Relation of absolute values inside integrals

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I have to prove the following:



$$||gu||_2 leq ||u||_2$$



where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.



Can I state the following?



$$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$



The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?







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    up vote
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    down vote

    favorite












    I have to prove the following:



    $$||gu||_2 leq ||u||_2$$



    where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.



    Can I state the following?



    $$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$



    The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?







    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have to prove the following:



      $$||gu||_2 leq ||u||_2$$



      where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.



      Can I state the following?



      $$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$



      The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?







      share|cite|improve this question











      I have to prove the following:



      $$||gu||_2 leq ||u||_2$$



      where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.



      Can I state the following?



      $$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$



      The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?









      share|cite|improve this question










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      asked Jul 24 at 8:49









      giovanni_13

      566




      566




















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          $1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.






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          • oh thank you very much!
            – giovanni_13
            Jul 24 at 8:57










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          up vote
          0
          down vote



          accepted










          $1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.






          share|cite|improve this answer





















          • oh thank you very much!
            – giovanni_13
            Jul 24 at 8:57














          up vote
          0
          down vote



          accepted










          $1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.






          share|cite|improve this answer





















          • oh thank you very much!
            – giovanni_13
            Jul 24 at 8:57












          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          $1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.






          share|cite|improve this answer













          $1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 24 at 8:54









          Kavi Rama Murthy

          20.2k2829




          20.2k2829











          • oh thank you very much!
            – giovanni_13
            Jul 24 at 8:57
















          • oh thank you very much!
            – giovanni_13
            Jul 24 at 8:57















          oh thank you very much!
          – giovanni_13
          Jul 24 at 8:57




          oh thank you very much!
          – giovanni_13
          Jul 24 at 8:57












           

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