Is this true that the sequences evenly sampled from functions in $L^p$ are in $l_p$?












0












$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







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$endgroup$












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12
















0












$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12














0












0








0





$begingroup$


Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces







share|cite|improve this question











$endgroup$




Is the following statement true ?




Given a function $f(x)$ in $L^p(mathbb{R})$, $1le p<infty$, the sequence ${f(x+kT)}_{kinmathbb{Z}}$ is in $l_p$ for any $x$ and $T$ if every number in the sequence is finite.







real-analysis sequences-and-series functional-analysis lp-spaces




real-analysis sequences-and-series functional-analysis lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 20:37









Davide Giraudo

126k16150261




126k16150261










asked Oct 13 '16 at 10:33









bridgerbridger

503315




503315












  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12


















  • $begingroup$
    $L_p$ consists of equivalence classes of a.e. equal functions...
    $endgroup$
    – user251257
    Oct 13 '16 at 10:40






  • 2




    $begingroup$
    No, "taking a sample" is unbounded operator in $L_p$.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 10:41












  • $begingroup$
    @user251257 True but what does this imply ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:04










  • $begingroup$
    @A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
    $endgroup$
    – bridger
    Oct 13 '16 at 11:06






  • 2




    $begingroup$
    @Hua It does not help. You may need $L_p$ bound on the derivative too.
    $endgroup$
    – A.Γ.
    Oct 13 '16 at 11:12
















$begingroup$
$L_p$ consists of equivalence classes of a.e. equal functions...
$endgroup$
– user251257
Oct 13 '16 at 10:40




$begingroup$
$L_p$ consists of equivalence classes of a.e. equal functions...
$endgroup$
– user251257
Oct 13 '16 at 10:40




2




2




$begingroup$
No, "taking a sample" is unbounded operator in $L_p$.
$endgroup$
– A.Γ.
Oct 13 '16 at 10:41






$begingroup$
No, "taking a sample" is unbounded operator in $L_p$.
$endgroup$
– A.Γ.
Oct 13 '16 at 10:41














$begingroup$
@user251257 True but what does this imply ?
$endgroup$
– bridger
Oct 13 '16 at 11:04




$begingroup$
@user251257 True but what does this imply ?
$endgroup$
– bridger
Oct 13 '16 at 11:04












$begingroup$
@A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
$endgroup$
– bridger
Oct 13 '16 at 11:06




$begingroup$
@A.G. Maybe I can add some mild conditions, e.g., continuous function in $L_p$ ?
$endgroup$
– bridger
Oct 13 '16 at 11:06




2




2




$begingroup$
@Hua It does not help. You may need $L_p$ bound on the derivative too.
$endgroup$
– A.Γ.
Oct 13 '16 at 11:12




$begingroup$
@Hua It does not help. You may need $L_p$ bound on the derivative too.
$endgroup$
– A.Γ.
Oct 13 '16 at 11:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30
















1












$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30














1












1








1





$begingroup$

This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.






share|cite|improve this answer









$endgroup$



This statement is not true for every $xinmathbb{R}$, however it is true for almost every $xinmathbb{R}$:



Suppose that there is a measurable $Esubsetmathbb{R}$ with $mathrm{Leb}(E)>0$ such that $(f(x+kT))_{k}notinell^p$ for all $xin E$. W.l.o.g. assume that $Ecap (T+E)=emptyset$. Then
$$
int_{mathbb R}|f(x)|^p,dxgeq sum_{kinmathbb Z}int_{kT+E}|f(x)|^p,dx=sum_{kinmathbb Z}int_E|f(x-kT)|^p,dx=int_Esum_{kinmathbb Z}|f(x-kT)|^p,dx=infty
$$
contradiciting the assumption $fin L^p$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 13 '16 at 13:30









MaoWaoMaoWao

3,153617




3,153617












  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30


















  • $begingroup$
    Why could you exchange the sum and integral in the second equality ?
    $endgroup$
    – bridger
    Oct 13 '16 at 14:08










  • $begingroup$
    That's an application of Fubini's theorem.
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:19










  • $begingroup$
    Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
    $endgroup$
    – bridger
    Oct 13 '16 at 16:24






  • 1




    $begingroup$
    There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
    $endgroup$
    – MaoWao
    Oct 13 '16 at 16:30
















$begingroup$
Why could you exchange the sum and integral in the second equality ?
$endgroup$
– bridger
Oct 13 '16 at 14:08




$begingroup$
Why could you exchange the sum and integral in the second equality ?
$endgroup$
– bridger
Oct 13 '16 at 14:08












$begingroup$
That's an application of Fubini's theorem.
$endgroup$
– MaoWao
Oct 13 '16 at 16:19




$begingroup$
That's an application of Fubini's theorem.
$endgroup$
– MaoWao
Oct 13 '16 at 16:19












$begingroup$
Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
$endgroup$
– bridger
Oct 13 '16 at 16:24




$begingroup$
Er, as far as I know, Fubini theorem applies when the two iterated integrals are bounded. But here you are suggesting they are not bounded. I don't think the theorem can apply here.
$endgroup$
– bridger
Oct 13 '16 at 16:24




1




1




$begingroup$
There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
$endgroup$
– MaoWao
Oct 13 '16 at 16:30




$begingroup$
There's also a version of Fubini's theorem for non-negative integrands (wikipedia calls it "Tonelli's theorem": en.wikipedia.org/wiki/…).
$endgroup$
– MaoWao
Oct 13 '16 at 16:30


















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