Mixing
calculating norm with cut off function
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Let $chi$ be a smooth cut-off function, defined on an interval $[0,infty)$ and it is equal 1 on $[0,X-1)$ and goes to 0 on $[X-1,X)$ then equal 0 on $[X,infty)$.. Let $v=exp(-ax)$ on $[0,infty)$..
I need to find the $L^2$-norm of $chi' v$ and here is my attempt:
$| chi' v |^2=int_{0}^{infty} chi'^2 v^2dx= int_{X-1}^{X} chi'^2 exp(-2ax) dx$ ..
Now, how can I finish the calculation because I don't know what is exactly $chi'$ on $[X-1,X]$ ?
I appreciate any help ..
functional-analysis operator-theory norm
asked yesterday
S.N.A
857
|
show 6 more comments
up vote
0
down vote
favorite
Let $chi$ be a smooth cut-off function, defined on an interval $[0,infty)$ and it is equal 1 on $[0,X-1)$ and goes to 0 on $[X-1,X)$ then equal 0 on $[X,infty)$.. Let $v=exp(-ax)$ on $[0,infty)$..
I need to find the $L^2$-norm of $chi' v$ and here is my attempt:
$| chi' v |^2=int_{0}^{infty} chi'^2 v^2dx= int_{X-1}^{X} chi'^2 exp(-2ax) dx$ ..
Now, how can I finish the calculation because I don't know what is exactly $chi'$ on $[X-1,X]$ ?
I appreciate any help ..
functional-analysis operator-theory norm
asked yesterday
S.N.A
857
Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $chi$ be a smooth cut-off function, defined on an interval $[0,infty)$ and it is equal 1 on $[0,X-1)$ and goes to 0 on $[X-1,X)$ then equal 0 on $[X,infty)$.. Let $v=exp(-ax)$ on $[0,infty)$..
I need to find the $L^2$-norm of $chi' v$ and here is my attempt:
$| chi' v |^2=int_{0}^{infty} chi'^2 v^2dx= int_{X-1}^{X} chi'^2 exp(-2ax) dx$ ..
Now, how can I finish the calculation because I don't know what is exactly $chi'$ on $[X-1,X]$ ?
I appreciate any help ..
functional-analysis operator-theory norm
asked yesterday
S.N.A
857
Let $chi$ be a smooth cut-off function, defined on an interval $[0,infty)$ and it is equal 1 on $[0,X-1)$ and goes to 0 on $[X-1,X)$ then equal 0 on $[X,infty)$.. Let $v=exp(-ax)$ on $[0,infty)$..
I need to find the $L^2$-norm of $chi' v$ and here is my attempt:
$| chi' v |^2=int_{0}^{infty} chi'^2 v^2dx= int_{X-1}^{X} chi'^2 exp(-2ax) dx$ ..
Now, how can I finish the calculation because I don't know what is exactly $chi'$ on $[X-1,X]$ ?
I appreciate any help ..
functional-analysis operator-theory norm
functional-analysis operator-theory norm
asked yesterday
S.N.A
857
asked yesterday
S.N.A
857
asked yesterday
S.N.A
857
asked yesterday
S.N.A
857
asked yesterday
S.N.A
857
857
Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday
|
show 6 more comments
Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday
Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday
|
show 6 more comments
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Basically you can't! Cut-off functions can be made in a lot of different ways, so the problem is not well posed.
– Marco
yesterday
I don't think that.. cut off are used a lot in situation like this ..
– S.N.A
yesterday
Yes, but usually to give some esteem.
– Marco
yesterday
Iam thinking about the boundedness of the cutoff function if it can help me in this integration.. the cut off on $[X-1,X]$ is continuous and a continuous function on a bounded set is bounded ..
– S.N.A
yesterday
For sure $chi$ and all its derivatives are bounded on $[X-1,X]$, what I mean is that you can't find a precise result in your integration because is not univocal the way you choose the cut-off functions. For instance if $chi(x)=phi(x)$ for $xin[X-1,X]$, you could also define another cut-off $bar{chi}$ that equals a "contraction" of $phi$ in $[X-1,X-1/2]$ and equals $0$ for each $xgeq X-1/2$. The cut-off functions are not equal so you can't expect the integral to be the same.
– Marco
yesterday