A class of weight function
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Let $sin [frac{1}{p-1},infty)cap(frac{N}{p},infty)$ for $1<p<N$. Assume $Omega$ is a smooth bounded domain in $mathbb{R}^N$ and $w(x)=|x|^alpha$. Then I want to prove that $w^{-s}in L^1(Omega)$ for any $alphain(-frac{N}{s},frac{N}{s})$.
I am using the following argument to proceed:
Let $0<alpha<frac{N}{s}$. Then we know $int_{Omega}w^{-s},dx=int_{Omega}|x|^{-alpha s},dx<infty$ if $alpha s<N$, which is assumed. Hence true.
Let $-frac{N}{s}<alphaleq 0$. Then $int_{Omega}|x|^{beta s},dx<infty$ if $beta s<N$ for $beta=-alpha geq 0$. That is $-alpha s<N$ implies $alpha>-frac{N}{s}$, which is also assumed. Hence true.
I think the argument is fine, my only confusion is that I am using the fact that $int_{Omega}|x|^alpha,dx<infty$ if $alpha<N$ for any bounded smooth domain (may contain $0$)) in $mathbb{R}^N$.
Can you kindly help to observe if the argument is correct or not.
Thanks in advance.
real-analysis complex-analysis functional-analysis analysis
asked Nov 18 at 7:46
Mathlover
1088
add a comment |
up vote
0
down vote
favorite
Let $sin [frac{1}{p-1},infty)cap(frac{N}{p},infty)$ for $1<p<N$. Assume $Omega$ is a smooth bounded domain in $mathbb{R}^N$ and $w(x)=|x|^alpha$. Then I want to prove that $w^{-s}in L^1(Omega)$ for any $alphain(-frac{N}{s},frac{N}{s})$.
I am using the following argument to proceed:
Let $0<alpha<frac{N}{s}$. Then we know $int_{Omega}w^{-s},dx=int_{Omega}|x|^{-alpha s},dx<infty$ if $alpha s<N$, which is assumed. Hence true.
Let $-frac{N}{s}<alphaleq 0$. Then $int_{Omega}|x|^{beta s},dx<infty$ if $beta s<N$ for $beta=-alpha geq 0$. That is $-alpha s<N$ implies $alpha>-frac{N}{s}$, which is also assumed. Hence true.
I think the argument is fine, my only confusion is that I am using the fact that $int_{Omega}|x|^alpha,dx<infty$ if $alpha<N$ for any bounded smooth domain (may contain $0$)) in $mathbb{R}^N$.
Can you kindly help to observe if the argument is correct or not.
Thanks in advance.
real-analysis complex-analysis functional-analysis analysis
asked Nov 18 at 7:46
Mathlover
1088
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $sin [frac{1}{p-1},infty)cap(frac{N}{p},infty)$ for $1<p<N$. Assume $Omega$ is a smooth bounded domain in $mathbb{R}^N$ and $w(x)=|x|^alpha$. Then I want to prove that $w^{-s}in L^1(Omega)$ for any $alphain(-frac{N}{s},frac{N}{s})$.
I am using the following argument to proceed:
Let $0<alpha<frac{N}{s}$. Then we know $int_{Omega}w^{-s},dx=int_{Omega}|x|^{-alpha s},dx<infty$ if $alpha s<N$, which is assumed. Hence true.
Let $-frac{N}{s}<alphaleq 0$. Then $int_{Omega}|x|^{beta s},dx<infty$ if $beta s<N$ for $beta=-alpha geq 0$. That is $-alpha s<N$ implies $alpha>-frac{N}{s}$, which is also assumed. Hence true.
I think the argument is fine, my only confusion is that I am using the fact that $int_{Omega}|x|^alpha,dx<infty$ if $alpha<N$ for any bounded smooth domain (may contain $0$)) in $mathbb{R}^N$.
Can you kindly help to observe if the argument is correct or not.
Thanks in advance.
real-analysis complex-analysis functional-analysis analysis
asked Nov 18 at 7:46
Mathlover
1088
Let $sin [frac{1}{p-1},infty)cap(frac{N}{p},infty)$ for $1<p<N$. Assume $Omega$ is a smooth bounded domain in $mathbb{R}^N$ and $w(x)=|x|^alpha$. Then I want to prove that $w^{-s}in L^1(Omega)$ for any $alphain(-frac{N}{s},frac{N}{s})$.
I am using the following argument to proceed:
Let $0<alpha<frac{N}{s}$. Then we know $int_{Omega}w^{-s},dx=int_{Omega}|x|^{-alpha s},dx<infty$ if $alpha s<N$, which is assumed. Hence true.
Let $-frac{N}{s}<alphaleq 0$. Then $int_{Omega}|x|^{beta s},dx<infty$ if $beta s<N$ for $beta=-alpha geq 0$. That is $-alpha s<N$ implies $alpha>-frac{N}{s}$, which is also assumed. Hence true.
I think the argument is fine, my only confusion is that I am using the fact that $int_{Omega}|x|^alpha,dx<infty$ if $alpha<N$ for any bounded smooth domain (may contain $0$)) in $mathbb{R}^N$.
Can you kindly help to observe if the argument is correct or not.
Thanks in advance.
real-analysis complex-analysis functional-analysis analysis
real-analysis complex-analysis functional-analysis analysis
asked Nov 18 at 7:46
Mathlover
1088
asked Nov 18 at 7:46
Mathlover
1088
asked Nov 18 at 7:46
Mathlover
1088
asked Nov 18 at 7:46
Mathlover
1088
asked Nov 18 at 7:46
Mathlover
1088
1088
add a comment |
add a comment |
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