An example of a function in weighted Lipschitz class but not in Lipschitz class The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Non uniform continuity of a function and almost periodicityIntegrate a periodic absolute value functionOn the weak closedness of a closed ball with fixed $L^2$-norm in a periodic Sobolev spaceProjection of periodic trajectoriesApproximation of Lipschitz function in uniform normProve that the set of functions $sin kx$ is closed and boundedExample of nonseparable partial Lipschitz continuous gradientDoes locally Lipschitz imply Lipschitz on closed balls?Is it true that $int_a^a+Tf(x),dx ,=, int_0^Tf(x),dx $ for function $f$ with period $T$?If a function is log-lipschitz, then it is $alpha$-Hölder for all 0<$alpha$<1 and isn't Lipschitz continuous

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An example of a function in weighted Lipschitz class but not in Lipschitz class



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Non uniform continuity of a function and almost periodicityIntegrate a periodic absolute value functionOn the weak closedness of a closed ball with fixed $L^2$-norm in a periodic Sobolev spaceProjection of periodic trajectoriesApproximation of Lipschitz function in uniform normProve that the set of functions $sin kx$ is closed and boundedExample of nonseparable partial Lipschitz continuous gradientDoes locally Lipschitz imply Lipschitz on closed balls?Is it true that $int_a^a+Tf(x),dx ,=, int_0^Tf(x),dx $ for function $f$ with period $T$?If a function is log-lipschitz, then it is $alpha$-Hölder for all 0<$alpha$<1 and isn't Lipschitz continuous










2












$begingroup$


For $p ge 1,$ let $L^p[0,2pi]$ be the space of $2pi$-periodic measurable real valued functions. The norm in $L^p[0,2pi]$ is defined as



$$|f|_p = left(frac12piint_0^2pi|f(x)|^pdxright)^1/p.$$



The $W(L_p[0,2pi],beta), beta >0$ spaces are defined to contain $2pi$-periodic real valued functions such that



$$frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdx<infty.$$



The norm in $W(L_p[0,2pi],beta)$ is defined as



$$|f|_p,beta = left(frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdxright)^1/p.$$



Now I have to find an example of a function $fin W(L_p[0,2pi],beta)$ such that $fnotin L_p[0,2pi]$ and
$$|f(x+t)-f(x)|_p,beta le M t^alpha, quad0 < alpha le 1.$$
An example for $p=beta=alpha = 1$ will also be helpful. For $p=beta=alpha = 1,$ I tried with



$$f(x)=fracxsin(x/2), x in [0, 2pi]$$



but was unsuccessful. I was unable to control $sin((x+t)/2)$ in the denominator of the first term of the integral in the weighted norm.










share|cite|improve this question









$endgroup$











  • $begingroup$
    I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
    $endgroup$
    – Kore-N
    Apr 2 at 9:15











  • $begingroup$
    @Kore-N I have a similar belief but without concrete proof.
    $endgroup$
    – Birendra Singh
    Apr 2 at 10:30










  • $begingroup$
    I will write a rigorous proof below.
    $endgroup$
    – Kore-N
    Apr 2 at 11:04
















2












$begingroup$


For $p ge 1,$ let $L^p[0,2pi]$ be the space of $2pi$-periodic measurable real valued functions. The norm in $L^p[0,2pi]$ is defined as



$$|f|_p = left(frac12piint_0^2pi|f(x)|^pdxright)^1/p.$$



The $W(L_p[0,2pi],beta), beta >0$ spaces are defined to contain $2pi$-periodic real valued functions such that



$$frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdx<infty.$$



The norm in $W(L_p[0,2pi],beta)$ is defined as



$$|f|_p,beta = left(frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdxright)^1/p.$$



Now I have to find an example of a function $fin W(L_p[0,2pi],beta)$ such that $fnotin L_p[0,2pi]$ and
$$|f(x+t)-f(x)|_p,beta le M t^alpha, quad0 < alpha le 1.$$
An example for $p=beta=alpha = 1$ will also be helpful. For $p=beta=alpha = 1,$ I tried with



$$f(x)=fracxsin(x/2), x in [0, 2pi]$$



but was unsuccessful. I was unable to control $sin((x+t)/2)$ in the denominator of the first term of the integral in the weighted norm.










share|cite|improve this question









$endgroup$











  • $begingroup$
    I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
    $endgroup$
    – Kore-N
    Apr 2 at 9:15











  • $begingroup$
    @Kore-N I have a similar belief but without concrete proof.
    $endgroup$
    – Birendra Singh
    Apr 2 at 10:30










  • $begingroup$
    I will write a rigorous proof below.
    $endgroup$
    – Kore-N
    Apr 2 at 11:04














2












2








2





$begingroup$


For $p ge 1,$ let $L^p[0,2pi]$ be the space of $2pi$-periodic measurable real valued functions. The norm in $L^p[0,2pi]$ is defined as



$$|f|_p = left(frac12piint_0^2pi|f(x)|^pdxright)^1/p.$$



The $W(L_p[0,2pi],beta), beta >0$ spaces are defined to contain $2pi$-periodic real valued functions such that



$$frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdx<infty.$$



The norm in $W(L_p[0,2pi],beta)$ is defined as



$$|f|_p,beta = left(frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdxright)^1/p.$$



Now I have to find an example of a function $fin W(L_p[0,2pi],beta)$ such that $fnotin L_p[0,2pi]$ and
$$|f(x+t)-f(x)|_p,beta le M t^alpha, quad0 < alpha le 1.$$
An example for $p=beta=alpha = 1$ will also be helpful. For $p=beta=alpha = 1,$ I tried with



$$f(x)=fracxsin(x/2), x in [0, 2pi]$$



but was unsuccessful. I was unable to control $sin((x+t)/2)$ in the denominator of the first term of the integral in the weighted norm.










share|cite|improve this question









$endgroup$




For $p ge 1,$ let $L^p[0,2pi]$ be the space of $2pi$-periodic measurable real valued functions. The norm in $L^p[0,2pi]$ is defined as



$$|f|_p = left(frac12piint_0^2pi|f(x)|^pdxright)^1/p.$$



The $W(L_p[0,2pi],beta), beta >0$ spaces are defined to contain $2pi$-periodic real valued functions such that



$$frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdx<infty.$$



The norm in $W(L_p[0,2pi],beta)$ is defined as



$$|f|_p,beta = left(frac12piint_0^2pileft|f(x) sin^betaleft(fracx2right)right|^pdxright)^1/p.$$



Now I have to find an example of a function $fin W(L_p[0,2pi],beta)$ such that $fnotin L_p[0,2pi]$ and
$$|f(x+t)-f(x)|_p,beta le M t^alpha, quad0 < alpha le 1.$$
An example for $p=beta=alpha = 1$ will also be helpful. For $p=beta=alpha = 1,$ I tried with



$$f(x)=fracxsin(x/2), x in [0, 2pi]$$



but was unsuccessful. I was unable to control $sin((x+t)/2)$ in the denominator of the first term of the integral in the weighted norm.







lp-spaces periodic-functions lipschitz-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 31 at 5:17









Birendra SinghBirendra Singh

939




939











  • $begingroup$
    I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
    $endgroup$
    – Kore-N
    Apr 2 at 9:15











  • $begingroup$
    @Kore-N I have a similar belief but without concrete proof.
    $endgroup$
    – Birendra Singh
    Apr 2 at 10:30










  • $begingroup$
    I will write a rigorous proof below.
    $endgroup$
    – Kore-N
    Apr 2 at 11:04

















  • $begingroup$
    I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
    $endgroup$
    – Kore-N
    Apr 2 at 9:15











  • $begingroup$
    @Kore-N I have a similar belief but without concrete proof.
    $endgroup$
    – Birendra Singh
    Apr 2 at 10:30










  • $begingroup$
    I will write a rigorous proof below.
    $endgroup$
    – Kore-N
    Apr 2 at 11:04
















$begingroup$
I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
$endgroup$
– Kore-N
Apr 2 at 9:15





$begingroup$
I believe that if $f(cdot + t) in W(L^p, beta)$ for any $0 <t<epsilon$ and some $epsilon >0$, then $f in L^p$, because you can always translate it so that the sinus is bounded from below. In particular, I do not think that the $t-$translation of the function you wrote lies in the required space.
$endgroup$
– Kore-N
Apr 2 at 9:15













$begingroup$
@Kore-N I have a similar belief but without concrete proof.
$endgroup$
– Birendra Singh
Apr 2 at 10:30




$begingroup$
@Kore-N I have a similar belief but without concrete proof.
$endgroup$
– Birendra Singh
Apr 2 at 10:30












$begingroup$
I will write a rigorous proof below.
$endgroup$
– Kore-N
Apr 2 at 11:04





$begingroup$
I will write a rigorous proof below.
$endgroup$
– Kore-N
Apr 2 at 11:04











1 Answer
1






active

oldest

votes


















0





+50







$begingroup$

Fix $p geq 1, beta geq0, epsilon >0$. Suppose that $f(cdot + t) in W(L^p, beta)$ for all $0<t<epsilon$. We prove that $f in L^p$.



Indeed, it is enough to prove that for some $delta>0$ and any $a$



$$int_a^a + delta |f(x)|^p dx < +infty $$



(we count $a$, $a+delta$ modulo $2pi$). Let us choose $delta$ small enough (w.r.t. $epsilon)$ such that for any $a$ there exists a $0<t_a < epsilon$ such that



$$ sin(x -t_a) neq 0, forall x in [a, a+delta].$$



Then we can estimate for a constant $c>0$:



$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
    $endgroup$
    – Birendra Singh
    Apr 2 at 13:09










  • $begingroup$
    Edited: there was a constant missing.
    $endgroup$
    – Kore-N
    Apr 2 at 13:17










  • $begingroup$
    can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
    $endgroup$
    – Birendra Singh
    Apr 6 at 15:21











  • $begingroup$
    It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
    $endgroup$
    – Kore-N
    Apr 8 at 7:15












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0





+50







$begingroup$

Fix $p geq 1, beta geq0, epsilon >0$. Suppose that $f(cdot + t) in W(L^p, beta)$ for all $0<t<epsilon$. We prove that $f in L^p$.



Indeed, it is enough to prove that for some $delta>0$ and any $a$



$$int_a^a + delta |f(x)|^p dx < +infty $$



(we count $a$, $a+delta$ modulo $2pi$). Let us choose $delta$ small enough (w.r.t. $epsilon)$ such that for any $a$ there exists a $0<t_a < epsilon$ such that



$$ sin(x -t_a) neq 0, forall x in [a, a+delta].$$



Then we can estimate for a constant $c>0$:



$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
    $endgroup$
    – Birendra Singh
    Apr 2 at 13:09










  • $begingroup$
    Edited: there was a constant missing.
    $endgroup$
    – Kore-N
    Apr 2 at 13:17










  • $begingroup$
    can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
    $endgroup$
    – Birendra Singh
    Apr 6 at 15:21











  • $begingroup$
    It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
    $endgroup$
    – Kore-N
    Apr 8 at 7:15
















0





+50







$begingroup$

Fix $p geq 1, beta geq0, epsilon >0$. Suppose that $f(cdot + t) in W(L^p, beta)$ for all $0<t<epsilon$. We prove that $f in L^p$.



Indeed, it is enough to prove that for some $delta>0$ and any $a$



$$int_a^a + delta |f(x)|^p dx < +infty $$



(we count $a$, $a+delta$ modulo $2pi$). Let us choose $delta$ small enough (w.r.t. $epsilon)$ such that for any $a$ there exists a $0<t_a < epsilon$ such that



$$ sin(x -t_a) neq 0, forall x in [a, a+delta].$$



Then we can estimate for a constant $c>0$:



$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
    $endgroup$
    – Birendra Singh
    Apr 2 at 13:09










  • $begingroup$
    Edited: there was a constant missing.
    $endgroup$
    – Kore-N
    Apr 2 at 13:17










  • $begingroup$
    can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
    $endgroup$
    – Birendra Singh
    Apr 6 at 15:21











  • $begingroup$
    It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
    $endgroup$
    – Kore-N
    Apr 8 at 7:15














0





+50







0





+50



0




+50



$begingroup$

Fix $p geq 1, beta geq0, epsilon >0$. Suppose that $f(cdot + t) in W(L^p, beta)$ for all $0<t<epsilon$. We prove that $f in L^p$.



Indeed, it is enough to prove that for some $delta>0$ and any $a$



$$int_a^a + delta |f(x)|^p dx < +infty $$



(we count $a$, $a+delta$ modulo $2pi$). Let us choose $delta$ small enough (w.r.t. $epsilon)$ such that for any $a$ there exists a $0<t_a < epsilon$ such that



$$ sin(x -t_a) neq 0, forall x in [a, a+delta].$$



Then we can estimate for a constant $c>0$:



$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$






share|cite|improve this answer











$endgroup$



Fix $p geq 1, beta geq0, epsilon >0$. Suppose that $f(cdot + t) in W(L^p, beta)$ for all $0<t<epsilon$. We prove that $f in L^p$.



Indeed, it is enough to prove that for some $delta>0$ and any $a$



$$int_a^a + delta |f(x)|^p dx < +infty $$



(we count $a$, $a+delta$ modulo $2pi$). Let us choose $delta$ small enough (w.r.t. $epsilon)$ such that for any $a$ there exists a $0<t_a < epsilon$ such that



$$ sin(x -t_a) neq 0, forall x in [a, a+delta].$$



Then we can estimate for a constant $c>0$:



$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 2 at 13:17

























answered Apr 2 at 11:04









Kore-NKore-N

2,6521026




2,6521026











  • $begingroup$
    I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
    $endgroup$
    – Birendra Singh
    Apr 2 at 13:09










  • $begingroup$
    Edited: there was a constant missing.
    $endgroup$
    – Kore-N
    Apr 2 at 13:17










  • $begingroup$
    can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
    $endgroup$
    – Birendra Singh
    Apr 6 at 15:21











  • $begingroup$
    It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
    $endgroup$
    – Kore-N
    Apr 8 at 7:15

















  • $begingroup$
    I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
    $endgroup$
    – Birendra Singh
    Apr 2 at 13:09










  • $begingroup$
    Edited: there was a constant missing.
    $endgroup$
    – Kore-N
    Apr 2 at 13:17










  • $begingroup$
    can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
    $endgroup$
    – Birendra Singh
    Apr 6 at 15:21











  • $begingroup$
    It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
    $endgroup$
    – Kore-N
    Apr 8 at 7:15
















$begingroup$
I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
$endgroup$
– Birendra Singh
Apr 2 at 13:09




$begingroup$
I think the inequality $int_a^a + delta |f(x)|^p dx leq int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx$ is not valid as $-1leq sin(x) leq 1.$
$endgroup$
– Birendra Singh
Apr 2 at 13:09












$begingroup$
Edited: there was a constant missing.
$endgroup$
– Kore-N
Apr 2 at 13:17




$begingroup$
Edited: there was a constant missing.
$endgroup$
– Kore-N
Apr 2 at 13:17












$begingroup$
can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
$endgroup$
– Birendra Singh
Apr 6 at 15:21





$begingroup$
can you please explain Indeed, it is enough to prove that for some δ>0 and any a?
$endgroup$
– Birendra Singh
Apr 6 at 15:21













$begingroup$
It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
$endgroup$
– Kore-N
Apr 8 at 7:15





$begingroup$
It is enough, because we can estimate the total integral by a finite sum of small integrals (since $delta$ is fixed and $a$ is arbitrary). Every small integral can then be estimated in the way I proposed.
$endgroup$
– Kore-N
Apr 8 at 7:15


















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