Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$












1














Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.



$(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and



$$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$



$(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$



Answer:



$(i)$



Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.



Thus,



$ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.



Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have



$M-epsilon leq |f(x)|$.



Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then



$ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.



From $(1)$ and $(2)$, we have



$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.



I need confirmation of my work.



Also help me with part $(ii)$.






real-analysis integration







share|cite|improve this question



























    1














    Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.



    $(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and



    $$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$



    $(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$



    Answer:



    $(i)$



    Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.



    Thus,



    $ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.



    Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have



    $M-epsilon leq |f(x)|$.



    Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then



    $ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.



    From $(1)$ and $(2)$, we have



    $ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.



    I need confirmation of my work.



    Also help me with part $(ii)$.






    real-analysis integration







    share|cite|improve this question

























      1












      1








      1


      2





      Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.



      $(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and



      $$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$



      $(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$



      Answer:



      $(i)$



      Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.



      Thus,



      $ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.



      Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have



      $M-epsilon leq |f(x)|$.



      Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then



      $ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.



      From $(1)$ and $(2)$, we have



      $ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.



      I need confirmation of my work.



      Also help me with part $(ii)$.






      real-analysis integration







      share|cite|improve this question













      Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.



      $(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and



      $$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$



      $(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$



      Answer:



      $(i)$



      Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.



      Thus,



      $ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.



      Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have



      $M-epsilon leq |f(x)|$.



      Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then



      $ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.



      From $(1)$ and $(2)$, we have



      $ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.



      I need confirmation of my work.



      Also help me with part $(ii)$.






      real-analysis integration




      real-analysis integration






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      asked Nov 22 '18 at 22:16









      arifamatharifamath

      1176




      1176






















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          It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).






          share|cite|improve this answer





















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            It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).






            share|cite|improve this answer


























              5














              It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).






              share|cite|improve this answer
























                5












                5








                5






                It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).






                share|cite|improve this answer












                It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 '18 at 23:24









                Kavi Rama MurthyKavi Rama Murthy

                51.8k32055




                51.8k32055






























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