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a norm that does not arise from an inner product
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In Pugh’s Real Mathematical Analysis, in order to bring an example that norms do not necessarily come from inner products it is stated that the unit sphere for every norm induced by an inner product is smooth but for the maximum norm $||.||_max$ the unit sphere is not smooth.
I know that intuitively the author by smooth means having no corners, but what is the mathematical definition of smooth in this context?
real-analysis
asked Jul 25 at 8:36
Selflearner
157110
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up vote
2
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In Pugh’s Real Mathematical Analysis, in order to bring an example that norms do not necessarily come from inner products it is stated that the unit sphere for every norm induced by an inner product is smooth but for the maximum norm $||.||_max$ the unit sphere is not smooth.
I know that intuitively the author by smooth means having no corners, but what is the mathematical definition of smooth in this context?
real-analysis
asked Jul 25 at 8:36
Selflearner
157110
@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In Pugh’s Real Mathematical Analysis, in order to bring an example that norms do not necessarily come from inner products it is stated that the unit sphere for every norm induced by an inner product is smooth but for the maximum norm $||.||_max$ the unit sphere is not smooth.
I know that intuitively the author by smooth means having no corners, but what is the mathematical definition of smooth in this context?
real-analysis
asked Jul 25 at 8:36
Selflearner
157110
In Pugh’s Real Mathematical Analysis, in order to bring an example that norms do not necessarily come from inner products it is stated that the unit sphere for every norm induced by an inner product is smooth but for the maximum norm $||.||_max$ the unit sphere is not smooth.
I know that intuitively the author by smooth means having no corners, but what is the mathematical definition of smooth in this context?
real-analysis
asked Jul 25 at 8:36
Selflearner
157110
asked Jul 25 at 8:36
Selflearner
157110
asked Jul 25 at 8:36
Selflearner
157110
asked Jul 25 at 8:36
Selflearner
157110
157110
@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54
add a comment |Â
@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54
@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54
add a comment |Â
1 Answer
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Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($star$) There is a smooth function $F:mathbbR^n rightarrow mathbbR$ with $S=xvert F(x) = 1$ and $nabla F(x)neq 0$ for all $xin S$.
Now if $S$ is the unit sphere with respect to an inner product $langle cdot , cdot rangle$, then you can take $F(x) := langle x , x rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $nabla F(x) = 2x$, hence ($star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($star$) is not satisfied. First note that $F(x) = Vert x Vert_max^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_nin S$ which approach $x_0$ from different faces. Since $nabla F(x) perp S$ (with respect to the usual dot-product) for all $xin S$, we get $$nabla F(x_0)= lim nabla F(a_n) perp lim nabla F(b_n) = nabla F(x_0)$$
and hence $nabla F(x_0)= 0$, a contradiction.
answered Jul 25 at 9:03
Jan Bohr
3,0991419
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($star$) There is a smooth function $F:mathbbR^n rightarrow mathbbR$ with $S=xvert F(x) = 1$ and $nabla F(x)neq 0$ for all $xin S$.
Now if $S$ is the unit sphere with respect to an inner product $langle cdot , cdot rangle$, then you can take $F(x) := langle x , x rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $nabla F(x) = 2x$, hence ($star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($star$) is not satisfied. First note that $F(x) = Vert x Vert_max^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_nin S$ which approach $x_0$ from different faces. Since $nabla F(x) perp S$ (with respect to the usual dot-product) for all $xin S$, we get $$nabla F(x_0)= lim nabla F(a_n) perp lim nabla F(b_n) = nabla F(x_0)$$
and hence $nabla F(x_0)= 0$, a contradiction.
answered Jul 25 at 9:03
Jan Bohr
3,0991419
add a comment |Â
up vote
3
down vote
accepted
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($star$) There is a smooth function $F:mathbbR^n rightarrow mathbbR$ with $S=xvert F(x) = 1$ and $nabla F(x)neq 0$ for all $xin S$.
Now if $S$ is the unit sphere with respect to an inner product $langle cdot , cdot rangle$, then you can take $F(x) := langle x , x rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $nabla F(x) = 2x$, hence ($star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($star$) is not satisfied. First note that $F(x) = Vert x Vert_max^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_nin S$ which approach $x_0$ from different faces. Since $nabla F(x) perp S$ (with respect to the usual dot-product) for all $xin S$, we get $$nabla F(x_0)= lim nabla F(a_n) perp lim nabla F(b_n) = nabla F(x_0)$$
and hence $nabla F(x_0)= 0$, a contradiction.
answered Jul 25 at 9:03
Jan Bohr
3,0991419
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($star$) There is a smooth function $F:mathbbR^n rightarrow mathbbR$ with $S=xvert F(x) = 1$ and $nabla F(x)neq 0$ for all $xin S$.
Now if $S$ is the unit sphere with respect to an inner product $langle cdot , cdot rangle$, then you can take $F(x) := langle x , x rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $nabla F(x) = 2x$, hence ($star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($star$) is not satisfied. First note that $F(x) = Vert x Vert_max^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_nin S$ which approach $x_0$ from different faces. Since $nabla F(x) perp S$ (with respect to the usual dot-product) for all $xin S$, we get $$nabla F(x_0)= lim nabla F(a_n) perp lim nabla F(b_n) = nabla F(x_0)$$
and hence $nabla F(x_0)= 0$, a contradiction.
answered Jul 25 at 9:03
Jan Bohr
3,0991419
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($star$) There is a smooth function $F:mathbbR^n rightarrow mathbbR$ with $S=xvert F(x) = 1$ and $nabla F(x)neq 0$ for all $xin S$.
Now if $S$ is the unit sphere with respect to an inner product $langle cdot , cdot rangle$, then you can take $F(x) := langle x , x rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $nabla F(x) = 2x$, hence ($star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($star$) is not satisfied. First note that $F(x) = Vert x Vert_max^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_nin S$ which approach $x_0$ from different faces. Since $nabla F(x) perp S$ (with respect to the usual dot-product) for all $xin S$, we get $$nabla F(x_0)= lim nabla F(a_n) perp lim nabla F(b_n) = nabla F(x_0)$$
and hence $nabla F(x_0)= 0$, a contradiction.
answered Jul 25 at 9:03
Jan Bohr
3,0991419
edited Jul 25 at 9:35
answered Jul 25 at 9:03
Jan Bohr
3,0991419
answered Jul 25 at 9:03
Jan Bohr
3,0991419
answered Jul 25 at 9:03
Jan Bohr
3,0991419
3,0991419
add a comment |Â
add a comment |Â
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@M.Winter I don’t think so because that statement comes in the beginning pages of the first chapter which is solely about real numbers
– Selflearner
Jul 25 at 8:43
You are probably right. This would also only work in 2D. But maybe the author just mentions this as an additional information for the interested reader and the prove is really out of scope for the book. I am not sure you can argue with the niveau of the chapter if its just a side note.
– M. Winter
Jul 25 at 8:54