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Getting “rid” of $p in [1,infty[$ to show that $f_{n} in mathcal{L}^{p}[0,infty[$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)$mathcal{L}^p$ spaces and convolution$mathcal L^{infty}$ space propertiesDoes $Vert f-s_n Vert_infty to 0$ still hold for $fin C^0[a,b]$?Did B[0,1] is a closed subspace of $L^2[0,1]$$mathcal{BC}^n(X,E)$ is not closed in $mathcal{BC}(X,E)$Show that $Phi$ is a contraction with a maximum norm.One Estimate about $int_{mathbb{R}^n} frac{g(x+z)f(x+z)-g(x)f(x)- mathbf{1}_{|z|<1}nabla g(x)f(x)cdot z}{|z|^{n+2s}}dz$Show that $Ain mathcal{B}(X_{2},X_{1}).$ and Compute the norm $Vert AVert.$$lim_{n to infty}left(int_{a}^{b}phi^{2n}mathrm{d} alpha right)^{frac{1}{2}} = sup_{a leq x leq b} phi^{2}(x)$What to use for $lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$












0












$begingroup$


Let $p in [1,infty[$ and show that $f_{n} in mathcal{L}^{p}[0,infty[$



where $f_{n}(x):=frac{nsqrt{x} sin{(x)}}{1+nx^{2}}$ for $x in [0,infty[$



My idea:



Let $n in mathbb N$,



$vert vert f_{n} vert vert_{p}^{p}=int_{0}^{infty}vert frac{nsqrt{x} sin{(x)}}{1+nx^{2}}vert^{p} dx=int_{0}^{infty}vert frac{sqrt{x} sin{(x)}}{frac{1}{n}+x^{2}}vert^{p} dx=int_{0}^{infty} frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} dx$



Now I know that



$frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} 1_{[0,infty[}leq x^{frac{p}{2}}1_{[0,1[}+frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}$. Note that $frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}leq frac{vertsin^{p}{(x)}vert}{x^{p}}1_{[1,infty[}$



Now I know that $x^{frac{p}{2}}1_{[0,1[} in L^{p}$



And I remember that $int_{0}^{infty}frac{sin{(x)}}{x}dx=frac{pi}{2}$ but $int_{0}^{infty}frac{vertsin{(x)}vert}{x}dx$ is divergent, so obviously my bounds do not help me further. Any tips are appreciated.






real-analysis convergence lp-spaces







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $p in [1,infty[$ and show that $f_{n} in mathcal{L}^{p}[0,infty[$



    where $f_{n}(x):=frac{nsqrt{x} sin{(x)}}{1+nx^{2}}$ for $x in [0,infty[$



    My idea:



    Let $n in mathbb N$,



    $vert vert f_{n} vert vert_{p}^{p}=int_{0}^{infty}vert frac{nsqrt{x} sin{(x)}}{1+nx^{2}}vert^{p} dx=int_{0}^{infty}vert frac{sqrt{x} sin{(x)}}{frac{1}{n}+x^{2}}vert^{p} dx=int_{0}^{infty} frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} dx$



    Now I know that



    $frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} 1_{[0,infty[}leq x^{frac{p}{2}}1_{[0,1[}+frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}$. Note that $frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}leq frac{vertsin^{p}{(x)}vert}{x^{p}}1_{[1,infty[}$



    Now I know that $x^{frac{p}{2}}1_{[0,1[} in L^{p}$



    And I remember that $int_{0}^{infty}frac{sin{(x)}}{x}dx=frac{pi}{2}$ but $int_{0}^{infty}frac{vertsin{(x)}vert}{x}dx$ is divergent, so obviously my bounds do not help me further. Any tips are appreciated.






    real-analysis convergence lp-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $p in [1,infty[$ and show that $f_{n} in mathcal{L}^{p}[0,infty[$



      where $f_{n}(x):=frac{nsqrt{x} sin{(x)}}{1+nx^{2}}$ for $x in [0,infty[$



      My idea:



      Let $n in mathbb N$,



      $vert vert f_{n} vert vert_{p}^{p}=int_{0}^{infty}vert frac{nsqrt{x} sin{(x)}}{1+nx^{2}}vert^{p} dx=int_{0}^{infty}vert frac{sqrt{x} sin{(x)}}{frac{1}{n}+x^{2}}vert^{p} dx=int_{0}^{infty} frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} dx$



      Now I know that



      $frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} 1_{[0,infty[}leq x^{frac{p}{2}}1_{[0,1[}+frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}$. Note that $frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}leq frac{vertsin^{p}{(x)}vert}{x^{p}}1_{[1,infty[}$



      Now I know that $x^{frac{p}{2}}1_{[0,1[} in L^{p}$



      And I remember that $int_{0}^{infty}frac{sin{(x)}}{x}dx=frac{pi}{2}$ but $int_{0}^{infty}frac{vertsin{(x)}vert}{x}dx$ is divergent, so obviously my bounds do not help me further. Any tips are appreciated.






      real-analysis convergence lp-spaces







      share|cite|improve this question









      $endgroup$




      Let $p in [1,infty[$ and show that $f_{n} in mathcal{L}^{p}[0,infty[$



      where $f_{n}(x):=frac{nsqrt{x} sin{(x)}}{1+nx^{2}}$ for $x in [0,infty[$



      My idea:



      Let $n in mathbb N$,



      $vert vert f_{n} vert vert_{p}^{p}=int_{0}^{infty}vert frac{nsqrt{x} sin{(x)}}{1+nx^{2}}vert^{p} dx=int_{0}^{infty}vert frac{sqrt{x} sin{(x)}}{frac{1}{n}+x^{2}}vert^{p} dx=int_{0}^{infty} frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} dx$



      Now I know that



      $frac{x^{frac{p}{2}} vertsin^{p}{(x)}vert}{(frac{1}{n}+x^{2})^{p}} 1_{[0,infty[}leq x^{frac{p}{2}}1_{[0,1[}+frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}$. Note that $frac{vertsin^{p}{(x)}vert}{x^{frac{3}{2}p}}1_{[1,infty[}leq frac{vertsin^{p}{(x)}vert}{x^{p}}1_{[1,infty[}$



      Now I know that $x^{frac{p}{2}}1_{[0,1[} in L^{p}$



      And I remember that $int_{0}^{infty}frac{sin{(x)}}{x}dx=frac{pi}{2}$ but $int_{0}^{infty}frac{vertsin{(x)}vert}{x}dx$ is divergent, so obviously my bounds do not help me further. Any tips are appreciated.






      real-analysis convergence lp-spaces




      real-analysis convergence lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 25 at 23:46









      SABOYSABOY

      600311




      600311






















          1 Answer
          1






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          1












          $begingroup$

          We have



          $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|le left | frac{sqrt{x}}{x^{2}}right|=|x^{-3/2}|$.



          On the other hand, if $0<xle 1,$ then



          $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|=left|frac{nsqrt xcdot (x+O(x^3))}{1+nx^2}right|=left|frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}right|$,



          and these two facts combine to prove the claim.






          share|cite|improve this answer









          $endgroup$














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            1 Answer
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            active

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            We have



            $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|le left | frac{sqrt{x}}{x^{2}}right|=|x^{-3/2}|$.



            On the other hand, if $0<xle 1,$ then



            $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|=left|frac{nsqrt xcdot (x+O(x^3))}{1+nx^2}right|=left|frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}right|$,



            and these two facts combine to prove the claim.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We have



              $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|le left | frac{sqrt{x}}{x^{2}}right|=|x^{-3/2}|$.



              On the other hand, if $0<xle 1,$ then



              $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|=left|frac{nsqrt xcdot (x+O(x^3))}{1+nx^2}right|=left|frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}right|$,



              and these two facts combine to prove the claim.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We have



                $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|le left | frac{sqrt{x}}{x^{2}}right|=|x^{-3/2}|$.



                On the other hand, if $0<xle 1,$ then



                $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|=left|frac{nsqrt xcdot (x+O(x^3))}{1+nx^2}right|=left|frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}right|$,



                and these two facts combine to prove the claim.






                share|cite|improve this answer









                $endgroup$



                We have



                $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|le left | frac{sqrt{x}}{x^{2}}right|=|x^{-3/2}|$.



                On the other hand, if $0<xle 1,$ then



                $left | frac{nsqrt{x} sin{(x)}}{1+nx^{2}}right|=left|frac{nsqrt xcdot (x+O(x^3))}{1+nx^2}right|=left|frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}right|$,



                and these two facts combine to prove the claim.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 26 at 0:30









                MatematletaMatematleta

                12.3k21020




                12.3k21020






























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