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
$begingroup$
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.
lp-spaces periodic-functions lipschitz-functions
$endgroup$
add a comment |
$begingroup$
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.
lp-spaces periodic-functions lipschitz-functions
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
lp-spaces periodic-functions lipschitz-functions
$endgroup$
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
lp-spaces periodic-functions lipschitz-functions
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $c>0$:
$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$
$endgroup$
$begingroup$
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 isenoughto 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
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $c>0$:
$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$
$endgroup$
$begingroup$
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 isenoughto 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
add a comment |
$begingroup$
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 $c>0$:
$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$
$endgroup$
$begingroup$
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 isenoughto 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
add a comment |
$begingroup$
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 $c>0$:
$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$
$endgroup$
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 $c>0$:
$$ int_a^a + delta |f(x)|^p dx leq c int_a^a + delta |f(x) sin^beta (x -t_a)|^p dx leq c int_0^2pi |f(x+t_a) sin^beta (x)|^p < +infty.$$
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 isenoughto 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
add a comment |
$begingroup$
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 isenoughto 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
add a comment |
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$begingroup$
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