Mixing
Two smooth functions glued together become a Sobolev function
Clash Royale CLAN TAG#URR8PPP
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Consider the square $S:=(0,1)times (0,1)subset mathbbR^2$ and smooth functions $f,g:Srightarrow (0,infty)$, which can be extended up to the boundary by zero (i.e. $g(x),f(x):=0$ for all $xinpartial S$ are continuous extensions). Hence $f,gin H_0^1(S)$.
The diagonal $y=x$ separates the square into two triangles. If we glue $f$ and $g$ along that diagonal, would the resulting function be in $H_0^1(S)$? More precisely, do we have $hin H_0^1(S)$, where
$$h(x,y):=begincases f(x,y), ,if,, xgeq y\
g(x,y), ,if,, x<y endcases$$
I do not know how to check if $hin H_0^1(S)$. I tried to look for a sequence $varphi_nin C^infty_c(S)$ which converges in $H^1_0(S)$ to $h$. I couldn't find a good candidate. Maybe, if $h$ is not continuous on the diagonal $lbrace x=yrbrace$, it is not possible (That's my feeling at least). Can someone help me?
Are there any conditions on $f,g$ which would guarantee that $h$ becomes a function in $H^1_0(S)$?
real-analysis functional-analysis sobolev-spaces
edited Jul 28 at 15:30
user357151
13.8k31140
asked Jul 28 at 14:59
Denilson Orr
1287
add a comment |Â
up vote
1
down vote
favorite
Consider the square $S:=(0,1)times (0,1)subset mathbbR^2$ and smooth functions $f,g:Srightarrow (0,infty)$, which can be extended up to the boundary by zero (i.e. $g(x),f(x):=0$ for all $xinpartial S$ are continuous extensions). Hence $f,gin H_0^1(S)$.
The diagonal $y=x$ separates the square into two triangles. If we glue $f$ and $g$ along that diagonal, would the resulting function be in $H_0^1(S)$? More precisely, do we have $hin H_0^1(S)$, where
$$h(x,y):=begincases f(x,y), ,if,, xgeq y\
g(x,y), ,if,, x<y endcases$$
I do not know how to check if $hin H_0^1(S)$. I tried to look for a sequence $varphi_nin C^infty_c(S)$ which converges in $H^1_0(S)$ to $h$. I couldn't find a good candidate. Maybe, if $h$ is not continuous on the diagonal $lbrace x=yrbrace$, it is not possible (That's my feeling at least). Can someone help me?
Are there any conditions on $f,g$ which would guarantee that $h$ becomes a function in $H^1_0(S)$?
real-analysis functional-analysis sobolev-spaces
edited Jul 28 at 15:30
user357151
13.8k31140
asked Jul 28 at 14:59
Denilson Orr
1287
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the square $S:=(0,1)times (0,1)subset mathbbR^2$ and smooth functions $f,g:Srightarrow (0,infty)$, which can be extended up to the boundary by zero (i.e. $g(x),f(x):=0$ for all $xinpartial S$ are continuous extensions). Hence $f,gin H_0^1(S)$.
The diagonal $y=x$ separates the square into two triangles. If we glue $f$ and $g$ along that diagonal, would the resulting function be in $H_0^1(S)$? More precisely, do we have $hin H_0^1(S)$, where
$$h(x,y):=begincases f(x,y), ,if,, xgeq y\
g(x,y), ,if,, x<y endcases$$
I do not know how to check if $hin H_0^1(S)$. I tried to look for a sequence $varphi_nin C^infty_c(S)$ which converges in $H^1_0(S)$ to $h$. I couldn't find a good candidate. Maybe, if $h$ is not continuous on the diagonal $lbrace x=yrbrace$, it is not possible (That's my feeling at least). Can someone help me?
Are there any conditions on $f,g$ which would guarantee that $h$ becomes a function in $H^1_0(S)$?
real-analysis functional-analysis sobolev-spaces
edited Jul 28 at 15:30
user357151
13.8k31140
asked Jul 28 at 14:59
Denilson Orr
1287
Consider the square $S:=(0,1)times (0,1)subset mathbbR^2$ and smooth functions $f,g:Srightarrow (0,infty)$, which can be extended up to the boundary by zero (i.e. $g(x),f(x):=0$ for all $xinpartial S$ are continuous extensions). Hence $f,gin H_0^1(S)$.
The diagonal $y=x$ separates the square into two triangles. If we glue $f$ and $g$ along that diagonal, would the resulting function be in $H_0^1(S)$? More precisely, do we have $hin H_0^1(S)$, where
$$h(x,y):=begincases f(x,y), ,if,, xgeq y\
g(x,y), ,if,, x<y endcases$$
I do not know how to check if $hin H_0^1(S)$. I tried to look for a sequence $varphi_nin C^infty_c(S)$ which converges in $H^1_0(S)$ to $h$. I couldn't find a good candidate. Maybe, if $h$ is not continuous on the diagonal $lbrace x=yrbrace$, it is not possible (That's my feeling at least). Can someone help me?
Are there any conditions on $f,g$ which would guarantee that $h$ becomes a function in $H^1_0(S)$?
real-analysis functional-analysis sobolev-spaces
edited Jul 28 at 15:30
user357151
13.8k31140
asked Jul 28 at 14:59
Denilson Orr
1287
edited Jul 28 at 15:30
user357151
13.8k31140
edited Jul 28 at 15:30
user357151
13.8k31140
edited Jul 28 at 15:30
user357151
13.8k31140
13.8k31140
asked Jul 28 at 14:59
Denilson Orr
1287
asked Jul 28 at 14:59
Denilson Orr
1287
asked Jul 28 at 14:59
Denilson Orr
1287
1287
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^1,p$ spaces.
answered Jul 28 at 15:29
user357151
13.8k31140
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^1,p$ spaces.
answered Jul 28 at 15:29
user357151
13.8k31140
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
add a comment |Â
up vote
1
down vote
accepted
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^1,p$ spaces.
answered Jul 28 at 15:29
user357151
13.8k31140
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^1,p$ spaces.
answered Jul 28 at 15:29
user357151
13.8k31140
The ACL characterization of Sobolev functions is useful here. A Sobolev function must have a representative that is absolutely continuous on a.e. horizontal line. If $f$ and $g$ do not agree on the diagonal, this condition fails, so we do not get a Sobolev function.
On the other hand, if $f=g$ on the boundary, then the ACL condition holds, as continuously gluing two absolutely continuous functions on an interval yields another absolutely continuous function. Since the integrability of derivatives is assured as well, we do get a Sobolev function here.
An alternative proof of the second paragraph is that $f$ and $g$ are Lipschitz continuous, and gluing (when the values on the cut match) preserves the Lipschitz property. A Lipschitz function is in all $W^1,p$ spaces.
answered Jul 28 at 15:29
user357151
13.8k31140
answered Jul 28 at 15:29
user357151
13.8k31140
answered Jul 28 at 15:29
user357151
13.8k31140
answered Jul 28 at 15:29
user357151
13.8k31140
13.8k31140
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
add a comment |Â
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
Wow! Your answer is so helpful!
– Denilson Orr
Jul 28 at 15:50
add a comment |Â
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