How to show that the following norms are equivalent?











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Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.



I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?






norm lp-spaces







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  • 1




    The first "norm" is not a norm, please put the modulus inside the integral.
    – dan_fulea
    Nov 23 at 10:54










  • The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
    – Arthur
    Nov 23 at 10:55










  • Looks like Mean Value Thm
    – Richard Martin
    Nov 23 at 10:55










  • what have you tried?
    – supinf
    Nov 23 at 10:57















up vote
0
down vote

favorite












Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.



I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?






norm lp-spaces







share|cite|improve this question




















  • 1




    The first "norm" is not a norm, please put the modulus inside the integral.
    – dan_fulea
    Nov 23 at 10:54










  • The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
    – Arthur
    Nov 23 at 10:55










  • Looks like Mean Value Thm
    – Richard Martin
    Nov 23 at 10:55










  • what have you tried?
    – supinf
    Nov 23 at 10:57













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.



I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?






norm lp-spaces







share|cite|improve this question















Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.



I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?






norm lp-spaces




norm lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 10:55









José Carlos Santos

147k22117218




147k22117218










asked Nov 23 at 10:51









Mittal G

1,192515




1,192515








  • 1




    The first "norm" is not a norm, please put the modulus inside the integral.
    – dan_fulea
    Nov 23 at 10:54










  • The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
    – Arthur
    Nov 23 at 10:55










  • Looks like Mean Value Thm
    – Richard Martin
    Nov 23 at 10:55










  • what have you tried?
    – supinf
    Nov 23 at 10:57














  • 1




    The first "norm" is not a norm, please put the modulus inside the integral.
    – dan_fulea
    Nov 23 at 10:54










  • The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
    – Arthur
    Nov 23 at 10:55










  • Looks like Mean Value Thm
    – Richard Martin
    Nov 23 at 10:55










  • what have you tried?
    – supinf
    Nov 23 at 10:57








1




1




The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54




The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54












The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55




The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55












Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55




Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55












what have you tried?
– supinf
Nov 23 at 10:57




what have you tried?
– supinf
Nov 23 at 10:57










2 Answers
2






active

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2
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accepted










Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?






share|cite|improve this answer





















  • What is the value of $K$?
    – Mittal G
    Nov 23 at 11:28






  • 1




    @MittalG $K=2$ will do.
    – José Carlos Santos
    Nov 23 at 11:29










  • @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
    – Mittal G
    Nov 23 at 11:30








  • 1




    If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
    – José Carlos Santos
    Nov 23 at 11:32










  • Great. Thanks a lot.
    – Mittal G
    Nov 23 at 11:38


















up vote
0
down vote













Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.



In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$






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    2 Answers
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    2 Answers
    2






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    up vote
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    down vote



    accepted










    Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?






    share|cite|improve this answer





















    • What is the value of $K$?
      – Mittal G
      Nov 23 at 11:28






    • 1




      @MittalG $K=2$ will do.
      – José Carlos Santos
      Nov 23 at 11:29










    • @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
      – Mittal G
      Nov 23 at 11:30








    • 1




      If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
      – José Carlos Santos
      Nov 23 at 11:32










    • Great. Thanks a lot.
      – Mittal G
      Nov 23 at 11:38















    up vote
    2
    down vote



    accepted










    Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?






    share|cite|improve this answer





















    • What is the value of $K$?
      – Mittal G
      Nov 23 at 11:28






    • 1




      @MittalG $K=2$ will do.
      – José Carlos Santos
      Nov 23 at 11:29










    • @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
      – Mittal G
      Nov 23 at 11:30








    • 1




      If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
      – José Carlos Santos
      Nov 23 at 11:32










    • Great. Thanks a lot.
      – Mittal G
      Nov 23 at 11:38













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?






    share|cite|improve this answer












    Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 23 at 11:04









    José Carlos Santos

    147k22117218




    147k22117218












    • What is the value of $K$?
      – Mittal G
      Nov 23 at 11:28






    • 1




      @MittalG $K=2$ will do.
      – José Carlos Santos
      Nov 23 at 11:29










    • @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
      – Mittal G
      Nov 23 at 11:30








    • 1




      If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
      – José Carlos Santos
      Nov 23 at 11:32










    • Great. Thanks a lot.
      – Mittal G
      Nov 23 at 11:38


















    • What is the value of $K$?
      – Mittal G
      Nov 23 at 11:28






    • 1




      @MittalG $K=2$ will do.
      – José Carlos Santos
      Nov 23 at 11:29










    • @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
      – Mittal G
      Nov 23 at 11:30








    • 1




      If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
      – José Carlos Santos
      Nov 23 at 11:32










    • Great. Thanks a lot.
      – Mittal G
      Nov 23 at 11:38
















    What is the value of $K$?
    – Mittal G
    Nov 23 at 11:28




    What is the value of $K$?
    – Mittal G
    Nov 23 at 11:28




    1




    1




    @MittalG $K=2$ will do.
    – José Carlos Santos
    Nov 23 at 11:29




    @MittalG $K=2$ will do.
    – José Carlos Santos
    Nov 23 at 11:29












    @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
    – Mittal G
    Nov 23 at 11:30






    @ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
    – Mittal G
    Nov 23 at 11:30






    1




    1




    If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
    – José Carlos Santos
    Nov 23 at 11:32




    If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
    – José Carlos Santos
    Nov 23 at 11:32












    Great. Thanks a lot.
    – Mittal G
    Nov 23 at 11:38




    Great. Thanks a lot.
    – Mittal G
    Nov 23 at 11:38










    up vote
    0
    down vote













    Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.



    In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
    $$ M
    =m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
    $$
    In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
    $$
    begin{aligned}
    |M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
    |m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
    end{aligned}
    $$
    Summing up:
    $$
    |f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.



      In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
      $$ M
      =m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
      $$
      In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
      $$
      begin{aligned}
      |M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
      |m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
      end{aligned}
      $$
      Summing up:
      $$
      |f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.



        In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
        $$ M
        =m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
        $$
        In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
        $$
        begin{aligned}
        |M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
        |m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
        end{aligned}
        $$
        Summing up:
        $$
        |f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
        $$






        share|cite|improve this answer












        Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.



        In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
        $$ M
        =m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
        $$
        In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
        $$
        begin{aligned}
        |M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
        |m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
        end{aligned}
        $$
        Summing up:
        $$
        |f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 17:40









        dan_fulea

        6,2401312




        6,2401312






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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