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Are $L_p$ norm and discrete $L_p$ norm comparable?

If $fin L_p,varepsilon >0$ then exists simple function $phi$ where $|f-phi|_p<varepsilon$Norm operator and compactnessShow that the operator is bounded in $L_p$Why define norm in $L_p$ in that way?Moment and integral equalityIs the $l_p$-direct sum of uncountably many separable Banach spaces is separable?Algebra $A$ and its Gelfand spectrumIf $f in L_p$ and $g in L_infty$, show that $fg in L_p$.$L_p$ space inclusion for Riemann-Stieltjes integralConfused about using Cauchy sequence $(x_n)_1^infty in l_p$ to show the sequence space $l_p$ complete

0

$begingroup$

Are there any estimates on how a $L_p$ norm (say for a compact set in $mathbbR$) is related to a discrete $L_p$ norm, where we could for example consider the Jackson integral on this compact set. If necessary, we could for example work on the interval $[0,1]$ and then we are looking for any relationship between the integral beginequation left(intlimits_0^1 |f(x)|^p dx right)^1/p endequation and the integral beginequation left(intlimits_0^1 |f(x)|^p d_q x right)^1/p := left(frac1q-1sumlimits_i=0^infty q^i |f(q^i)|^p right)^1/p. endequation

functional-analysis measure-theory

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asked Mar 21 at 14:34

122333122333

275

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add a comment | 

0

$begingroup$

Are there any estimates on how a $L_p$ norm (say for a compact set in $mathbbR$) is related to a discrete $L_p$ norm, where we could for example consider the Jackson integral on this compact set. If necessary, we could for example work on the interval $[0,1]$ and then we are looking for any relationship between the integral beginequation left(intlimits_0^1 |f(x)|^p dx right)^1/p endequation and the integral beginequation left(intlimits_0^1 |f(x)|^p d_q x right)^1/p := left(frac1q-1sumlimits_i=0^infty q^i |f(q^i)|^p right)^1/p. endequation

functional-analysis measure-theory

share|cite|improve this question

asked Mar 21 at 14:34

122333122333

275

$endgroup$

add a comment | 

$begingroup$

Are there any estimates on how a $L_p$ norm (say for a compact set in $mathbbR$) is related to a discrete $L_p$ norm, where we could for example consider the Jackson integral on this compact set. If necessary, we could for example work on the interval $[0,1]$ and then we are looking for any relationship between the integral beginequation left(intlimits_0^1 |f(x)|^p dx right)^1/p endequation and the integral beginequation left(intlimits_0^1 |f(x)|^p d_q x right)^1/p := left(frac1q-1sumlimits_i=0^infty q^i |f(q^i)|^p right)^1/p. endequation

functional-analysis measure-theory

share|cite|improve this question

asked Mar 21 at 14:34

122333122333

275

$endgroup$

share|cite|improve this question

asked Mar 21 at 14:34

122333122333

275

share|cite|improve this question

asked Mar 21 at 14:34

122333122333

275

asked Mar 21 at 14:34

122333122333

275

asked Mar 21 at 14:34

122333122333

275

1 Answer
1

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1

$begingroup$

Let $f(x) = k sin(pi x)$, where $k in mathbbCsmallsetminus 0$. Then $ell_p(f) = 0$ and $L_p(f)$ can be as large as you like by adjusting $k$. On $[0,1]$, use $k chi_[0,1] smallsetminus mathbbQ$ (since $chi_S$ is the indicator function for the set $S$, this is $0$ on rationals, $k$ on irrationals).

To reverse the inequality, let $f$ be bumps of height $k$ and width $mathrme^-n$ for all $n in mathbbZ$ and zero otherwise. (Isosceles triangles are fine bumps, as is any other bump with compact support. It really doesn't matter how you deal with the overlaps near $n=0$.) On $[0,1]$, reverse the inequality with $f = k chi_mathbbQ$.

In other words, we can always arrange for a function to be zero on whatever set you like and nonzero off of it, so we can always "fool" the discrete norm, no matter what (uncountable) set you take as your domain, since that norm can only sample countably many points. (This requires the quantifiers of our game to be in the order "you tell me your norm(s), then I get to construct a function to fool it/them.")

share|cite|improve this answer

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this logic of "fooling" the discrete norm in the other way around? By this I mean that suppose that we have a function say $w$ that is given on this discrete sets of points but not in the intervals connecting them. If we then consider a polynomial say $p$, is it then possible to define our function w in such a way such that the $L_p$ norm of $wcdot p$ is close to the sup-norm of $wcdot p$ on the real line for example?
  $endgroup$
  – 122333
  Mar 26 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (Let's write $L_q$, so our notation isn't confused.) Let $S$ be the (countable) set of points sampled by your discrete metric. Since $mu(S) = 0$, $||w cdot p||_q = ||p||_q$ and $||w cdot p||_infty = ||p||_infty$. (If $p$ is a polynomial, these norms are both $infty$, which I guess makes them "close".) These two norms don't detect that we have altered $p$ on a set of measure zero, so nothing we do with $w$ matters.
  $endgroup$
  – Eric Towers
  Mar 27 at 10:57
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $f(x) = k sin(pi x)$, where $k in mathbbCsmallsetminus 0$. Then $ell_p(f) = 0$ and $L_p(f)$ can be as large as you like by adjusting $k$. On $[0,1]$, use $k chi_[0,1] smallsetminus mathbbQ$ (since $chi_S$ is the indicator function for the set $S$, this is $0$ on rationals, $k$ on irrationals).

To reverse the inequality, let $f$ be bumps of height $k$ and width $mathrme^-n$ for all $n in mathbbZ$ and zero otherwise. (Isosceles triangles are fine bumps, as is any other bump with compact support. It really doesn't matter how you deal with the overlaps near $n=0$.) On $[0,1]$, reverse the inequality with $f = k chi_mathbbQ$.

In other words, we can always arrange for a function to be zero on whatever set you like and nonzero off of it, so we can always "fool" the discrete norm, no matter what (uncountable) set you take as your domain, since that norm can only sample countably many points. (This requires the quantifiers of our game to be in the order "you tell me your norm(s), then I get to construct a function to fool it/them.")

share|cite|improve this answer

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this logic of "fooling" the discrete norm in the other way around? By this I mean that suppose that we have a function say $w$ that is given on this discrete sets of points but not in the intervals connecting them. If we then consider a polynomial say $p$, is it then possible to define our function w in such a way such that the $L_p$ norm of $wcdot p$ is close to the sup-norm of $wcdot p$ on the real line for example?
  $endgroup$
  – 122333
  Mar 26 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (Let's write $L_q$, so our notation isn't confused.) Let $S$ be the (countable) set of points sampled by your discrete metric. Since $mu(S) = 0$, $||w cdot p||_q = ||p||_q$ and $||w cdot p||_infty = ||p||_infty$. (If $p$ is a polynomial, these norms are both $infty$, which I guess makes them "close".) These two norms don't detect that we have altered $p$ on a set of measure zero, so nothing we do with $w$ matters.
  $endgroup$
  – Eric Towers
  Mar 27 at 10:57
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f(x) = k sin(pi x)$, where $k in mathbbCsmallsetminus 0$. Then $ell_p(f) = 0$ and $L_p(f)$ can be as large as you like by adjusting $k$. On $[0,1]$, use $k chi_[0,1] smallsetminus mathbbQ$ (since $chi_S$ is the indicator function for the set $S$, this is $0$ on rationals, $k$ on irrationals).

To reverse the inequality, let $f$ be bumps of height $k$ and width $mathrme^-n$ for all $n in mathbbZ$ and zero otherwise. (Isosceles triangles are fine bumps, as is any other bump with compact support. It really doesn't matter how you deal with the overlaps near $n=0$.) On $[0,1]$, reverse the inequality with $f = k chi_mathbbQ$.

In other words, we can always arrange for a function to be zero on whatever set you like and nonzero off of it, so we can always "fool" the discrete norm, no matter what (uncountable) set you take as your domain, since that norm can only sample countably many points. (This requires the quantifiers of our game to be in the order "you tell me your norm(s), then I get to construct a function to fool it/them.")

share|cite|improve this answer

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this logic of "fooling" the discrete norm in the other way around? By this I mean that suppose that we have a function say $w$ that is given on this discrete sets of points but not in the intervals connecting them. If we then consider a polynomial say $p$, is it then possible to define our function w in such a way such that the $L_p$ norm of $wcdot p$ is close to the sup-norm of $wcdot p$ on the real line for example?
  $endgroup$
  – 122333
  Mar 26 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (Let's write $L_q$, so our notation isn't confused.) Let $S$ be the (countable) set of points sampled by your discrete metric. Since $mu(S) = 0$, $||w cdot p||_q = ||p||_q$ and $||w cdot p||_infty = ||p||_infty$. (If $p$ is a polynomial, these norms are both $infty$, which I guess makes them "close".) These two norms don't detect that we have altered $p$ on a set of measure zero, so nothing we do with $w$ matters.
  $endgroup$
  – Eric Towers
  Mar 27 at 10:57
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x) = k sin(pi x)$, where $k in mathbbCsmallsetminus 0$. Then $ell_p(f) = 0$ and $L_p(f)$ can be as large as you like by adjusting $k$. On $[0,1]$, use $k chi_[0,1] smallsetminus mathbbQ$ (since $chi_S$ is the indicator function for the set $S$, this is $0$ on rationals, $k$ on irrationals).

To reverse the inequality, let $f$ be bumps of height $k$ and width $mathrme^-n$ for all $n in mathbbZ$ and zero otherwise. (Isosceles triangles are fine bumps, as is any other bump with compact support. It really doesn't matter how you deal with the overlaps near $n=0$.) On $[0,1]$, reverse the inequality with $f = k chi_mathbbQ$.

In other words, we can always arrange for a function to be zero on whatever set you like and nonzero off of it, so we can always "fool" the discrete norm, no matter what (uncountable) set you take as your domain, since that norm can only sample countably many points. (This requires the quantifiers of our game to be in the order "you tell me your norm(s), then I get to construct a function to fool it/them.")

share|cite|improve this answer

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

$endgroup$

share|cite|improve this answer

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370

answered Mar 21 at 14:52

Eric TowersEric Towers

33.4k22370