Mixing
Problem about strictly normed spaces.
$begingroup$
A normed vector space $(V,Vert cdot Vert)$ is strictly normed if
$$Vert x + yVert = Vert x Vert + Vert y Vert$$
with $x,yneq 0$ only if $y = lambda x$ where $lambda >0$.
(a) Prove that $V$ is strictly normed if only if the sphere
$$sigma_{1}(0) = {x in V mid Vert x Vert = 1}$$
contains no segments.
(b) Give examples of strictly normed spaces and not strictly normed spaces.
My attempt.
(a) Suppose that $V$ is strictly normed. Take $x, y in sigma_{1}(0)$ with $x neq y$. If $y = alpha x$ with $alpha > 0$, then
$$1 = Vert y Vert = alpha Vert x Vert = alpha Longrightarrow y = x.$$
Moreover
$$Vert lambda x + (1-lambda)yVert = Vert lambda x Vert + Vert(1-lambda)yVert$$
only if $(1-lambda)y = alpha lambda x$, that is, $displaystyle y = left(frac{lambdaalpha}{1-lambda}right)x$ (we can take $lambda in (0,1)$), then
$$Vert lambda x + (1-lambda)yVert < lambdaVert x Vert + (1-lambda)Vert y Vert = 1$$
and if $lambda x + (1-lambda)y in sigma_{1}(0)$, $Vert lambda x + (1-lambda)y Vert = 1$ and so, $1<1$, an absurd!
For converse, I take $x,y in V$ with $x neq y$. So, $displaystyle frac{x}{Vert x Vert},frac{y}{Vert y Vert} in sigma_{1}(0)$. But I dont know how to use the hypothesis. Can someone help me?
Edit.
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert leq lambda frac{Vert x Vert}{Vert x Vert} + (1-lambda)frac{Vert y Vert}{Vert y Vert} = 1,$$
but, if $lambda in (0,1)$,
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert notin sigma_{1}(0),$$
then
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert < 1.$$
(b) Consider the Euclidean norm $Vert cdot Vert_{E}$ and the sum norm $Vert cdot Vert_{infty}$. So, $(mathbb{R}^{n},Vert cdot Vert_{E})$ is strictly normed and $(mathbb{R}^{n}, Vert cdot Vert_{infty})$ is not strictly normed.
Here, I didnt write proof of this, I'm just using the previous equivalence. Please, correct me if Im wrong.
Can someone knows another examples of not strictly normed?
metric-spaces normed-spaces
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
$endgroup$
add a comment |
$begingroup$
A normed vector space $(V,Vert cdot Vert)$ is strictly normed if
$$Vert x + yVert = Vert x Vert + Vert y Vert$$
with $x,yneq 0$ only if $y = lambda x$ where $lambda >0$.
(a) Prove that $V$ is strictly normed if only if the sphere
$$sigma_{1}(0) = {x in V mid Vert x Vert = 1}$$
contains no segments.
(b) Give examples of strictly normed spaces and not strictly normed spaces.
My attempt.
(a) Suppose that $V$ is strictly normed. Take $x, y in sigma_{1}(0)$ with $x neq y$. If $y = alpha x$ with $alpha > 0$, then
$$1 = Vert y Vert = alpha Vert x Vert = alpha Longrightarrow y = x.$$
Moreover
$$Vert lambda x + (1-lambda)yVert = Vert lambda x Vert + Vert(1-lambda)yVert$$
only if $(1-lambda)y = alpha lambda x$, that is, $displaystyle y = left(frac{lambdaalpha}{1-lambda}right)x$ (we can take $lambda in (0,1)$), then
$$Vert lambda x + (1-lambda)yVert < lambdaVert x Vert + (1-lambda)Vert y Vert = 1$$
and if $lambda x + (1-lambda)y in sigma_{1}(0)$, $Vert lambda x + (1-lambda)y Vert = 1$ and so, $1<1$, an absurd!
For converse, I take $x,y in V$ with $x neq y$. So, $displaystyle frac{x}{Vert x Vert},frac{y}{Vert y Vert} in sigma_{1}(0)$. But I dont know how to use the hypothesis. Can someone help me?
Edit.
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert leq lambda frac{Vert x Vert}{Vert x Vert} + (1-lambda)frac{Vert y Vert}{Vert y Vert} = 1,$$
but, if $lambda in (0,1)$,
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert notin sigma_{1}(0),$$
then
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert < 1.$$
(b) Consider the Euclidean norm $Vert cdot Vert_{E}$ and the sum norm $Vert cdot Vert_{infty}$. So, $(mathbb{R}^{n},Vert cdot Vert_{E})$ is strictly normed and $(mathbb{R}^{n}, Vert cdot Vert_{infty})$ is not strictly normed.
Here, I didnt write proof of this, I'm just using the previous equivalence. Please, correct me if Im wrong.
Can someone knows another examples of not strictly normed?
metric-spaces normed-spaces
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
$endgroup$
$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
1
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34
add a comment |
$begingroup$
A normed vector space $(V,Vert cdot Vert)$ is strictly normed if
$$Vert x + yVert = Vert x Vert + Vert y Vert$$
with $x,yneq 0$ only if $y = lambda x$ where $lambda >0$.
(a) Prove that $V$ is strictly normed if only if the sphere
$$sigma_{1}(0) = {x in V mid Vert x Vert = 1}$$
contains no segments.
(b) Give examples of strictly normed spaces and not strictly normed spaces.
My attempt.
(a) Suppose that $V$ is strictly normed. Take $x, y in sigma_{1}(0)$ with $x neq y$. If $y = alpha x$ with $alpha > 0$, then
$$1 = Vert y Vert = alpha Vert x Vert = alpha Longrightarrow y = x.$$
Moreover
$$Vert lambda x + (1-lambda)yVert = Vert lambda x Vert + Vert(1-lambda)yVert$$
only if $(1-lambda)y = alpha lambda x$, that is, $displaystyle y = left(frac{lambdaalpha}{1-lambda}right)x$ (we can take $lambda in (0,1)$), then
$$Vert lambda x + (1-lambda)yVert < lambdaVert x Vert + (1-lambda)Vert y Vert = 1$$
and if $lambda x + (1-lambda)y in sigma_{1}(0)$, $Vert lambda x + (1-lambda)y Vert = 1$ and so, $1<1$, an absurd!
For converse, I take $x,y in V$ with $x neq y$. So, $displaystyle frac{x}{Vert x Vert},frac{y}{Vert y Vert} in sigma_{1}(0)$. But I dont know how to use the hypothesis. Can someone help me?
Edit.
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert leq lambda frac{Vert x Vert}{Vert x Vert} + (1-lambda)frac{Vert y Vert}{Vert y Vert} = 1,$$
but, if $lambda in (0,1)$,
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert notin sigma_{1}(0),$$
then
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert < 1.$$
(b) Consider the Euclidean norm $Vert cdot Vert_{E}$ and the sum norm $Vert cdot Vert_{infty}$. So, $(mathbb{R}^{n},Vert cdot Vert_{E})$ is strictly normed and $(mathbb{R}^{n}, Vert cdot Vert_{infty})$ is not strictly normed.
Here, I didnt write proof of this, I'm just using the previous equivalence. Please, correct me if Im wrong.
Can someone knows another examples of not strictly normed?
metric-spaces normed-spaces
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
$endgroup$
A normed vector space $(V,Vert cdot Vert)$ is strictly normed if
$$Vert x + yVert = Vert x Vert + Vert y Vert$$
with $x,yneq 0$ only if $y = lambda x$ where $lambda >0$.
(a) Prove that $V$ is strictly normed if only if the sphere
$$sigma_{1}(0) = {x in V mid Vert x Vert = 1}$$
contains no segments.
(b) Give examples of strictly normed spaces and not strictly normed spaces.
My attempt.
(a) Suppose that $V$ is strictly normed. Take $x, y in sigma_{1}(0)$ with $x neq y$. If $y = alpha x$ with $alpha > 0$, then
$$1 = Vert y Vert = alpha Vert x Vert = alpha Longrightarrow y = x.$$
Moreover
$$Vert lambda x + (1-lambda)yVert = Vert lambda x Vert + Vert(1-lambda)yVert$$
only if $(1-lambda)y = alpha lambda x$, that is, $displaystyle y = left(frac{lambdaalpha}{1-lambda}right)x$ (we can take $lambda in (0,1)$), then
$$Vert lambda x + (1-lambda)yVert < lambdaVert x Vert + (1-lambda)Vert y Vert = 1$$
and if $lambda x + (1-lambda)y in sigma_{1}(0)$, $Vert lambda x + (1-lambda)y Vert = 1$ and so, $1<1$, an absurd!
For converse, I take $x,y in V$ with $x neq y$. So, $displaystyle frac{x}{Vert x Vert},frac{y}{Vert y Vert} in sigma_{1}(0)$. But I dont know how to use the hypothesis. Can someone help me?
Edit.
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert leq lambda frac{Vert x Vert}{Vert x Vert} + (1-lambda)frac{Vert y Vert}{Vert y Vert} = 1,$$
but, if $lambda in (0,1)$,
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert notin sigma_{1}(0),$$
then
$$leftVert lambda frac{x}{Vert x Vert} + (1-lambda)frac{y}{Vert y Vert}rightVert < 1.$$
(b) Consider the Euclidean norm $Vert cdot Vert_{E}$ and the sum norm $Vert cdot Vert_{infty}$. So, $(mathbb{R}^{n},Vert cdot Vert_{E})$ is strictly normed and $(mathbb{R}^{n}, Vert cdot Vert_{infty})$ is not strictly normed.
Here, I didnt write proof of this, I'm just using the previous equivalence. Please, correct me if Im wrong.
Can someone knows another examples of not strictly normed?
metric-spaces normed-spaces
metric-spaces normed-spaces
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
edited Jan 5 at 2:58
Lucas Corrêa
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
asked Jan 5 at 2:13
Lucas CorrêaLucas Corrêa
1,6051321
1,6051321
$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
1
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34
add a comment |
$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
1
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34
$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
1
1
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
frac{x+y}{|x|+|y|}=frac{|x|}{|x|+|y|}Big(frac{x}{|x|}Big)+frac{|y|}{|x|+|y|}Big(frac{y}{|y|}Big)
$$
This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|le 1$.
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
$endgroup$
add a comment |
$begingroup$
Suppose $V$'s unit sphere contains no line segments, and $x, y in V$ such that
$$|x + y| = |x| + |y|.$$
Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $|x|$ from $0$ and distance $|y|$ from $x + y$. Working this out, you'll get
$$z = frac{|x|(x + y)}{|x + y|}.$$
Note that $z$ lies on the spheres $S[0; |x|]$ and $S[x + y; |y|]$.
Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; |x|]$, it follows from the convexity of the ball that $frac{x + z}{2}$ must lie in the open ball $B(0; |x|)$, which is to say $left|frac{x + z}{2}right| < |x|$. On the same token, we have $frac{x + z}{2} in B(x + y, |y|)$. Hence,
$$|x + y| le left|frac{x + z}{2}right| + left|x + y - frac{x + z}{2}right| < |x| + |y| = |x + y|,$$
which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.
As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
frac{x+y}{|x|+|y|}=frac{|x|}{|x|+|y|}Big(frac{x}{|x|}Big)+frac{|y|}{|x|+|y|}Big(frac{y}{|y|}Big)
$$
This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|le 1$.
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
$endgroup$
add a comment |
$begingroup$
$$
frac{x+y}{|x|+|y|}=frac{|x|}{|x|+|y|}Big(frac{x}{|x|}Big)+frac{|y|}{|x|+|y|}Big(frac{y}{|y|}Big)
$$
This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|le 1$.
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
$endgroup$
add a comment |
$begingroup$
$$
frac{x+y}{|x|+|y|}=frac{|x|}{|x|+|y|}Big(frac{x}{|x|}Big)+frac{|y|}{|x|+|y|}Big(frac{y}{|y|}Big)
$$
This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|le 1$.
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
$endgroup$
$$
frac{x+y}{|x|+|y|}=frac{|x|}{|x|+|y|}Big(frac{x}{|x|}Big)+frac{|y|}{|x|+|y|}Big(frac{y}{|y|}Big)
$$
This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|le 1$.
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
answered Jan 5 at 2:56
Mike EarnestMike Earnest
21.4k11951
21.4k11951
add a comment |
add a comment |
$begingroup$
Suppose $V$'s unit sphere contains no line segments, and $x, y in V$ such that
$$|x + y| = |x| + |y|.$$
Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $|x|$ from $0$ and distance $|y|$ from $x + y$. Working this out, you'll get
$$z = frac{|x|(x + y)}{|x + y|}.$$
Note that $z$ lies on the spheres $S[0; |x|]$ and $S[x + y; |y|]$.
Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; |x|]$, it follows from the convexity of the ball that $frac{x + z}{2}$ must lie in the open ball $B(0; |x|)$, which is to say $left|frac{x + z}{2}right| < |x|$. On the same token, we have $frac{x + z}{2} in B(x + y, |y|)$. Hence,
$$|x + y| le left|frac{x + z}{2}right| + left|x + y - frac{x + z}{2}right| < |x| + |y| = |x + y|,$$
which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.
As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
$endgroup$
add a comment |
$begingroup$
Suppose $V$'s unit sphere contains no line segments, and $x, y in V$ such that
$$|x + y| = |x| + |y|.$$
Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $|x|$ from $0$ and distance $|y|$ from $x + y$. Working this out, you'll get
$$z = frac{|x|(x + y)}{|x + y|}.$$
Note that $z$ lies on the spheres $S[0; |x|]$ and $S[x + y; |y|]$.
Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; |x|]$, it follows from the convexity of the ball that $frac{x + z}{2}$ must lie in the open ball $B(0; |x|)$, which is to say $left|frac{x + z}{2}right| < |x|$. On the same token, we have $frac{x + z}{2} in B(x + y, |y|)$. Hence,
$$|x + y| le left|frac{x + z}{2}right| + left|x + y - frac{x + z}{2}right| < |x| + |y| = |x + y|,$$
which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.
As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
$endgroup$
add a comment |
$begingroup$
Suppose $V$'s unit sphere contains no line segments, and $x, y in V$ such that
$$|x + y| = |x| + |y|.$$
Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $|x|$ from $0$ and distance $|y|$ from $x + y$. Working this out, you'll get
$$z = frac{|x|(x + y)}{|x + y|}.$$
Note that $z$ lies on the spheres $S[0; |x|]$ and $S[x + y; |y|]$.
Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; |x|]$, it follows from the convexity of the ball that $frac{x + z}{2}$ must lie in the open ball $B(0; |x|)$, which is to say $left|frac{x + z}{2}right| < |x|$. On the same token, we have $frac{x + z}{2} in B(x + y, |y|)$. Hence,
$$|x + y| le left|frac{x + z}{2}right| + left|x + y - frac{x + z}{2}right| < |x| + |y| = |x + y|,$$
which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.
As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
$endgroup$
Suppose $V$'s unit sphere contains no line segments, and $x, y in V$ such that
$$|x + y| = |x| + |y|.$$
Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $|x|$ from $0$ and distance $|y|$ from $x + y$. Working this out, you'll get
$$z = frac{|x|(x + y)}{|x + y|}.$$
Note that $z$ lies on the spheres $S[0; |x|]$ and $S[x + y; |y|]$.
Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; |x|]$, it follows from the convexity of the ball that $frac{x + z}{2}$ must lie in the open ball $B(0; |x|)$, which is to say $left|frac{x + z}{2}right| < |x|$. On the same token, we have $frac{x + z}{2} in B(x + y, |y|)$. Hence,
$$|x + y| le left|frac{x + z}{2}right| + left|x + y - frac{x + z}{2}right| < |x| + |y| = |x + y|,$$
which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.
As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
answered Jan 5 at 2:56
Theo BenditTheo Bendit
17.3k12149
17.3k12149
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$begingroup$
The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/|x|$ and $y/|y|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$.
$endgroup$
– Mike Earnest
Jan 5 at 2:48
$begingroup$
@MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it.
$endgroup$
– Lucas Corrêa
Jan 5 at 2:52
1
$begingroup$
Your work was correct, you just had to choose $lambda$ so that $|x+y|$ somehow entered the picture. It turns out the correct choice is $lambda=|x|/(|x|+|y|)$.
$endgroup$
– Mike Earnest
Jan 5 at 3:01
$begingroup$
Also, the $L_1$ norm is not strict.
$endgroup$
– Mike Earnest
Jan 5 at 15:34