Mixing
Brezis exercise, under what condition does $f$ belongs to $L_p(mathbb{R^n})$
1
$begingroup$
Brezis excercise 4.1: I am asked to show under what conditions does $f(x)={1+|x|^alpha}^{-1}{1+|log|x||^beta}^{-1}$ belong to $L_p(mathbb{R}^n)$ where $alpha,beta>0$..
I know that for $f$ to belong is must show that $int_{(mathbb{R^n})}|f|^p<infty$, so far I got this:
$$|f(x)|^p=dfrac{1}{|1+|x|^alpha|^p|1+|log|x||^beta|^p}$$ however I don't know what to do after this. Thanks in advance.
functional-analysis lp-spaces
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
$endgroup$
add a comment |
1
$begingroup$
Brezis excercise 4.1: I am asked to show under what conditions does $f(x)={1+|x|^alpha}^{-1}{1+|log|x||^beta}^{-1}$ belong to $L_p(mathbb{R}^n)$ where $alpha,beta>0$..
I know that for $f$ to belong is must show that $int_{(mathbb{R^n})}|f|^p<infty$, so far I got this:
$$|f(x)|^p=dfrac{1}{|1+|x|^alpha|^p|1+|log|x||^beta|^p}$$ however I don't know what to do after this. Thanks in advance.
functional-analysis lp-spaces
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
$endgroup$
2
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
1
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
1
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43
add a comment |
1
1
1
$begingroup$
Brezis excercise 4.1: I am asked to show under what conditions does $f(x)={1+|x|^alpha}^{-1}{1+|log|x||^beta}^{-1}$ belong to $L_p(mathbb{R}^n)$ where $alpha,beta>0$..
I know that for $f$ to belong is must show that $int_{(mathbb{R^n})}|f|^p<infty$, so far I got this:
$$|f(x)|^p=dfrac{1}{|1+|x|^alpha|^p|1+|log|x||^beta|^p}$$ however I don't know what to do after this. Thanks in advance.
functional-analysis lp-spaces
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
$endgroup$
Brezis excercise 4.1: I am asked to show under what conditions does $f(x)={1+|x|^alpha}^{-1}{1+|log|x||^beta}^{-1}$ belong to $L_p(mathbb{R}^n)$ where $alpha,beta>0$..
I know that for $f$ to belong is must show that $int_{(mathbb{R^n})}|f|^p<infty$, so far I got this:
$$|f(x)|^p=dfrac{1}{|1+|x|^alpha|^p|1+|log|x||^beta|^p}$$ however I don't know what to do after this. Thanks in advance.
functional-analysis lp-spaces
functional-analysis lp-spaces
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
edited Jan 8 at 18:00
Davide Giraudo
126k16150261
126k16150261
asked Jan 7 at 20:21
AlfdavAlfdav
787
asked Jan 7 at 20:21
AlfdavAlfdav
787
asked Jan 7 at 20:21
AlfdavAlfdav
787
787
2
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
1
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
1
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43
add a comment |
2
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
1
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
1
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43
2
2
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
1
1
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
1
1
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43
add a comment |
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2
$begingroup$
It's a radial function, so after changing to polar coordinates, it comes down to whether a 1-dimensional integral converges. My hint: work out the case $beta=0$ first.
$endgroup$
– Ian
Jan 7 at 20:27
$begingroup$
I will give it a try, however, two points. 1. $beta$ is supposed to be nonzero. 2. In order to apply the polar coordinates I know, $f$ must be "summable" meaning the integral of it mod is finite. which is what I must show i think.
$endgroup$
– Alfdav
Jan 7 at 20:32
1
$begingroup$
1. I'm aware that in your actual problem $beta>0$, but the situation here is actually mostly controlled by $alpha$. 2. As long as the integral makes sense (the integrand is measurable and the integral isn't a $infty - infty$ form), you can definitely change variables freely.
$endgroup$
– Ian
Jan 7 at 20:37
$begingroup$
I used polar coordinates and got: $$int_{mathbb{R^n}}f=w_{n-1}int_{0}^{infty}dfrac{1}{r^{alpha p -n+1}(log (r))^{beta p}}$$ I assume that this converges for some special values, where can I read more about this?
$endgroup$
– Alfdav
Jan 7 at 21:19
1
$begingroup$
If you set $beta=0$ then the situation should be quite obvious. Now for $beta>0$, you can use a comparison argument to reduce it to the $beta=0$ case, except when $alpha p - n + 1$ is one particular value. If you show you can get to this point, then I'll explain what the deal is with $beta$.
$endgroup$
– Ian
Jan 7 at 21:43