Bounds on function from $[a,b]$ to $mathbb{R}^m$
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1
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Let $phi : [a,b] to mathbb{R}^m$ be continues and differentiable on $( a,b)$ and let $lVert xrVert = sqrt{langle x,xrangle}$ the euclidean norm in $mathbb{R}^m$, show that $lVertphi(b)-phi(a)rVert leq sup_{t in (a,b)} lVertphi'(t)rVert (b-a)$ ?
I did used Lagrange mean value theorem on $g(t) = langlephi(t),phi(b)-phi(a)rangle$ but ended up with two sup on the interval $(a,b)$
Any hints or proof or ideas are appreciated
calculus multivariable-calculus derivatives partial-derivative
edited yesterday
user10354138
6,199523
asked yesterday
Ahmad
2,4401625
add a comment |
up vote
1
down vote
favorite
Let $phi : [a,b] to mathbb{R}^m$ be continues and differentiable on $( a,b)$ and let $lVert xrVert = sqrt{langle x,xrangle}$ the euclidean norm in $mathbb{R}^m$, show that $lVertphi(b)-phi(a)rVert leq sup_{t in (a,b)} lVertphi'(t)rVert (b-a)$ ?
I did used Lagrange mean value theorem on $g(t) = langlephi(t),phi(b)-phi(a)rangle$ but ended up with two sup on the interval $(a,b)$
Any hints or proof or ideas are appreciated
calculus multivariable-calculus derivatives partial-derivative
edited yesterday
user10354138
6,199523
asked yesterday
Ahmad
2,4401625
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $phi : [a,b] to mathbb{R}^m$ be continues and differentiable on $( a,b)$ and let $lVert xrVert = sqrt{langle x,xrangle}$ the euclidean norm in $mathbb{R}^m$, show that $lVertphi(b)-phi(a)rVert leq sup_{t in (a,b)} lVertphi'(t)rVert (b-a)$ ?
I did used Lagrange mean value theorem on $g(t) = langlephi(t),phi(b)-phi(a)rangle$ but ended up with two sup on the interval $(a,b)$
Any hints or proof or ideas are appreciated
calculus multivariable-calculus derivatives partial-derivative
edited yesterday
user10354138
6,199523
asked yesterday
Ahmad
2,4401625
Let $phi : [a,b] to mathbb{R}^m$ be continues and differentiable on $( a,b)$ and let $lVert xrVert = sqrt{langle x,xrangle}$ the euclidean norm in $mathbb{R}^m$, show that $lVertphi(b)-phi(a)rVert leq sup_{t in (a,b)} lVertphi'(t)rVert (b-a)$ ?
I did used Lagrange mean value theorem on $g(t) = langlephi(t),phi(b)-phi(a)rangle$ but ended up with two sup on the interval $(a,b)$
Any hints or proof or ideas are appreciated
calculus multivariable-calculus derivatives partial-derivative
calculus multivariable-calculus derivatives partial-derivative
edited yesterday
user10354138
6,199523
asked yesterday
Ahmad
2,4401625
edited yesterday
user10354138
6,199523
asked yesterday
Ahmad
2,4401625
edited yesterday
user10354138
6,199523
edited yesterday
user10354138
6,199523
edited yesterday
user10354138
6,199523
6,199523
asked yesterday
Ahmad
2,4401625
asked yesterday
Ahmad
2,4401625
asked yesterday
Ahmad
2,4401625
2,4401625
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday
add a comment |
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday
1
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You have just to apply the Mean Value Theorem to your function
$$
g(t) := langle phi(t), phi(b) - phi(a)rangle,
qquad t in [a,b].
$$
Namely,
$$
g(b) - g(a) = |phi(b) - phi(a)|^2,
$$
whereas, for $tin (a,b)$, by the Cauchy-Schwarz inequality we get
$$
g'(t) = langle phi'(t), phi(b) - phi(a)rangle
leq |phi'(t)|, |phi(b)-phi(a)|
leq sup_{sin (a,b)}|phi'(s)|, |phi(b) - phi(a)|.
$$
On the other hand, by the MVT, there exists $xi in (a,b)$ such that
$$
g(b) - g(a) = g'(xi) (b-a)
$$
so that
$$
|phi(b) - phi(a)|^2 = g'(xi) (b-a)
leq sup|phi'(t)|, |phi(b) - phi(a)|, (b-a).
$$
answered yesterday
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have just to apply the Mean Value Theorem to your function
$$
g(t) := langle phi(t), phi(b) - phi(a)rangle,
qquad t in [a,b].
$$
Namely,
$$
g(b) - g(a) = |phi(b) - phi(a)|^2,
$$
whereas, for $tin (a,b)$, by the Cauchy-Schwarz inequality we get
$$
g'(t) = langle phi'(t), phi(b) - phi(a)rangle
leq |phi'(t)|, |phi(b)-phi(a)|
leq sup_{sin (a,b)}|phi'(s)|, |phi(b) - phi(a)|.
$$
On the other hand, by the MVT, there exists $xi in (a,b)$ such that
$$
g(b) - g(a) = g'(xi) (b-a)
$$
so that
$$
|phi(b) - phi(a)|^2 = g'(xi) (b-a)
leq sup|phi'(t)|, |phi(b) - phi(a)|, (b-a).
$$
answered yesterday
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
add a comment |
up vote
2
down vote
accepted
You have just to apply the Mean Value Theorem to your function
$$
g(t) := langle phi(t), phi(b) - phi(a)rangle,
qquad t in [a,b].
$$
Namely,
$$
g(b) - g(a) = |phi(b) - phi(a)|^2,
$$
whereas, for $tin (a,b)$, by the Cauchy-Schwarz inequality we get
$$
g'(t) = langle phi'(t), phi(b) - phi(a)rangle
leq |phi'(t)|, |phi(b)-phi(a)|
leq sup_{sin (a,b)}|phi'(s)|, |phi(b) - phi(a)|.
$$
On the other hand, by the MVT, there exists $xi in (a,b)$ such that
$$
g(b) - g(a) = g'(xi) (b-a)
$$
so that
$$
|phi(b) - phi(a)|^2 = g'(xi) (b-a)
leq sup|phi'(t)|, |phi(b) - phi(a)|, (b-a).
$$
answered yesterday
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have just to apply the Mean Value Theorem to your function
$$
g(t) := langle phi(t), phi(b) - phi(a)rangle,
qquad t in [a,b].
$$
Namely,
$$
g(b) - g(a) = |phi(b) - phi(a)|^2,
$$
whereas, for $tin (a,b)$, by the Cauchy-Schwarz inequality we get
$$
g'(t) = langle phi'(t), phi(b) - phi(a)rangle
leq |phi'(t)|, |phi(b)-phi(a)|
leq sup_{sin (a,b)}|phi'(s)|, |phi(b) - phi(a)|.
$$
On the other hand, by the MVT, there exists $xi in (a,b)$ such that
$$
g(b) - g(a) = g'(xi) (b-a)
$$
so that
$$
|phi(b) - phi(a)|^2 = g'(xi) (b-a)
leq sup|phi'(t)|, |phi(b) - phi(a)|, (b-a).
$$
answered yesterday
Rigel
10.6k11319
You have just to apply the Mean Value Theorem to your function
$$
g(t) := langle phi(t), phi(b) - phi(a)rangle,
qquad t in [a,b].
$$
Namely,
$$
g(b) - g(a) = |phi(b) - phi(a)|^2,
$$
whereas, for $tin (a,b)$, by the Cauchy-Schwarz inequality we get
$$
g'(t) = langle phi'(t), phi(b) - phi(a)rangle
leq |phi'(t)|, |phi(b)-phi(a)|
leq sup_{sin (a,b)}|phi'(s)|, |phi(b) - phi(a)|.
$$
On the other hand, by the MVT, there exists $xi in (a,b)$ such that
$$
g(b) - g(a) = g'(xi) (b-a)
$$
so that
$$
|phi(b) - phi(a)|^2 = g'(xi) (b-a)
leq sup|phi'(t)|, |phi(b) - phi(a)|, (b-a).
$$
answered yesterday
Rigel
10.6k11319
edited yesterday
answered yesterday
Rigel
10.6k11319
answered yesterday
Rigel
10.6k11319
answered yesterday
Rigel
10.6k11319
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
add a comment |
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
you used Cauchy-Schwartz inequality ?
– Ahmad
yesterday
Yes, in the third formula.
– Rigel
yesterday
Yes, in the third formula.
– Rigel
yesterday
add a comment |
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1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
yesterday
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
yesterday
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
yesterday