diego Vargas
                Dec 19 '18 at 23:46






              • 1




                $begingroup$
                Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
                $endgroup$
                – John Hughes
                Dec 19 '18 at 23:48
















              3












              $begingroup$

              HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
                $endgroup$
                – Diego Vargas
                Dec 19 '18 at 23:46






              • 1




                $begingroup$
                Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
                $endgroup$
                – John Hughes
                Dec 19 '18 at 23:48














              3












              3








              3





              $begingroup$

              HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?






              share|cite|improve this answer









              $endgroup$



              HINT: Suppose $alpha(x_0)ne 0$, and so $alpha$ is nonzero (say positive) on some interval $(x_0-delta,x_0+delta)$. Can you make up an appropriate function $h$ that will give you a nonzero integral?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 19 '18 at 23:42









              Ted ShifrinTed Shifrin

              64.7k44692




              64.7k44692












              • $begingroup$
                I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
                $endgroup$
                – Diego Vargas
                Dec 19 '18 at 23:46






              • 1




                $begingroup$
                Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
                $endgroup$
                – John Hughes
                Dec 19 '18 at 23:48


















              • $begingroup$
                I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
                $endgroup$
                – Diego Vargas
                Dec 19 '18 at 23:46






              • 1




                $begingroup$
                Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
                $endgroup$
                – John Hughes
                Dec 19 '18 at 23:48
















              $begingroup$
              I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
              $endgroup$
              – Diego Vargas
              Dec 19 '18 at 23:46




              $begingroup$
              I was thiking in that option but I can not find a continuous funcion $h$. The big trouble is that I can not find a function such that $h(a)=h(b)=0$
              $endgroup$
              – Diego Vargas
              Dec 19 '18 at 23:46




              1




              1




              $begingroup$
              Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
              $endgroup$
              – John Hughes
              Dec 19 '18 at 23:48




              $begingroup$
              Try taking any function you like and multiplying it by $(x-a)(x-b)$. :)
              $endgroup$
              – John Hughes
              Dec 19 '18 at 23:48











              2












              $begingroup$

              Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].






                  share|cite|improve this answer









                  $endgroup$



                  Let $h(x)=alpha (x)$ for $a+epsilon <x <b-epsilon$, $h(a)=h(b)=0$ and let $h$ have straight line graphs in $[a,a+epsilon]$ as well as $[b-epsilon,b]$. Write down the given equation and let $epsilon to 0$. You will get $int (alpha (x))^{2}), dx=0$ so $alpha =0$. [ Note that $h$ is bounded by $sup {|alpha (x)|: aleq x leq b}=M$ say. Hence $int _a ^{a+epsilon} |alpha (x) h(x)| , dxleq M^{2} epsilon to 0$. Similar argument for integral over $[b-epsilon,b]$].







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 23:50









                  Kavi Rama MurthyKavi Rama Murthy

                  72.6k53170




                  72.6k53170























                      0












                      $begingroup$

                      Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.






                          share|cite|improve this answer









                          $endgroup$



                          Consider proving this by contradiction, where you assume that $alpha (x_0) neq 0$ for some $x_0 in [a,b]$, then due to $alpha$ being continuous, this means there is some region (often called a "ball") around $x_0$ where $alpha$ is non-zero, even if $x_0$ is at a boundary, i.e., $a$ or $b$. Next, think about how that will affect the integration using any continuous function $h(x)$, especially if its values are very large or small in that region, especially compared to its values elsewhere in the range $[a,b]$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 19 '18 at 23:48









                          John OmielanJohn Omielan

                          4,6312215




                          4,6312215






























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