Mixing
I have written up a proof on why norms on $mathbb{R}^n$ are equivalent
$begingroup$
Just wanted to share my proof with you smart people to have some feedback and to share each other's ideas. Some disclaimers: this is my first post, English is not my first language, and I know that there are simpler ways to prove this. Here it goes:
Suppose $f$ and $g$ are two norms on $mathbb{R}^n$, and define the following function for each non-zero vector in $mathbb{R}^n$:
$$h(x) := dfrac{f(x)}{g(x)}$$
$h$ is well defined because $g$ is zero iff $x$ is zero. Now since $h$ is positively homogeneous of degree zero (because $f$ and $g$ both are positively homogeneous of degree one), it is constant on all half-lines passing through the origin (excluding the origin). With this I mean that it is constant on every set of the type ${tv}$, with $v$ non-zero vector of $mathbb{R}^n$ and $t in mathbb{R}_+^*$. Because of this, if we want to find the sup and the inf of $h$ on its domain, we can restrict ourselves to finding its sup and inf on any set that contains at least one point for each half-line described above.
Since the set $ G:={x in mathbb{R}^n , | , g(x) = 1}$ has this property, we can safely say that sup $h(x)$ = sup${h(x) , | , x in G}$ , and same for the inf.
Now, for every point $x$ in $G$, $h(x) = f(x)$ , so it suffices to find the sup and the inf of $f$. But $G$ is a compact set, and therefore from the continuity of $f$ (norms are continuous) and Weierstrass' theorem, we conclude that sup $f$ and inf $f$ on $G$ are both in $mathbb{R}_+^*$ (they can't be zero because they are actually a min and a max, and $f$ is never zero outside of the origin). This means that there are two real positive constants $c_1$ and $c_2$ such that $$c_1 leq frac{f(x)}{g(x)} , leq , c_2 quad Longrightarrow quad c_1 , g(x) leq f(x) leq c_2 , g(x) $$
for any non-zero $x$. Since for $x=0$ the inequality is true, we have the equivalence of the two norms.
TLDR: let $h$ be the ratio of two norms and show that $h$ is bounded above and below by two positive real numbers.
subproof 1 Let's show that $G$ has the property that it crosses each half line through the origin at least once.
Let $tv$ be a half-line. Now, we want to see if there's a $t$ for which it crosses $G$, so we can just plug the points of the half-line into $g$ and use homogeneity: $g(tv) = t g(v)$. So, if we choose $t$ to be $1/g(v)$ , which can be done because g is non-zero outside the origin, we can be sure that the half line will cross our set. Actually for any other $t$ we are not touching $G$ so each half-line crosses exactly once the level sets of a norm :) pretty neat.
subproof 2 I wanted to show that $G$ is closed and bounded, again just because.
The fact that it's closed follows from the fact that it is the counter image of the closed subset of the real numbers ${1}$ through a continuous function $g$.
Let's see why it's bounded (my reasoning is pretty convoluted; I would be curious to see other ways to show this).
Let $S_1$ be the euclidean unit sphere around the origin. Let's call $h$ the minimum of $g$ on $S_1$. This number exists since $g$ is continuous and $S_1$ is compact.
case $h<1$: Let $a$ be positive, real, and strictly greater than 1. Let $S_{a/h}$ be the euclidean sphere of radius $a/h$ around the origin. Since the halflines through the origin (let's say $v$ is a unit vector) provide a bijection between the two sets, we can say that for each $x$ in $S_{a/h}$ , there is a $y$ in $S_1$ such that
$$g(x) = gleft(frac{a}{h} , yright) = frac{a}{h} , g(y) geq frac{a}{h} , h = a > 1 , .$$
Thus for each $x$ with an euclidean distance more than $1/h$ from the origin, its $g$-norm will be strictly greater than 1, making $G$ bounded.
Case $h geq 1 $: Each $x$ with euclidean norm $a>1$ will be such that there is a corresponding $y$ on $S_1$ for which:
$$ g(x) = g(a , y) = a ,g(y) geq a cdot 1 > 1 , .$$
Thus $G$ is bounded.
I am curious to have some feedback about all this. Especially regarding ways in which this can be generalized, or things that have crossed your mind while reading this, as well as other proofs.
This idea came to me while i was solving this problem: find, for each even $n in mathbb{N}$ the best constant $a$ such that, for all real $x$ and $y$,
$$ (x + y)^n leq a (x^n + y^n) , .$$
Other related questions/ideas that came up and I have not yet answered, mainly because I know pretty much nothing about convex functions apart from the definition:
1) are sublevels of convex functions convex? Is this an if and only if?
2) is it true that if a function is convex then the level sets have the property that they intersect every halfline trough the origin exactly once? And is this an if and only if?
3) are sublevels of convex continuous functions compact?
Thanks for reading :)
real-analysis proof-verification proof-writing norm equivalent-metrics
asked Jan 31 at 14:08
gent96gent96
414
$endgroup$
add a comment |
$begingroup$
Just wanted to share my proof with you smart people to have some feedback and to share each other's ideas. Some disclaimers: this is my first post, English is not my first language, and I know that there are simpler ways to prove this. Here it goes:
Suppose $f$ and $g$ are two norms on $mathbb{R}^n$, and define the following function for each non-zero vector in $mathbb{R}^n$:
$$h(x) := dfrac{f(x)}{g(x)}$$
$h$ is well defined because $g$ is zero iff $x$ is zero. Now since $h$ is positively homogeneous of degree zero (because $f$ and $g$ both are positively homogeneous of degree one), it is constant on all half-lines passing through the origin (excluding the origin). With this I mean that it is constant on every set of the type ${tv}$, with $v$ non-zero vector of $mathbb{R}^n$ and $t in mathbb{R}_+^*$. Because of this, if we want to find the sup and the inf of $h$ on its domain, we can restrict ourselves to finding its sup and inf on any set that contains at least one point for each half-line described above.
Since the set $ G:={x in mathbb{R}^n , | , g(x) = 1}$ has this property, we can safely say that sup $h(x)$ = sup${h(x) , | , x in G}$ , and same for the inf.
Now, for every point $x$ in $G$, $h(x) = f(x)$ , so it suffices to find the sup and the inf of $f$. But $G$ is a compact set, and therefore from the continuity of $f$ (norms are continuous) and Weierstrass' theorem, we conclude that sup $f$ and inf $f$ on $G$ are both in $mathbb{R}_+^*$ (they can't be zero because they are actually a min and a max, and $f$ is never zero outside of the origin). This means that there are two real positive constants $c_1$ and $c_2$ such that $$c_1 leq frac{f(x)}{g(x)} , leq , c_2 quad Longrightarrow quad c_1 , g(x) leq f(x) leq c_2 , g(x) $$
for any non-zero $x$. Since for $x=0$ the inequality is true, we have the equivalence of the two norms.
TLDR: let $h$ be the ratio of two norms and show that $h$ is bounded above and below by two positive real numbers.
subproof 1 Let's show that $G$ has the property that it crosses each half line through the origin at least once.
Let $tv$ be a half-line. Now, we want to see if there's a $t$ for which it crosses $G$, so we can just plug the points of the half-line into $g$ and use homogeneity: $g(tv) = t g(v)$. So, if we choose $t$ to be $1/g(v)$ , which can be done because g is non-zero outside the origin, we can be sure that the half line will cross our set. Actually for any other $t$ we are not touching $G$ so each half-line crosses exactly once the level sets of a norm :) pretty neat.
subproof 2 I wanted to show that $G$ is closed and bounded, again just because.
The fact that it's closed follows from the fact that it is the counter image of the closed subset of the real numbers ${1}$ through a continuous function $g$.
Let's see why it's bounded (my reasoning is pretty convoluted; I would be curious to see other ways to show this).
Let $S_1$ be the euclidean unit sphere around the origin. Let's call $h$ the minimum of $g$ on $S_1$. This number exists since $g$ is continuous and $S_1$ is compact.
case $h<1$: Let $a$ be positive, real, and strictly greater than 1. Let $S_{a/h}$ be the euclidean sphere of radius $a/h$ around the origin. Since the halflines through the origin (let's say $v$ is a unit vector) provide a bijection between the two sets, we can say that for each $x$ in $S_{a/h}$ , there is a $y$ in $S_1$ such that
$$g(x) = gleft(frac{a}{h} , yright) = frac{a}{h} , g(y) geq frac{a}{h} , h = a > 1 , .$$
Thus for each $x$ with an euclidean distance more than $1/h$ from the origin, its $g$-norm will be strictly greater than 1, making $G$ bounded.
Case $h geq 1 $: Each $x$ with euclidean norm $a>1$ will be such that there is a corresponding $y$ on $S_1$ for which:
$$ g(x) = g(a , y) = a ,g(y) geq a cdot 1 > 1 , .$$
Thus $G$ is bounded.
I am curious to have some feedback about all this. Especially regarding ways in which this can be generalized, or things that have crossed your mind while reading this, as well as other proofs.
This idea came to me while i was solving this problem: find, for each even $n in mathbb{N}$ the best constant $a$ such that, for all real $x$ and $y$,
$$ (x + y)^n leq a (x^n + y^n) , .$$
Other related questions/ideas that came up and I have not yet answered, mainly because I know pretty much nothing about convex functions apart from the definition:
1) are sublevels of convex functions convex? Is this an if and only if?
2) is it true that if a function is convex then the level sets have the property that they intersect every halfline trough the origin exactly once? And is this an if and only if?
3) are sublevels of convex continuous functions compact?
Thanks for reading :)
real-analysis proof-verification proof-writing norm equivalent-metrics
asked Jan 31 at 14:08
gent96gent96
414
$endgroup$
add a comment |
$begingroup$
Just wanted to share my proof with you smart people to have some feedback and to share each other's ideas. Some disclaimers: this is my first post, English is not my first language, and I know that there are simpler ways to prove this. Here it goes:
Suppose $f$ and $g$ are two norms on $mathbb{R}^n$, and define the following function for each non-zero vector in $mathbb{R}^n$:
$$h(x) := dfrac{f(x)}{g(x)}$$
$h$ is well defined because $g$ is zero iff $x$ is zero. Now since $h$ is positively homogeneous of degree zero (because $f$ and $g$ both are positively homogeneous of degree one), it is constant on all half-lines passing through the origin (excluding the origin). With this I mean that it is constant on every set of the type ${tv}$, with $v$ non-zero vector of $mathbb{R}^n$ and $t in mathbb{R}_+^*$. Because of this, if we want to find the sup and the inf of $h$ on its domain, we can restrict ourselves to finding its sup and inf on any set that contains at least one point for each half-line described above.
Since the set $ G:={x in mathbb{R}^n , | , g(x) = 1}$ has this property, we can safely say that sup $h(x)$ = sup${h(x) , | , x in G}$ , and same for the inf.
Now, for every point $x$ in $G$, $h(x) = f(x)$ , so it suffices to find the sup and the inf of $f$. But $G$ is a compact set, and therefore from the continuity of $f$ (norms are continuous) and Weierstrass' theorem, we conclude that sup $f$ and inf $f$ on $G$ are both in $mathbb{R}_+^*$ (they can't be zero because they are actually a min and a max, and $f$ is never zero outside of the origin). This means that there are two real positive constants $c_1$ and $c_2$ such that $$c_1 leq frac{f(x)}{g(x)} , leq , c_2 quad Longrightarrow quad c_1 , g(x) leq f(x) leq c_2 , g(x) $$
for any non-zero $x$. Since for $x=0$ the inequality is true, we have the equivalence of the two norms.
TLDR: let $h$ be the ratio of two norms and show that $h$ is bounded above and below by two positive real numbers.
subproof 1 Let's show that $G$ has the property that it crosses each half line through the origin at least once.
Let $tv$ be a half-line. Now, we want to see if there's a $t$ for which it crosses $G$, so we can just plug the points of the half-line into $g$ and use homogeneity: $g(tv) = t g(v)$. So, if we choose $t$ to be $1/g(v)$ , which can be done because g is non-zero outside the origin, we can be sure that the half line will cross our set. Actually for any other $t$ we are not touching $G$ so each half-line crosses exactly once the level sets of a norm :) pretty neat.
subproof 2 I wanted to show that $G$ is closed and bounded, again just because.
The fact that it's closed follows from the fact that it is the counter image of the closed subset of the real numbers ${1}$ through a continuous function $g$.
Let's see why it's bounded (my reasoning is pretty convoluted; I would be curious to see other ways to show this).
Let $S_1$ be the euclidean unit sphere around the origin. Let's call $h$ the minimum of $g$ on $S_1$. This number exists since $g$ is continuous and $S_1$ is compact.
case $h<1$: Let $a$ be positive, real, and strictly greater than 1. Let $S_{a/h}$ be the euclidean sphere of radius $a/h$ around the origin. Since the halflines through the origin (let's say $v$ is a unit vector) provide a bijection between the two sets, we can say that for each $x$ in $S_{a/h}$ , there is a $y$ in $S_1$ such that
$$g(x) = gleft(frac{a}{h} , yright) = frac{a}{h} , g(y) geq frac{a}{h} , h = a > 1 , .$$
Thus for each $x$ with an euclidean distance more than $1/h$ from the origin, its $g$-norm will be strictly greater than 1, making $G$ bounded.
Case $h geq 1 $: Each $x$ with euclidean norm $a>1$ will be such that there is a corresponding $y$ on $S_1$ for which:
$$ g(x) = g(a , y) = a ,g(y) geq a cdot 1 > 1 , .$$
Thus $G$ is bounded.
I am curious to have some feedback about all this. Especially regarding ways in which this can be generalized, or things that have crossed your mind while reading this, as well as other proofs.
This idea came to me while i was solving this problem: find, for each even $n in mathbb{N}$ the best constant $a$ such that, for all real $x$ and $y$,
$$ (x + y)^n leq a (x^n + y^n) , .$$
Other related questions/ideas that came up and I have not yet answered, mainly because I know pretty much nothing about convex functions apart from the definition:
1) are sublevels of convex functions convex? Is this an if and only if?
2) is it true that if a function is convex then the level sets have the property that they intersect every halfline trough the origin exactly once? And is this an if and only if?
3) are sublevels of convex continuous functions compact?
Thanks for reading :)
real-analysis proof-verification proof-writing norm equivalent-metrics
asked Jan 31 at 14:08
gent96gent96
414
$endgroup$
Just wanted to share my proof with you smart people to have some feedback and to share each other's ideas. Some disclaimers: this is my first post, English is not my first language, and I know that there are simpler ways to prove this. Here it goes:
Suppose $f$ and $g$ are two norms on $mathbb{R}^n$, and define the following function for each non-zero vector in $mathbb{R}^n$:
$$h(x) := dfrac{f(x)}{g(x)}$$
$h$ is well defined because $g$ is zero iff $x$ is zero. Now since $h$ is positively homogeneous of degree zero (because $f$ and $g$ both are positively homogeneous of degree one), it is constant on all half-lines passing through the origin (excluding the origin). With this I mean that it is constant on every set of the type ${tv}$, with $v$ non-zero vector of $mathbb{R}^n$ and $t in mathbb{R}_+^*$. Because of this, if we want to find the sup and the inf of $h$ on its domain, we can restrict ourselves to finding its sup and inf on any set that contains at least one point for each half-line described above.
Since the set $ G:={x in mathbb{R}^n , | , g(x) = 1}$ has this property, we can safely say that sup $h(x)$ = sup${h(x) , | , x in G}$ , and same for the inf.
Now, for every point $x$ in $G$, $h(x) = f(x)$ , so it suffices to find the sup and the inf of $f$. But $G$ is a compact set, and therefore from the continuity of $f$ (norms are continuous) and Weierstrass' theorem, we conclude that sup $f$ and inf $f$ on $G$ are both in $mathbb{R}_+^*$ (they can't be zero because they are actually a min and a max, and $f$ is never zero outside of the origin). This means that there are two real positive constants $c_1$ and $c_2$ such that $$c_1 leq frac{f(x)}{g(x)} , leq , c_2 quad Longrightarrow quad c_1 , g(x) leq f(x) leq c_2 , g(x) $$
for any non-zero $x$. Since for $x=0$ the inequality is true, we have the equivalence of the two norms.
TLDR: let $h$ be the ratio of two norms and show that $h$ is bounded above and below by two positive real numbers.
subproof 1 Let's show that $G$ has the property that it crosses each half line through the origin at least once.
Let $tv$ be a half-line. Now, we want to see if there's a $t$ for which it crosses $G$, so we can just plug the points of the half-line into $g$ and use homogeneity: $g(tv) = t g(v)$. So, if we choose $t$ to be $1/g(v)$ , which can be done because g is non-zero outside the origin, we can be sure that the half line will cross our set. Actually for any other $t$ we are not touching $G$ so each half-line crosses exactly once the level sets of a norm :) pretty neat.
subproof 2 I wanted to show that $G$ is closed and bounded, again just because.
The fact that it's closed follows from the fact that it is the counter image of the closed subset of the real numbers ${1}$ through a continuous function $g$.
Let's see why it's bounded (my reasoning is pretty convoluted; I would be curious to see other ways to show this).
Let $S_1$ be the euclidean unit sphere around the origin. Let's call $h$ the minimum of $g$ on $S_1$. This number exists since $g$ is continuous and $S_1$ is compact.
case $h<1$: Let $a$ be positive, real, and strictly greater than 1. Let $S_{a/h}$ be the euclidean sphere of radius $a/h$ around the origin. Since the halflines through the origin (let's say $v$ is a unit vector) provide a bijection between the two sets, we can say that for each $x$ in $S_{a/h}$ , there is a $y$ in $S_1$ such that
$$g(x) = gleft(frac{a}{h} , yright) = frac{a}{h} , g(y) geq frac{a}{h} , h = a > 1 , .$$
Thus for each $x$ with an euclidean distance more than $1/h$ from the origin, its $g$-norm will be strictly greater than 1, making $G$ bounded.
Case $h geq 1 $: Each $x$ with euclidean norm $a>1$ will be such that there is a corresponding $y$ on $S_1$ for which:
$$ g(x) = g(a , y) = a ,g(y) geq a cdot 1 > 1 , .$$
Thus $G$ is bounded.
I am curious to have some feedback about all this. Especially regarding ways in which this can be generalized, or things that have crossed your mind while reading this, as well as other proofs.
This idea came to me while i was solving this problem: find, for each even $n in mathbb{N}$ the best constant $a$ such that, for all real $x$ and $y$,
$$ (x + y)^n leq a (x^n + y^n) , .$$
Other related questions/ideas that came up and I have not yet answered, mainly because I know pretty much nothing about convex functions apart from the definition:
1) are sublevels of convex functions convex? Is this an if and only if?
2) is it true that if a function is convex then the level sets have the property that they intersect every halfline trough the origin exactly once? And is this an if and only if?
3) are sublevels of convex continuous functions compact?
Thanks for reading :)
real-analysis proof-verification proof-writing norm equivalent-metrics
real-analysis proof-verification proof-writing norm equivalent-metrics
asked Jan 31 at 14:08
gent96gent96
414
asked Jan 31 at 14:08
gent96gent96
414
edited Jan 31 at 14:25
gent96
asked Jan 31 at 14:08
gent96gent96
414
asked Jan 31 at 14:08
gent96gent96
414
asked Jan 31 at 14:08
gent96gent96
414
414
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Your proof is incorrect. Or rather, it is missing something crucial: that $f$ and $g$ are continuous with respect to the standard topology on $mathbb{R}^n$, i.e. the one that comes the usual Euclidean norm. If you know that $f$ is equivalent to the usual Euclidean norm, then it's easy enough to show that $f$ is continuous for the usual topology (and similarly for $g$); but this is precisely what you want to show! So you cannot assume that $f$ is equivalent to the usual Euclidean norm, and then, how do you show that it it continuous for the usual topology?
You have used this fact three times. In logical order:
In "subproof 2", you show that $G$ is closed, as it is the preimage of ${1}$ under $g$. But you actually show that it is closed for the topology induced by $g$, not the usual topology. From this, you cannot deduce that $G$ is compact under the usual topology using the Heine–Borel theorem; this theorem requires $G$ to be closed wrt the usual topology (and bounded wrt the usual norm).
Also in "subproof 2", you say that $g$ attains a minimum on $S_1$ because $S_1$ is compact (implicitly, wrt the usual topology). But you do not know that $g$ is continuous wrt the usual topology.
In your main proof, you say that $G$ is compact (implicitly, with the usual topology) and therefore that $f$ attains a maximum and a minimum on $G$. But you do not know that $f$ is continuous wrt the usual topology, so you cannot deduce this.
tl;dr If you do not know that the norms are equivalent, how would you show that they induce the same topology on $mathbb{R}^n$?
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
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– gent96
Jan 31 at 14:54
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Your proof is incorrect. Or rather, it is missing something crucial: that $f$ and $g$ are continuous with respect to the standard topology on $mathbb{R}^n$, i.e. the one that comes the usual Euclidean norm. If you know that $f$ is equivalent to the usual Euclidean norm, then it's easy enough to show that $f$ is continuous for the usual topology (and similarly for $g$); but this is precisely what you want to show! So you cannot assume that $f$ is equivalent to the usual Euclidean norm, and then, how do you show that it it continuous for the usual topology?
You have used this fact three times. In logical order:
In "subproof 2", you show that $G$ is closed, as it is the preimage of ${1}$ under $g$. But you actually show that it is closed for the topology induced by $g$, not the usual topology. From this, you cannot deduce that $G$ is compact under the usual topology using the Heine–Borel theorem; this theorem requires $G$ to be closed wrt the usual topology (and bounded wrt the usual norm).
Also in "subproof 2", you say that $g$ attains a minimum on $S_1$ because $S_1$ is compact (implicitly, wrt the usual topology). But you do not know that $g$ is continuous wrt the usual topology.
In your main proof, you say that $G$ is compact (implicitly, with the usual topology) and therefore that $f$ attains a maximum and a minimum on $G$. But you do not know that $f$ is continuous wrt the usual topology, so you cannot deduce this.
tl;dr If you do not know that the norms are equivalent, how would you show that they induce the same topology on $mathbb{R}^n$?
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
add a comment |
$begingroup$
Your proof is incorrect. Or rather, it is missing something crucial: that $f$ and $g$ are continuous with respect to the standard topology on $mathbb{R}^n$, i.e. the one that comes the usual Euclidean norm. If you know that $f$ is equivalent to the usual Euclidean norm, then it's easy enough to show that $f$ is continuous for the usual topology (and similarly for $g$); but this is precisely what you want to show! So you cannot assume that $f$ is equivalent to the usual Euclidean norm, and then, how do you show that it it continuous for the usual topology?
You have used this fact three times. In logical order:
In "subproof 2", you show that $G$ is closed, as it is the preimage of ${1}$ under $g$. But you actually show that it is closed for the topology induced by $g$, not the usual topology. From this, you cannot deduce that $G$ is compact under the usual topology using the Heine–Borel theorem; this theorem requires $G$ to be closed wrt the usual topology (and bounded wrt the usual norm).
Also in "subproof 2", you say that $g$ attains a minimum on $S_1$ because $S_1$ is compact (implicitly, wrt the usual topology). But you do not know that $g$ is continuous wrt the usual topology.
In your main proof, you say that $G$ is compact (implicitly, with the usual topology) and therefore that $f$ attains a maximum and a minimum on $G$. But you do not know that $f$ is continuous wrt the usual topology, so you cannot deduce this.
tl;dr If you do not know that the norms are equivalent, how would you show that they induce the same topology on $mathbb{R}^n$?
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
add a comment |
$begingroup$
Your proof is incorrect. Or rather, it is missing something crucial: that $f$ and $g$ are continuous with respect to the standard topology on $mathbb{R}^n$, i.e. the one that comes the usual Euclidean norm. If you know that $f$ is equivalent to the usual Euclidean norm, then it's easy enough to show that $f$ is continuous for the usual topology (and similarly for $g$); but this is precisely what you want to show! So you cannot assume that $f$ is equivalent to the usual Euclidean norm, and then, how do you show that it it continuous for the usual topology?
You have used this fact three times. In logical order:
In "subproof 2", you show that $G$ is closed, as it is the preimage of ${1}$ under $g$. But you actually show that it is closed for the topology induced by $g$, not the usual topology. From this, you cannot deduce that $G$ is compact under the usual topology using the Heine–Borel theorem; this theorem requires $G$ to be closed wrt the usual topology (and bounded wrt the usual norm).
Also in "subproof 2", you say that $g$ attains a minimum on $S_1$ because $S_1$ is compact (implicitly, wrt the usual topology). But you do not know that $g$ is continuous wrt the usual topology.
In your main proof, you say that $G$ is compact (implicitly, with the usual topology) and therefore that $f$ attains a maximum and a minimum on $G$. But you do not know that $f$ is continuous wrt the usual topology, so you cannot deduce this.
tl;dr If you do not know that the norms are equivalent, how would you show that they induce the same topology on $mathbb{R}^n$?
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
Your proof is incorrect. Or rather, it is missing something crucial: that $f$ and $g$ are continuous with respect to the standard topology on $mathbb{R}^n$, i.e. the one that comes the usual Euclidean norm. If you know that $f$ is equivalent to the usual Euclidean norm, then it's easy enough to show that $f$ is continuous for the usual topology (and similarly for $g$); but this is precisely what you want to show! So you cannot assume that $f$ is equivalent to the usual Euclidean norm, and then, how do you show that it it continuous for the usual topology?
You have used this fact three times. In logical order:
In "subproof 2", you show that $G$ is closed, as it is the preimage of ${1}$ under $g$. But you actually show that it is closed for the topology induced by $g$, not the usual topology. From this, you cannot deduce that $G$ is compact under the usual topology using the Heine–Borel theorem; this theorem requires $G$ to be closed wrt the usual topology (and bounded wrt the usual norm).
Also in "subproof 2", you say that $g$ attains a minimum on $S_1$ because $S_1$ is compact (implicitly, wrt the usual topology). But you do not know that $g$ is continuous wrt the usual topology.
In your main proof, you say that $G$ is compact (implicitly, with the usual topology) and therefore that $f$ attains a maximum and a minimum on $G$. But you do not know that $f$ is continuous wrt the usual topology, so you cannot deduce this.
tl;dr If you do not know that the norms are equivalent, how would you show that they induce the same topology on $mathbb{R}^n$?
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
edited Jan 31 at 14:34
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
answered Jan 31 at 14:28
Najib IdrissiNajib Idrissi
41.9k473143
41.9k473143
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
add a comment |
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
$begingroup$
I think you are right. I thought I had proven that any norm is continuous with sequences but I didn't pay attention to a crucial step (let $x_k$ be a sequence converging to the vector $x$, then with the second triangle inequality we show that $f(x_k)$ converges to $f(x)$. But this cannot be said if we don't say that the $f$ norm is smaller than the euclidean norm multiplied by some constant). Can it not be shown, that norms are all continuous with the euclidean definition of continuity, without saying that they are equivalent? I guess if it cannot be done then the proof can't work..
$endgroup$
– gent96
Jan 31 at 14:54
add a comment |
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