How can a discontinuous function belong to $C_B^1(Omega)$, the space of continuous functions $u$ with bounded...
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Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
$$
u(x,y) =
begin{cases}
1, quad x > 0, \
0, quad x < 0.
end{cases}
$$
On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.
But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?
pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions
asked Nov 21 at 10:58
eurocoder
1,095315
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up vote
0
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Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
$$
u(x,y) =
begin{cases}
1, quad x > 0, \
0, quad x < 0.
end{cases}
$$
On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.
But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?
pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions
asked Nov 21 at 10:58
eurocoder
1,095315
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
$$
u(x,y) =
begin{cases}
1, quad x > 0, \
0, quad x < 0.
end{cases}
$$
On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.
But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?
pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions
asked Nov 21 at 10:58
eurocoder
1,095315
Let $Omega = {(x,y) in mathbb{R}^2 : 0 < |x| < 1, 0 < y < 1}$ and consider the function $u$ defined on $Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10)
$$
u(x,y) =
begin{cases}
1, quad x > 0, \
0, quad x < 0.
end{cases}
$$
On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(Omega)$ which consists of function in $C^1(Omega)$ such that $D^alpha u$ is bounded for $0 le alpha le 1$.
But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?
pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions
pde continuity sobolev-spaces regularity-theory-of-pdes discontinuous-functions
asked Nov 21 at 10:58
eurocoder
1,095315
asked Nov 21 at 10:58
eurocoder
1,095315
asked Nov 21 at 10:58
eurocoder
1,095315
asked Nov 21 at 10:58
eurocoder
1,095315
asked Nov 21 at 10:58
eurocoder
1,095315
1,095315
add a comment |
add a comment |
1 Answer
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Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
$$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
$f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
$$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
$f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
add a comment |
up vote
3
down vote
accepted
Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
$$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
$f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
$$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
$f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is
$$ f: [-1,0)cup (0,1] to mathbb R, quad f(x) = frac{operatorname{sgn(x)+1}}2$$
$f$ is continuous (even $C^infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
answered Nov 21 at 11:18
Calvin Khor
10.8k21437
10.8k21437
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
add a comment |
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(Omega)$, it is not in $C^1(bar Omega)$, which is the closed subspace of $C_1^j(Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $Omega$? ..how can $C^1(bar Omega)$ even be a subspace of $C_B^1(Omega)$ as $C^1(bar Omega)$ is defined on $bar Omega$ which is bigger than $Omega$?
– eurocoder
Nov 21 at 11:23
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
Nevermind, he says on page that 10 that this is an abuse of notation.
– eurocoder
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A subset B$ implies $C^1_B(B) subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0({0})$ contains every function $f:mathbb Rto mathbb R$. @eurocoder
– Calvin Khor
Nov 21 at 11:26
add a comment |
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