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Prove that Laplace density (standard) lies in Sobolev space of order $leq 3/2$
Clash Royale CLAN TAG#URR8PPP
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I'm trying to prove, that the function $f(x) = e^$ lies in $H^s(mathbbR^2)$ for $sleq frac32$.
Therefore I calculate the functions sobolev norm $$|f|_s^2 = frac14pi^2int_mathbbR^2(1+| u|)^s |mathscrFf(u)|^2du$$
Since for its fourier transform holds $mathscrFf(u) = (1+|u|^2)^-1$, we consider $$|f|_s^2 = int_mathbbR^2(1+|u|^2)^s-2du$$.
Now I want to show two things: for $s=frac32$, this integral is not finite (and therefore obviously not for bigger $s$ and secondly, that this is finite for $s<frac32$.
I proved the first part as following:
$$|f|_s^2 = int_mathbbR^2(1+|u|^2)^-frac12du
\ geq int_mathbbR^2frac11+du geq int_mathbbR^2frac1sum_i=1^2du > infty
$$, using $(a+b)^frac12 leq a^frac12 + b^frac12$ and the behaviour of $frac1x$ for $xrightarrow 0$.
For the second point I still miss any idea of how to calculate an upper bound for the integral.
I hope there is somebody seeing the point I'm missing actually in this case.
calculus multivariable-calculus fourier-analysis sobolev-spaces
edited Jul 27 at 9:58
mathreadler
13.6k71857
asked Jul 27 at 9:01
Nico Albers
1184
add a comment |Â
up vote
1
down vote
favorite
I'm trying to prove, that the function $f(x) = e^$ lies in $H^s(mathbbR^2)$ for $sleq frac32$.
Therefore I calculate the functions sobolev norm $$|f|_s^2 = frac14pi^2int_mathbbR^2(1+| u|)^s |mathscrFf(u)|^2du$$
Since for its fourier transform holds $mathscrFf(u) = (1+|u|^2)^-1$, we consider $$|f|_s^2 = int_mathbbR^2(1+|u|^2)^s-2du$$.
Now I want to show two things: for $s=frac32$, this integral is not finite (and therefore obviously not for bigger $s$ and secondly, that this is finite for $s<frac32$.
I proved the first part as following:
$$|f|_s^2 = int_mathbbR^2(1+|u|^2)^-frac12du
\ geq int_mathbbR^2frac11+du geq int_mathbbR^2frac1sum_i=1^2du > infty
$$, using $(a+b)^frac12 leq a^frac12 + b^frac12$ and the behaviour of $frac1x$ for $xrightarrow 0$.
For the second point I still miss any idea of how to calculate an upper bound for the integral.
I hope there is somebody seeing the point I'm missing actually in this case.
calculus multivariable-calculus fourier-analysis sobolev-spaces
edited Jul 27 at 9:58
mathreadler
13.6k71857
asked Jul 27 at 9:01
Nico Albers
1184
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to prove, that the function $f(x) = e^$ lies in $H^s(mathbbR^2)$ for $sleq frac32$.
Therefore I calculate the functions sobolev norm $$|f|_s^2 = frac14pi^2int_mathbbR^2(1+| u|)^s |mathscrFf(u)|^2du$$
Since for its fourier transform holds $mathscrFf(u) = (1+|u|^2)^-1$, we consider $$|f|_s^2 = int_mathbbR^2(1+|u|^2)^s-2du$$.
Now I want to show two things: for $s=frac32$, this integral is not finite (and therefore obviously not for bigger $s$ and secondly, that this is finite for $s<frac32$.
I proved the first part as following:
$$|f|_s^2 = int_mathbbR^2(1+|u|^2)^-frac12du
\ geq int_mathbbR^2frac11+du geq int_mathbbR^2frac1sum_i=1^2du > infty
$$, using $(a+b)^frac12 leq a^frac12 + b^frac12$ and the behaviour of $frac1x$ for $xrightarrow 0$.
For the second point I still miss any idea of how to calculate an upper bound for the integral.
I hope there is somebody seeing the point I'm missing actually in this case.
calculus multivariable-calculus fourier-analysis sobolev-spaces
edited Jul 27 at 9:58
mathreadler
13.6k71857
asked Jul 27 at 9:01
Nico Albers
1184
I'm trying to prove, that the function $f(x) = e^$ lies in $H^s(mathbbR^2)$ for $sleq frac32$.
Therefore I calculate the functions sobolev norm $$|f|_s^2 = frac14pi^2int_mathbbR^2(1+| u|)^s |mathscrFf(u)|^2du$$
Since for its fourier transform holds $mathscrFf(u) = (1+|u|^2)^-1$, we consider $$|f|_s^2 = int_mathbbR^2(1+|u|^2)^s-2du$$.
Now I want to show two things: for $s=frac32$, this integral is not finite (and therefore obviously not for bigger $s$ and secondly, that this is finite for $s<frac32$.
I proved the first part as following:
$$|f|_s^2 = int_mathbbR^2(1+|u|^2)^-frac12du
\ geq int_mathbbR^2frac11+du geq int_mathbbR^2frac1sum_i=1^2du > infty
$$, using $(a+b)^frac12 leq a^frac12 + b^frac12$ and the behaviour of $frac1x$ for $xrightarrow 0$.
For the second point I still miss any idea of how to calculate an upper bound for the integral.
I hope there is somebody seeing the point I'm missing actually in this case.
calculus multivariable-calculus fourier-analysis sobolev-spaces
edited Jul 27 at 9:58
mathreadler
13.6k71857
asked Jul 27 at 9:01
Nico Albers
1184
edited Jul 27 at 9:58
mathreadler
13.6k71857
edited Jul 27 at 9:58
mathreadler
13.6k71857
edited Jul 27 at 9:58
mathreadler
13.6k71857
13.6k71857
asked Jul 27 at 9:01
Nico Albers
1184
asked Jul 27 at 9:01
Nico Albers
1184
asked Jul 27 at 9:01
Nico Albers
1184
1184
add a comment |Â
add a comment |Â
1 Answer
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So, finally, threw the help of a fellow student I realized transforming to polar coordinates would make sense.
This yields $$int_0^2piint_0^infty r;(1+r^2)^s-2drdtheta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $rleq 1+r$, we find $$|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr = fracpi1-s ; forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).
answered Jul 27 at 13:10
Nico Albers
1184
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
So, finally, threw the help of a fellow student I realized transforming to polar coordinates would make sense.
This yields $$int_0^2piint_0^infty r;(1+r^2)^s-2drdtheta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $rleq 1+r$, we find $$|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr = fracpi1-s ; forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).
answered Jul 27 at 13:10
Nico Albers
1184
add a comment |Â
up vote
0
down vote
So, finally, threw the help of a fellow student I realized transforming to polar coordinates would make sense.
This yields $$int_0^2piint_0^infty r;(1+r^2)^s-2drdtheta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $rleq 1+r$, we find $$|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr = fracpi1-s ; forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).
answered Jul 27 at 13:10
Nico Albers
1184
add a comment |Â
up vote
0
down vote
up vote
0
down vote
So, finally, threw the help of a fellow student I realized transforming to polar coordinates would make sense.
This yields $$int_0^2piint_0^infty r;(1+r^2)^s-2drdtheta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $rleq 1+r$, we find $$|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr = fracpi1-s ; forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).
answered Jul 27 at 13:10
Nico Albers
1184
So, finally, threw the help of a fellow student I realized transforming to polar coordinates would make sense.
This yields $$int_0^2piint_0^infty r;(1+r^2)^s-2drdtheta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $rleq 1+r$, we find $$|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $|f|_s^2 leq 2pi int_0^infty (1+r)^2s-3dr = fracpi1-s ; forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).
answered Jul 27 at 13:10
Nico Albers
1184
answered Jul 27 at 13:10
Nico Albers
1184
answered Jul 27 at 13:10
Nico Albers
1184
answered Jul 27 at 13:10
Nico Albers
1184
1184
add a comment |Â
add a comment |Â
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