Continuous embedding of weighted Lebesgue space












0














Let $w$ belong to the class of Muckenhoupt weight $A_p$ for some $1<p<infty$ and define the weighted Lebsgue space $L^p(Omega,w):={u:Omegatomathbb{R} text{ measurable }: ||u||=(int_{Omega}|u|^pw(x),dx)^frac{1}{p}<infty}$.



One can prove that for any cube $Q$ in $mathbb{R}^N$ ($Ngeq 2$), there exist $s=s(A_p(w))>1$ such that for $q=frac{ps}{p+s-1}$, we have
$$
(frac{1}{|Q|}int_{Q}|f|^q,dx)^frac{1}{q}leq C(A_p(w))(frac{1}{w(Q)}int_{Q}|f|^p w,dx)^frac{1}{p},
$$
for every $fin L^p(Q,w)$ and for some constant $C=C(A_p(w))$ depending on the $A_p$ constant $A_p(w)$ of $w$.



My question is that does the above inequality imply that for any bounded smooth domain $Omega$ in $mathbb{R}^N$, one has
$$
(int_{Omega}|f|^q,dx)^frac{1}{q}leq C(int_{Omega}|f|^p w,dx)^frac{1}{p}
$$
for every $fin L^p(Q,w)$ with some constant $C$ independent of $f$.



Can you please help me regarding this.



This is mentioned in the paper (page 7) in the link : https://arxiv.org/pdf/1810.05061.pdf






functional-analysis analysis pde sobolev-spaces harmonic-analysis







share|cite|improve this question




















  • 1




    Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
    – Giuseppe Negro
    Nov 21 '18 at 8:18






  • 1




    By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
    – Mathlover
    Nov 21 '18 at 9:26






  • 1




    $(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
    – Mathlover
    Nov 21 '18 at 9:27








  • 1




    Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
    – Mathlover
    Nov 21 '18 at 9:33






  • 1




    Yes, it is. Best write it as an answer.
    – MaoWao
    Nov 21 '18 at 14:05
















0














Let $w$ belong to the class of Muckenhoupt weight $A_p$ for some $1<p<infty$ and define the weighted Lebsgue space $L^p(Omega,w):={u:Omegatomathbb{R} text{ measurable }: ||u||=(int_{Omega}|u|^pw(x),dx)^frac{1}{p}<infty}$.



One can prove that for any cube $Q$ in $mathbb{R}^N$ ($Ngeq 2$), there exist $s=s(A_p(w))>1$ such that for $q=frac{ps}{p+s-1}$, we have
$$
(frac{1}{|Q|}int_{Q}|f|^q,dx)^frac{1}{q}leq C(A_p(w))(frac{1}{w(Q)}int_{Q}|f|^p w,dx)^frac{1}{p},
$$
for every $fin L^p(Q,w)$ and for some constant $C=C(A_p(w))$ depending on the $A_p$ constant $A_p(w)$ of $w$.



My question is that does the above inequality imply that for any bounded smooth domain $Omega$ in $mathbb{R}^N$, one has
$$
(int_{Omega}|f|^q,dx)^frac{1}{q}leq C(int_{Omega}|f|^p w,dx)^frac{1}{p}
$$
for every $fin L^p(Q,w)$ with some constant $C$ independent of $f$.



Can you please help me regarding this.



This is mentioned in the paper (page 7) in the link : https://arxiv.org/pdf/1810.05061.pdf






functional-analysis analysis pde sobolev-spaces harmonic-analysis







share|cite|improve this question




















  • 1




    Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
    – Giuseppe Negro
    Nov 21 '18 at 8:18






  • 1




    By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
    – Mathlover
    Nov 21 '18 at 9:26






  • 1




    $(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
    – Mathlover
    Nov 21 '18 at 9:27








  • 1




    Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
    – Mathlover
    Nov 21 '18 at 9:33






  • 1




    Yes, it is. Best write it as an answer.
    – MaoWao
    Nov 21 '18 at 14:05














0












0








0







Let $w$ belong to the class of Muckenhoupt weight $A_p$ for some $1<p<infty$ and define the weighted Lebsgue space $L^p(Omega,w):={u:Omegatomathbb{R} text{ measurable }: ||u||=(int_{Omega}|u|^pw(x),dx)^frac{1}{p}<infty}$.



One can prove that for any cube $Q$ in $mathbb{R}^N$ ($Ngeq 2$), there exist $s=s(A_p(w))>1$ such that for $q=frac{ps}{p+s-1}$, we have
$$
(frac{1}{|Q|}int_{Q}|f|^q,dx)^frac{1}{q}leq C(A_p(w))(frac{1}{w(Q)}int_{Q}|f|^p w,dx)^frac{1}{p},
$$
for every $fin L^p(Q,w)$ and for some constant $C=C(A_p(w))$ depending on the $A_p$ constant $A_p(w)$ of $w$.



My question is that does the above inequality imply that for any bounded smooth domain $Omega$ in $mathbb{R}^N$, one has
$$
(int_{Omega}|f|^q,dx)^frac{1}{q}leq C(int_{Omega}|f|^p w,dx)^frac{1}{p}
$$
for every $fin L^p(Q,w)$ with some constant $C$ independent of $f$.



Can you please help me regarding this.



This is mentioned in the paper (page 7) in the link : https://arxiv.org/pdf/1810.05061.pdf






functional-analysis analysis pde sobolev-spaces harmonic-analysis







share|cite|improve this question















Let $w$ belong to the class of Muckenhoupt weight $A_p$ for some $1<p<infty$ and define the weighted Lebsgue space $L^p(Omega,w):={u:Omegatomathbb{R} text{ measurable }: ||u||=(int_{Omega}|u|^pw(x),dx)^frac{1}{p}<infty}$.



One can prove that for any cube $Q$ in $mathbb{R}^N$ ($Ngeq 2$), there exist $s=s(A_p(w))>1$ such that for $q=frac{ps}{p+s-1}$, we have
$$
(frac{1}{|Q|}int_{Q}|f|^q,dx)^frac{1}{q}leq C(A_p(w))(frac{1}{w(Q)}int_{Q}|f|^p w,dx)^frac{1}{p},
$$
for every $fin L^p(Q,w)$ and for some constant $C=C(A_p(w))$ depending on the $A_p$ constant $A_p(w)$ of $w$.



My question is that does the above inequality imply that for any bounded smooth domain $Omega$ in $mathbb{R}^N$, one has
$$
(int_{Omega}|f|^q,dx)^frac{1}{q}leq C(int_{Omega}|f|^p w,dx)^frac{1}{p}
$$
for every $fin L^p(Q,w)$ with some constant $C$ independent of $f$.



Can you please help me regarding this.



This is mentioned in the paper (page 7) in the link : https://arxiv.org/pdf/1810.05061.pdf






functional-analysis analysis pde sobolev-spaces harmonic-analysis




functional-analysis analysis pde sobolev-spaces harmonic-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 9:25

























asked Nov 21 '18 at 8:07









Mathlover

1388




1388








  • 1




    Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
    – Giuseppe Negro
    Nov 21 '18 at 8:18






  • 1




    By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
    – Mathlover
    Nov 21 '18 at 9:26






  • 1




    $(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
    – Mathlover
    Nov 21 '18 at 9:27








  • 1




    Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
    – Mathlover
    Nov 21 '18 at 9:33






  • 1




    Yes, it is. Best write it as an answer.
    – MaoWao
    Nov 21 '18 at 14:05














  • 1




    Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
    – Giuseppe Negro
    Nov 21 '18 at 8:18






  • 1




    By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
    – Mathlover
    Nov 21 '18 at 9:26






  • 1




    $(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
    – Mathlover
    Nov 21 '18 at 9:27








  • 1




    Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
    – Mathlover
    Nov 21 '18 at 9:33






  • 1




    Yes, it is. Best write it as an answer.
    – MaoWao
    Nov 21 '18 at 14:05








1




1




Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
– Giuseppe Negro
Nov 21 '18 at 8:18




Just encase your domain $Omega$ in a big cube, extend $f$ by zero and apply the inequality above, the one on cubes.
– Giuseppe Negro
Nov 21 '18 at 8:18




1




1




By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
– Mathlover
Nov 21 '18 at 9:26




By entending $f$ by zero in $QsetminusOmega$, I have got the following inequality,
– Mathlover
Nov 21 '18 at 9:26




1




1




$(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
– Mathlover
Nov 21 '18 at 9:27






$(int_{Omega}|f|^q w,dx)^frac{1}{q}leq C(A_p(w))frac{|Q|^frac{1}{q}}{w(Q)^frac{1}{p}}(int_{Omega}|f|^p w,dx)^frac{1}{p}$, where $w(Q)=int_{Q}w,dx$.
– Mathlover
Nov 21 '18 at 9:27






1




1




Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
– Mathlover
Nov 21 '18 at 9:33




Now, the right side constant is fixed for any fixed $w$ and positive, so I have the requried embedding. Is the argument correct? Thank you very much.
– Mathlover
Nov 21 '18 at 9:33




1




1




Yes, it is. Best write it as an answer.
– MaoWao
Nov 21 '18 at 14:05




Yes, it is. Best write it as an answer.
– MaoWao
Nov 21 '18 at 14:05















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