How to prove $vertbulletvertcircpi^{-1}$ is a norm
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Suppose the seminorm on X is real valued, $U=Xbigcaplbrace x:vert xvert =0rbrace$, Y=X/U, and $pi:Xrightarrow Y$ is the canonical projection.
I want to show that $vertbulletvertcircpi^{-1}$ is a norm on Y.
But I do not know how to prove that $vertbulletvertcircpi^{-1}(y_1+y_2)le vertbulletvertcircpi^{-1}(y_1)+vertbulletvertcircpi^{-1}(y_2)$, but it is one of the necessary condition, right?
functional-analysis
$endgroup$
add a comment |
$begingroup$
Suppose the seminorm on X is real valued, $U=Xbigcaplbrace x:vert xvert =0rbrace$, Y=X/U, and $pi:Xrightarrow Y$ is the canonical projection.
I want to show that $vertbulletvertcircpi^{-1}$ is a norm on Y.
But I do not know how to prove that $vertbulletvertcircpi^{-1}(y_1+y_2)le vertbulletvertcircpi^{-1}(y_1)+vertbulletvertcircpi^{-1}(y_2)$, but it is one of the necessary condition, right?
functional-analysis
$endgroup$
$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19
add a comment |
$begingroup$
Suppose the seminorm on X is real valued, $U=Xbigcaplbrace x:vert xvert =0rbrace$, Y=X/U, and $pi:Xrightarrow Y$ is the canonical projection.
I want to show that $vertbulletvertcircpi^{-1}$ is a norm on Y.
But I do not know how to prove that $vertbulletvertcircpi^{-1}(y_1+y_2)le vertbulletvertcircpi^{-1}(y_1)+vertbulletvertcircpi^{-1}(y_2)$, but it is one of the necessary condition, right?
functional-analysis
$endgroup$
Suppose the seminorm on X is real valued, $U=Xbigcaplbrace x:vert xvert =0rbrace$, Y=X/U, and $pi:Xrightarrow Y$ is the canonical projection.
I want to show that $vertbulletvertcircpi^{-1}$ is a norm on Y.
But I do not know how to prove that $vertbulletvertcircpi^{-1}(y_1+y_2)le vertbulletvertcircpi^{-1}(y_1)+vertbulletvertcircpi^{-1}(y_2)$, but it is one of the necessary condition, right?
functional-analysis
functional-analysis
asked Dec 17 '18 at 14:09
user627221
$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19
add a comment |
$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19
$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19
$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19
add a comment |
1 Answer
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$begingroup$
You have a seminorm $|cdot|$ on $X$ and define
$$
|pi(x)|=|x|
$$
over $Y=X/U$, where $U={xin X:|x|=0}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $pi(x)$ for some $xin X$.
First of all, you have to prove that $|cdot|$ is well defined on $Y$. If $pi(x)=pi(x')$, then $x-x'in U$, so
$$
|x|le |x-x'|+|x'|=|x'|
$$
and, similarly, $|x'|le|x|$. Thus $|x|=|x'|$ and the proof is complete.
The proof of the seminorm properties is essentially trivial; for the triangle inequality,
$$
|pi(x)+pi(y)|=|pi(x+y)|=|x+y|le|x|+|y|=|pi(x)|+|pi(y)|
$$
and similarly for the other properties. We have a norm because
$$
|pi(x)|=0 iff |x|=0 iff xin U iff pi(x)=pi(0)
$$
so $|pi(x)|=0$ implies $pi(x)=pi(0)$.
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have a seminorm $|cdot|$ on $X$ and define
$$
|pi(x)|=|x|
$$
over $Y=X/U$, where $U={xin X:|x|=0}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $pi(x)$ for some $xin X$.
First of all, you have to prove that $|cdot|$ is well defined on $Y$. If $pi(x)=pi(x')$, then $x-x'in U$, so
$$
|x|le |x-x'|+|x'|=|x'|
$$
and, similarly, $|x'|le|x|$. Thus $|x|=|x'|$ and the proof is complete.
The proof of the seminorm properties is essentially trivial; for the triangle inequality,
$$
|pi(x)+pi(y)|=|pi(x+y)|=|x+y|le|x|+|y|=|pi(x)|+|pi(y)|
$$
and similarly for the other properties. We have a norm because
$$
|pi(x)|=0 iff |x|=0 iff xin U iff pi(x)=pi(0)
$$
so $|pi(x)|=0$ implies $pi(x)=pi(0)$.
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
You have a seminorm $|cdot|$ on $X$ and define
$$
|pi(x)|=|x|
$$
over $Y=X/U$, where $U={xin X:|x|=0}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $pi(x)$ for some $xin X$.
First of all, you have to prove that $|cdot|$ is well defined on $Y$. If $pi(x)=pi(x')$, then $x-x'in U$, so
$$
|x|le |x-x'|+|x'|=|x'|
$$
and, similarly, $|x'|le|x|$. Thus $|x|=|x'|$ and the proof is complete.
The proof of the seminorm properties is essentially trivial; for the triangle inequality,
$$
|pi(x)+pi(y)|=|pi(x+y)|=|x+y|le|x|+|y|=|pi(x)|+|pi(y)|
$$
and similarly for the other properties. We have a norm because
$$
|pi(x)|=0 iff |x|=0 iff xin U iff pi(x)=pi(0)
$$
so $|pi(x)|=0$ implies $pi(x)=pi(0)$.
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
You have a seminorm $|cdot|$ on $X$ and define
$$
|pi(x)|=|x|
$$
over $Y=X/U$, where $U={xin X:|x|=0}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $pi(x)$ for some $xin X$.
First of all, you have to prove that $|cdot|$ is well defined on $Y$. If $pi(x)=pi(x')$, then $x-x'in U$, so
$$
|x|le |x-x'|+|x'|=|x'|
$$
and, similarly, $|x'|le|x|$. Thus $|x|=|x'|$ and the proof is complete.
The proof of the seminorm properties is essentially trivial; for the triangle inequality,
$$
|pi(x)+pi(y)|=|pi(x+y)|=|x+y|le|x|+|y|=|pi(x)|+|pi(y)|
$$
and similarly for the other properties. We have a norm because
$$
|pi(x)|=0 iff |x|=0 iff xin U iff pi(x)=pi(0)
$$
so $|pi(x)|=0$ implies $pi(x)=pi(0)$.
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
$endgroup$
You have a seminorm $|cdot|$ on $X$ and define
$$
|pi(x)|=|x|
$$
over $Y=X/U$, where $U={xin X:|x|=0}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $pi(x)$ for some $xin X$.
First of all, you have to prove that $|cdot|$ is well defined on $Y$. If $pi(x)=pi(x')$, then $x-x'in U$, so
$$
|x|le |x-x'|+|x'|=|x'|
$$
and, similarly, $|x'|le|x|$. Thus $|x|=|x'|$ and the proof is complete.
The proof of the seminorm properties is essentially trivial; for the triangle inequality,
$$
|pi(x)+pi(y)|=|pi(x+y)|=|x+y|le|x|+|y|=|pi(x)|+|pi(y)|
$$
and similarly for the other properties. We have a norm because
$$
|pi(x)|=0 iff |x|=0 iff xin U iff pi(x)=pi(0)
$$
so $|pi(x)|=0$ implies $pi(x)=pi(0)$.
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
answered Dec 17 '18 at 14:25
egregegreg
183k1486204
183k1486204
add a comment |
add a comment |
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$begingroup$
The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space.
$endgroup$
– user328442
Dec 17 '18 at 14:19