Is $mathbb{R}^n$ always an inner product space?
This may be a silly question but every two norms on $mathbb{R}^n$ are equivalent and $VertcdotVert_2$ comes from the usual dot product so $(mathbb{R}^n,VertcdotVert_2)$ is (at least) an inner product space (pre-Hilbert).
Why can't we say that $(mathbb{R}^n,VertcdotVert_{infty})$ is also pre-Hilbert? I know the uniform norm doesn't come from an inner product but two spaces are (topologically) the same if their norms are equivalent and up until now (at least in my course) it is only the topology that gives the space its "uniqueness".
The fact that a norm comes from an inner product doesn't change the topology nor the algebraic (vector space) structure. I get the product may be useful but what unique, defining, core property ... does it have??
norm normed-spaces inner-product-space
asked Jan 4 at 20:17
PedroPedro
517212
add a comment |
This may be a silly question but every two norms on $mathbb{R}^n$ are equivalent and $VertcdotVert_2$ comes from the usual dot product so $(mathbb{R}^n,VertcdotVert_2)$ is (at least) an inner product space (pre-Hilbert).
Why can't we say that $(mathbb{R}^n,VertcdotVert_{infty})$ is also pre-Hilbert? I know the uniform norm doesn't come from an inner product but two spaces are (topologically) the same if their norms are equivalent and up until now (at least in my course) it is only the topology that gives the space its "uniqueness".
The fact that a norm comes from an inner product doesn't change the topology nor the algebraic (vector space) structure. I get the product may be useful but what unique, defining, core property ... does it have??
norm normed-spaces inner-product-space
asked Jan 4 at 20:17
PedroPedro
517212
1
Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
1
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35
add a comment |
This may be a silly question but every two norms on $mathbb{R}^n$ are equivalent and $VertcdotVert_2$ comes from the usual dot product so $(mathbb{R}^n,VertcdotVert_2)$ is (at least) an inner product space (pre-Hilbert).
Why can't we say that $(mathbb{R}^n,VertcdotVert_{infty})$ is also pre-Hilbert? I know the uniform norm doesn't come from an inner product but two spaces are (topologically) the same if their norms are equivalent and up until now (at least in my course) it is only the topology that gives the space its "uniqueness".
The fact that a norm comes from an inner product doesn't change the topology nor the algebraic (vector space) structure. I get the product may be useful but what unique, defining, core property ... does it have??
norm normed-spaces inner-product-space
asked Jan 4 at 20:17
PedroPedro
517212
This may be a silly question but every two norms on $mathbb{R}^n$ are equivalent and $VertcdotVert_2$ comes from the usual dot product so $(mathbb{R}^n,VertcdotVert_2)$ is (at least) an inner product space (pre-Hilbert).
Why can't we say that $(mathbb{R}^n,VertcdotVert_{infty})$ is also pre-Hilbert? I know the uniform norm doesn't come from an inner product but two spaces are (topologically) the same if their norms are equivalent and up until now (at least in my course) it is only the topology that gives the space its "uniqueness".
The fact that a norm comes from an inner product doesn't change the topology nor the algebraic (vector space) structure. I get the product may be useful but what unique, defining, core property ... does it have??
norm normed-spaces inner-product-space
norm normed-spaces inner-product-space
asked Jan 4 at 20:17
PedroPedro
517212
asked Jan 4 at 20:17
PedroPedro
517212
asked Jan 4 at 20:17
PedroPedro
517212
asked Jan 4 at 20:17
PedroPedro
517212
asked Jan 4 at 20:17
PedroPedro
517212
517212
1
Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
1
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35
add a comment |
1
Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
1
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35
1
1
Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
1
1
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35
add a comment |
1 Answer
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The question is how do you interpret $mathbb R^n$. Indeed,
- if you consider it a real vector space, then it carries a natural inner product.
- if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
- if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $ell_p$-norms for $pneq 2$ are counterexamples too.
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
add a comment |
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1 Answer
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The question is how do you interpret $mathbb R^n$. Indeed,
- if you consider it a real vector space, then it carries a natural inner product.
- if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
- if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $ell_p$-norms for $pneq 2$ are counterexamples too.
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
add a comment |
The question is how do you interpret $mathbb R^n$. Indeed,
- if you consider it a real vector space, then it carries a natural inner product.
- if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
- if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $ell_p$-norms for $pneq 2$ are counterexamples too.
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
add a comment |
The question is how do you interpret $mathbb R^n$. Indeed,
- if you consider it a real vector space, then it carries a natural inner product.
- if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
- if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $ell_p$-norms for $pneq 2$ are counterexamples too.
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
The question is how do you interpret $mathbb R^n$. Indeed,
- if you consider it a real vector space, then it carries a natural inner product.
- if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
- if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $ell_p$-norms for $pneq 2$ are counterexamples too.
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 20:37
Tomek KaniaTomek Kania
12.1k11943
12.1k11943
add a comment |
add a comment |
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Because the $infty$ norm is not induced by an inner product.
– ncmathsadist
Jan 4 at 20:26
@ncmathsadist the question is why would $(mathbb{R}^n,VertcdotVert_2)$ and $(mathbb{R}^n,VertcdotVert_{infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product)
– Pedro
Jan 4 at 20:30
1
They are topologically equivalent but, for instance, they are not isometric as Banach spaces.
– ncmathsadist
Jan 4 at 20:35