Mixing
Norm of element from dual space
$begingroup$
I am reading the following book:
The Mathematical Tools for the Study of the Incompressible NS
And here is one of the propositions (Proposition II.2.1):
Let $E$ be a normed vector space. Then $forall x in E$ and $forall f in E^{'}$ (dual space) one can have:
$||x||_{E} = sup_{substack{fin E^{'}} \ f neq 0}frac{|<f,x>|_{E^{'},E}}{||f||}=mathrm{sup}_{substack{||f||_{E^{'}}}leq 1}|<f,x>|_{E^{'},E}$
in which $<f,x>_{E,E^{'}} = f(x)$ is the dulaity bracket.
My question is the second equal sign. I think it should be $||f||_{E^{'}} geq 1$?
real-analysis functional-analysis
asked Jan 2 at 2:37
mkomko
85
$endgroup$
add a comment |
$begingroup$
I am reading the following book:
The Mathematical Tools for the Study of the Incompressible NS
And here is one of the propositions (Proposition II.2.1):
Let $E$ be a normed vector space. Then $forall x in E$ and $forall f in E^{'}$ (dual space) one can have:
$||x||_{E} = sup_{substack{fin E^{'}} \ f neq 0}frac{|<f,x>|_{E^{'},E}}{||f||}=mathrm{sup}_{substack{||f||_{E^{'}}}leq 1}|<f,x>|_{E^{'},E}$
in which $<f,x>_{E,E^{'}} = f(x)$ is the dulaity bracket.
My question is the second equal sign. I think it should be $||f||_{E^{'}} geq 1$?
real-analysis functional-analysis
asked Jan 2 at 2:37
mkomko
85
$endgroup$
$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45
add a comment |
$begingroup$
I am reading the following book:
The Mathematical Tools for the Study of the Incompressible NS
And here is one of the propositions (Proposition II.2.1):
Let $E$ be a normed vector space. Then $forall x in E$ and $forall f in E^{'}$ (dual space) one can have:
$||x||_{E} = sup_{substack{fin E^{'}} \ f neq 0}frac{|<f,x>|_{E^{'},E}}{||f||}=mathrm{sup}_{substack{||f||_{E^{'}}}leq 1}|<f,x>|_{E^{'},E}$
in which $<f,x>_{E,E^{'}} = f(x)$ is the dulaity bracket.
My question is the second equal sign. I think it should be $||f||_{E^{'}} geq 1$?
real-analysis functional-analysis
asked Jan 2 at 2:37
mkomko
85
$endgroup$
I am reading the following book:
The Mathematical Tools for the Study of the Incompressible NS
And here is one of the propositions (Proposition II.2.1):
Let $E$ be a normed vector space. Then $forall x in E$ and $forall f in E^{'}$ (dual space) one can have:
$||x||_{E} = sup_{substack{fin E^{'}} \ f neq 0}frac{|<f,x>|_{E^{'},E}}{||f||}=mathrm{sup}_{substack{||f||_{E^{'}}}leq 1}|<f,x>|_{E^{'},E}$
in which $<f,x>_{E,E^{'}} = f(x)$ is the dulaity bracket.
My question is the second equal sign. I think it should be $||f||_{E^{'}} geq 1$?
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 2 at 2:37
mkomko
85
asked Jan 2 at 2:37
mkomko
85
asked Jan 2 at 2:37
mkomko
85
asked Jan 2 at 2:37
mkomko
85
asked Jan 2 at 2:37
mkomko
85
85
$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45
add a comment |
$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45
$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45
$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement is fine as it is written. Note that the map
$$Etomathbb [0,infty]qquad xmapstosup_{||f||geq 1}|langle f,xrangle|$$
takes $0$ to $0$ , and every nonzero $xin E$ to $infty$. Indeed, if $xneq 0$, then by the Hahn-Banach theorem, there is some nonzero $fin E'$ with $|f|=1$ and $f(x)=|x|$. Then for all $M>1$, we have
$$sup_{||f||geq 1}|langle f,xrangle|geqlangle Mf,xrangle=M|x|,$$
whence $sup_{||f||geq 1}|langle f,xrangle|=infty$.
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
$endgroup$
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is fine as it is written. Note that the map
$$Etomathbb [0,infty]qquad xmapstosup_{||f||geq 1}|langle f,xrangle|$$
takes $0$ to $0$ , and every nonzero $xin E$ to $infty$. Indeed, if $xneq 0$, then by the Hahn-Banach theorem, there is some nonzero $fin E'$ with $|f|=1$ and $f(x)=|x|$. Then for all $M>1$, we have
$$sup_{||f||geq 1}|langle f,xrangle|geqlangle Mf,xrangle=M|x|,$$
whence $sup_{||f||geq 1}|langle f,xrangle|=infty$.
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
$endgroup$
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
|
show 2 more comments
$begingroup$
The statement is fine as it is written. Note that the map
$$Etomathbb [0,infty]qquad xmapstosup_{||f||geq 1}|langle f,xrangle|$$
takes $0$ to $0$ , and every nonzero $xin E$ to $infty$. Indeed, if $xneq 0$, then by the Hahn-Banach theorem, there is some nonzero $fin E'$ with $|f|=1$ and $f(x)=|x|$. Then for all $M>1$, we have
$$sup_{||f||geq 1}|langle f,xrangle|geqlangle Mf,xrangle=M|x|,$$
whence $sup_{||f||geq 1}|langle f,xrangle|=infty$.
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
$endgroup$
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
|
show 2 more comments
$begingroup$
The statement is fine as it is written. Note that the map
$$Etomathbb [0,infty]qquad xmapstosup_{||f||geq 1}|langle f,xrangle|$$
takes $0$ to $0$ , and every nonzero $xin E$ to $infty$. Indeed, if $xneq 0$, then by the Hahn-Banach theorem, there is some nonzero $fin E'$ with $|f|=1$ and $f(x)=|x|$. Then for all $M>1$, we have
$$sup_{||f||geq 1}|langle f,xrangle|geqlangle Mf,xrangle=M|x|,$$
whence $sup_{||f||geq 1}|langle f,xrangle|=infty$.
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
$endgroup$
The statement is fine as it is written. Note that the map
$$Etomathbb [0,infty]qquad xmapstosup_{||f||geq 1}|langle f,xrangle|$$
takes $0$ to $0$ , and every nonzero $xin E$ to $infty$. Indeed, if $xneq 0$, then by the Hahn-Banach theorem, there is some nonzero $fin E'$ with $|f|=1$ and $f(x)=|x|$. Then for all $M>1$, we have
$$sup_{||f||geq 1}|langle f,xrangle|geqlangle Mf,xrangle=M|x|,$$
whence $sup_{||f||geq 1}|langle f,xrangle|=infty$.
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
answered Jan 2 at 2:59
AweyganAweygan
13.6k21441
13.6k21441
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
|
show 2 more comments
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
sorry if it is trivial: Let's say (I found this example online) $f(x)=frac{1}{2}(3x^2-1)$ in $[-1,1]$. Then $||f||_2=sqrt{2/5}<1$. And $|<f,x>|=|frac{1}{2}(3x^2-1)|$. Then I think the first equal sign give me $1/sqrt{2/5}$ and the second one just the number $1$. So we lower the upper bound. Sorry if it is stupid.
$endgroup$
– mko
Jan 2 at 3:51
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
two things: $[-1,1]$ is not a normed vector space, and $f$ is not linear.
$endgroup$
– Aweygan
Jan 2 at 3:58
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
and another thing: I don't know how you used these equations, you have to take the supremum over all $fin E'$ with the specified properties.
$endgroup$
– Aweygan
Jan 2 at 4:00
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
Thanks I got it now. Thanks again.
$endgroup$
– mko
Jan 2 at 4:10
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
$begingroup$
You're welcome, glad to help. If you're satisfied with my answer, feel free to accept it, so that no one gets the wrong idea and tries to answer.
$endgroup$
– Aweygan
Jan 2 at 4:14
|
show 2 more comments
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$begingroup$
No, the definition is OK.
$endgroup$
– Dante Grevino
Jan 2 at 2:45