Prove that any linear functional on $mathbb{R^{n}}$ with the Euclidean norm is bounded
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I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.
real-analysis functional-analysis
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up vote
1
down vote
favorite
I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.
real-analysis functional-analysis
1
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15
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show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.
real-analysis functional-analysis
I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.
real-analysis functional-analysis
real-analysis functional-analysis
edited Nov 17 at 17:05
asked Nov 17 at 16:58
Roger
545
545
1
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15
|
show 2 more comments
1
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15
1
1
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15
|
show 2 more comments
1 Answer
1
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For
$vec v in Bbb R^n, tag 1$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$
where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with
$phi: Bbb R^n to Bbb R tag 3$
linear, we have
$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$
thus
$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$
let
$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$
then
$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$
also,
$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$
whence
$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$
so that (7) becomes
$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$
which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.
Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if
$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$
we have
$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$
from which
$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$
the vectors $vec e_i$ are chosen orthonormal with respect to this basis:
$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$
End of Note.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For
$vec v in Bbb R^n, tag 1$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$
where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with
$phi: Bbb R^n to Bbb R tag 3$
linear, we have
$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$
thus
$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$
let
$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$
then
$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$
also,
$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$
whence
$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$
so that (7) becomes
$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$
which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.
Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if
$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$
we have
$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$
from which
$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$
the vectors $vec e_i$ are chosen orthonormal with respect to this basis:
$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$
End of Note.
add a comment |
up vote
1
down vote
accepted
For
$vec v in Bbb R^n, tag 1$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$
where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with
$phi: Bbb R^n to Bbb R tag 3$
linear, we have
$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$
thus
$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$
let
$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$
then
$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$
also,
$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$
whence
$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$
so that (7) becomes
$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$
which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.
Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if
$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$
we have
$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$
from which
$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$
the vectors $vec e_i$ are chosen orthonormal with respect to this basis:
$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$
End of Note.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For
$vec v in Bbb R^n, tag 1$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$
where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with
$phi: Bbb R^n to Bbb R tag 3$
linear, we have
$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$
thus
$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$
let
$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$
then
$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$
also,
$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$
whence
$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$
so that (7) becomes
$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$
which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.
Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if
$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$
we have
$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$
from which
$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$
the vectors $vec e_i$ are chosen orthonormal with respect to this basis:
$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$
End of Note.
For
$vec v in Bbb R^n, tag 1$
we may write
$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$
where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with
$phi: Bbb R^n to Bbb R tag 3$
linear, we have
$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$
thus
$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$
let
$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$
then
$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$
also,
$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$
whence
$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$
so that (7) becomes
$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$
which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.
Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if
$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$
we have
$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$
from which
$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$
the vectors $vec e_i$ are chosen orthonormal with respect to this basis:
$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$
End of Note.
answered Nov 17 at 18:27
Robert Lewis
42k22760
42k22760
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1
Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03
Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05
Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06
@AnuragA Thanks
– Roger
Nov 17 at 17:08
@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15