prove that the space $C^{infty}(R_{+}^n)cap W^{1,2}(R_{+}^n)$ is dense in $W^{1,2}(R_{+}^n)$
2
I already know that as $C^{infty}_c(R^n)$ is dense in $W^{1,2}(R^n)$, but I don't know why the space $C^{infty}(R_{+}^n)cap W^{1,2}(R_{+}^n)$ is dense in $W^{1,2}(R_{+}^n)$
Is the statement true and why?
Thank you!
real-analysis functional-analysis pde sobolev-spaces
asked Nov 18 at 15:04
chloe hj
626
|
show 2 more comments
2
I already know that as $C^{infty}_c(R^n)$ is dense in $W^{1,2}(R^n)$, but I don't know why the space $C^{infty}(R_{+}^n)cap W^{1,2}(R_{+}^n)$ is dense in $W^{1,2}(R_{+}^n)$
Is the statement true and why?
Thank you!
real-analysis functional-analysis pde sobolev-spaces
asked Nov 18 at 15:04
chloe hj
626
I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00
|
show 2 more comments
2
2
2
I already know that as $C^{infty}_c(R^n)$ is dense in $W^{1,2}(R^n)$, but I don't know why the space $C^{infty}(R_{+}^n)cap W^{1,2}(R_{+}^n)$ is dense in $W^{1,2}(R_{+}^n)$
Is the statement true and why?
Thank you!
real-analysis functional-analysis pde sobolev-spaces
asked Nov 18 at 15:04
chloe hj
626
I already know that as $C^{infty}_c(R^n)$ is dense in $W^{1,2}(R^n)$, but I don't know why the space $C^{infty}(R_{+}^n)cap W^{1,2}(R_{+}^n)$ is dense in $W^{1,2}(R_{+}^n)$
Is the statement true and why?
Thank you!
real-analysis functional-analysis pde sobolev-spaces
real-analysis functional-analysis pde sobolev-spaces
asked Nov 18 at 15:04
chloe hj
626
asked Nov 18 at 15:04
chloe hj
626
edited Nov 18 at 15:54
asked Nov 18 at 15:04
chloe hj
626
asked Nov 18 at 15:04
chloe hj
626
asked Nov 18 at 15:04
chloe hj
626
626
I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00
|
show 2 more comments
I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00
I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00
|
show 2 more comments
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I guess by $R^n_+$ you mean a half space of the form $R^{n-1}times (0,infty)$? In that case $C^infty(R^n_+)$ is not even contained in $H^1(R^n_+)$. But their intersection is indeed dense in $H^1(R^n_+)$ by the usual convolution argument.
– MaoWao
Nov 18 at 15:16
yes, $R_{+}^n$ means $R^{n-1}times(0,infty)$. Why their intersection is dense in $H^1(R_{+}^n)$, can you give me an explanation? thank you !
– chloe hj
Nov 18 at 15:30
and $H^1$ means $W^{1,2}$ in this book I read. :)
– chloe hj
Nov 18 at 15:32
Look up the Meyers-Serrin theorem. The original paper is quite short and approachable: pnas.org/content/pnas/51/6/1055.full.pdf
– MaoWao
Nov 18 at 15:53
thank you! does this statement can be extended to a more general case? and is this a general approach to solve the density problem? I'm very confused with those definitions and statements when I'm learning, and I don't know where to get the right answer or exact resources.
– chloe hj
Nov 18 at 16:00