Trig Integral Inequality












1












$begingroup$


Show that if $f$ is Riemann integrable on $[a,b]$, then



$$left(int_{a}^{b}f(x)sin x dxright)^2+left(int_{a}^{b}f(x)cos x dxright)^2le(b-a)int_{a}^{b}f^2(x) dx.$$



I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $left(int_{a}^{b}sin^2x+cos^2x dxright)left(int_{a}^{b}f^2(x) dxright)$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!






real-analysis integration inequality cauchy-schwarz-inequality holder-inequality







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  • $begingroup$
    What's $sin(x)^{2}+cos(x)^{2}$ equal to?
    $endgroup$
    – Brian Borchers
    Jan 21 at 16:13










  • $begingroup$
    The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
    $endgroup$
    – Song
    Jan 21 at 16:14


















1












$begingroup$


Show that if $f$ is Riemann integrable on $[a,b]$, then



$$left(int_{a}^{b}f(x)sin x dxright)^2+left(int_{a}^{b}f(x)cos x dxright)^2le(b-a)int_{a}^{b}f^2(x) dx.$$



I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $left(int_{a}^{b}sin^2x+cos^2x dxright)left(int_{a}^{b}f^2(x) dxright)$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!






real-analysis integration inequality cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's $sin(x)^{2}+cos(x)^{2}$ equal to?
    $endgroup$
    – Brian Borchers
    Jan 21 at 16:13










  • $begingroup$
    The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
    $endgroup$
    – Song
    Jan 21 at 16:14
















1












1








1





$begingroup$


Show that if $f$ is Riemann integrable on $[a,b]$, then



$$left(int_{a}^{b}f(x)sin x dxright)^2+left(int_{a}^{b}f(x)cos x dxright)^2le(b-a)int_{a}^{b}f^2(x) dx.$$



I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $left(int_{a}^{b}sin^2x+cos^2x dxright)left(int_{a}^{b}f^2(x) dxright)$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!






real-analysis integration inequality cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$




Show that if $f$ is Riemann integrable on $[a,b]$, then



$$left(int_{a}^{b}f(x)sin x dxright)^2+left(int_{a}^{b}f(x)cos x dxright)^2le(b-a)int_{a}^{b}f^2(x) dx.$$



I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $left(int_{a}^{b}sin^2x+cos^2x dxright)left(int_{a}^{b}f^2(x) dxright)$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!






real-analysis integration inequality cauchy-schwarz-inequality holder-inequality




real-analysis integration inequality cauchy-schwarz-inequality holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 16:54







ECON10105

















asked Jan 21 at 16:11









ECON10105ECON10105

62




62












  • $begingroup$
    What's $sin(x)^{2}+cos(x)^{2}$ equal to?
    $endgroup$
    – Brian Borchers
    Jan 21 at 16:13










  • $begingroup$
    The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
    $endgroup$
    – Song
    Jan 21 at 16:14




















  • $begingroup$
    What's $sin(x)^{2}+cos(x)^{2}$ equal to?
    $endgroup$
    – Brian Borchers
    Jan 21 at 16:13










  • $begingroup$
    The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
    $endgroup$
    – Song
    Jan 21 at 16:14


















$begingroup$
What's $sin(x)^{2}+cos(x)^{2}$ equal to?
$endgroup$
– Brian Borchers
Jan 21 at 16:13




$begingroup$
What's $sin(x)^{2}+cos(x)^{2}$ equal to?
$endgroup$
– Brian Borchers
Jan 21 at 16:13












$begingroup$
The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
$endgroup$
– Song
Jan 21 at 16:14






$begingroup$
The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz.
$endgroup$
– Song
Jan 21 at 16:14












2 Answers
2






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$begingroup$

By C-S $$(b-a)intlimits_a^bf(x)^2dx=intlimits_a^bf(x)^2sin^2xdxint_a^b1dx+intlimits_a^bf(x)^2cos^2xdxint_a^b1dxgeq$$
$$geqleft(intlimits_a^bf(x)sin{x}dxright)^2+left(intlimits_a^bf(x)cos{x}dxright)^2.$$
Actually, $f^2(x)=f(f(x))$ and $f(x)^2=left(f(x)right)^2.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    1.Prove that $<f,g>:=int_a^bf(x)g(x)dx$ is an inner product.



    2.Observe that the norm it induces is $|f|=big{(}int_a^bf^2(x)dxbig{)}^{1/2}$.



    3.Rewrite your inequality as $<f,sin>^2+<f,cos>^2leq(b-a)cdot|f|^2$



    4.Using Cauchy-Schwarz, show that $<f,sin>^2+<f,cos>^2 leq |f|^2(|sin|^2+|cos|^2)$



    5.Observe that $|sin|^2+|cos|^2=int_a^bsin(x)^2dx+int_a^bcos(x)^2dx=int_a^b1dx=b-a$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      By C-S $$(b-a)intlimits_a^bf(x)^2dx=intlimits_a^bf(x)^2sin^2xdxint_a^b1dx+intlimits_a^bf(x)^2cos^2xdxint_a^b1dxgeq$$
      $$geqleft(intlimits_a^bf(x)sin{x}dxright)^2+left(intlimits_a^bf(x)cos{x}dxright)^2.$$
      Actually, $f^2(x)=f(f(x))$ and $f(x)^2=left(f(x)right)^2.$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        By C-S $$(b-a)intlimits_a^bf(x)^2dx=intlimits_a^bf(x)^2sin^2xdxint_a^b1dx+intlimits_a^bf(x)^2cos^2xdxint_a^b1dxgeq$$
        $$geqleft(intlimits_a^bf(x)sin{x}dxright)^2+left(intlimits_a^bf(x)cos{x}dxright)^2.$$
        Actually, $f^2(x)=f(f(x))$ and $f(x)^2=left(f(x)right)^2.$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          By C-S $$(b-a)intlimits_a^bf(x)^2dx=intlimits_a^bf(x)^2sin^2xdxint_a^b1dx+intlimits_a^bf(x)^2cos^2xdxint_a^b1dxgeq$$
          $$geqleft(intlimits_a^bf(x)sin{x}dxright)^2+left(intlimits_a^bf(x)cos{x}dxright)^2.$$
          Actually, $f^2(x)=f(f(x))$ and $f(x)^2=left(f(x)right)^2.$






          share|cite|improve this answer









          $endgroup$



          By C-S $$(b-a)intlimits_a^bf(x)^2dx=intlimits_a^bf(x)^2sin^2xdxint_a^b1dx+intlimits_a^bf(x)^2cos^2xdxint_a^b1dxgeq$$
          $$geqleft(intlimits_a^bf(x)sin{x}dxright)^2+left(intlimits_a^bf(x)cos{x}dxright)^2.$$
          Actually, $f^2(x)=f(f(x))$ and $f(x)^2=left(f(x)right)^2.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 17:06









          Michael RozenbergMichael Rozenberg

          107k1894199




          107k1894199























              0












              $begingroup$

              1.Prove that $<f,g>:=int_a^bf(x)g(x)dx$ is an inner product.



              2.Observe that the norm it induces is $|f|=big{(}int_a^bf^2(x)dxbig{)}^{1/2}$.



              3.Rewrite your inequality as $<f,sin>^2+<f,cos>^2leq(b-a)cdot|f|^2$



              4.Using Cauchy-Schwarz, show that $<f,sin>^2+<f,cos>^2 leq |f|^2(|sin|^2+|cos|^2)$



              5.Observe that $|sin|^2+|cos|^2=int_a^bsin(x)^2dx+int_a^bcos(x)^2dx=int_a^b1dx=b-a$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                1.Prove that $<f,g>:=int_a^bf(x)g(x)dx$ is an inner product.



                2.Observe that the norm it induces is $|f|=big{(}int_a^bf^2(x)dxbig{)}^{1/2}$.



                3.Rewrite your inequality as $<f,sin>^2+<f,cos>^2leq(b-a)cdot|f|^2$



                4.Using Cauchy-Schwarz, show that $<f,sin>^2+<f,cos>^2 leq |f|^2(|sin|^2+|cos|^2)$



                5.Observe that $|sin|^2+|cos|^2=int_a^bsin(x)^2dx+int_a^bcos(x)^2dx=int_a^b1dx=b-a$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  1.Prove that $<f,g>:=int_a^bf(x)g(x)dx$ is an inner product.



                  2.Observe that the norm it induces is $|f|=big{(}int_a^bf^2(x)dxbig{)}^{1/2}$.



                  3.Rewrite your inequality as $<f,sin>^2+<f,cos>^2leq(b-a)cdot|f|^2$



                  4.Using Cauchy-Schwarz, show that $<f,sin>^2+<f,cos>^2 leq |f|^2(|sin|^2+|cos|^2)$



                  5.Observe that $|sin|^2+|cos|^2=int_a^bsin(x)^2dx+int_a^bcos(x)^2dx=int_a^b1dx=b-a$.






                  share|cite|improve this answer









                  $endgroup$



                  1.Prove that $<f,g>:=int_a^bf(x)g(x)dx$ is an inner product.



                  2.Observe that the norm it induces is $|f|=big{(}int_a^bf^2(x)dxbig{)}^{1/2}$.



                  3.Rewrite your inequality as $<f,sin>^2+<f,cos>^2leq(b-a)cdot|f|^2$



                  4.Using Cauchy-Schwarz, show that $<f,sin>^2+<f,cos>^2 leq |f|^2(|sin|^2+|cos|^2)$



                  5.Observe that $|sin|^2+|cos|^2=int_a^bsin(x)^2dx+int_a^bcos(x)^2dx=int_a^b1dx=b-a$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 17:11









                  JustDroppedInJustDroppedIn

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