Bounds on function from $[a,b]$ to $mathbb{R}^m$
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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 Nov 13 at 6:40
user10354138
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
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 Nov 13 at 6:40
user10354138
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42
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 Nov 13 at 6:40
user10354138
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
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 Nov 13 at 6:40
user10354138
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
edited Nov 13 at 6:40
user10354138
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
edited Nov 13 at 6:40
user10354138
6,294623
edited Nov 13 at 6:40
user10354138
6,294623
edited Nov 13 at 6:40
user10354138
6,294623
6,294623
asked Nov 13 at 6:32
Ahmad
2,4551625
asked Nov 13 at 6:32
Ahmad
2,4551625
asked Nov 13 at 6:32
Ahmad
2,4551625
2,4551625
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42
add a comment |
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42
1
1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42
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 Nov 13 at 6:46
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
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 Nov 13 at 6:46
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
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 Nov 13 at 6:46
Rigel
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
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 Nov 13 at 6:46
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 Nov 13 at 6:46
Rigel
10.6k11319
edited Nov 13 at 7:16
answered Nov 13 at 6:46
Rigel
10.6k11319
answered Nov 13 at 6:46
Rigel
10.6k11319
answered Nov 13 at 6:46
Rigel
10.6k11319
10.6k11319
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
add a comment |
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
you used Cauchy-Schwartz inequality ?
– Ahmad
Nov 13 at 6:49
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
Yes, in the third formula.
– Rigel
Nov 13 at 7:15
add a comment |
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1
I might be wrong, but I think this is an application of Mean Value Theorem.
– IAmNoOne
Nov 13 at 6:35
@IAmNoOne i think so, but i used it twice so my bound is weaker than what i have to show!
– Ahmad
Nov 13 at 6:38
Use MVT for the function $t to phi (at+(1-t)b)$ defined on $[0,1]$.
– Kavi Rama Murthy
Nov 13 at 6:42