If $langle f'(x) cdot v , v rangle > 0$ then $f$ is injective
$begingroup$
Question:
Let $f: U to mathbb R^m$ differentiable at the convex set $U subseteq mathbb R^m$. If $$langle f'(x) cdot v , v rangle > 0 , ,,, forall,, x in U, v neq 0 in mathbb R^m $$
then $f$ is injective. If $f in C^1$ then $f$ is a diffeomorphism of $U$ over a subset of $mathbb R^m$. Give an example such that $U = mathbb R^m$, but $f$ is not surjective.
Attempt: The idea is to show $$|f(x+v) - f(x)| > 0$$
As $U$ is convex and $f$ id differentiable in $U$ then $[x,x+v] subseteq U$ for any $x, x + v in U$, by the Mean Value Theorem there exists $theta in (0,1)$ such that $$f(x + v) - f(x) = frac{partial f}{partial v}(x + theta v) = f'(x + theta v) cdot v$$
Then $$langle f(x + v) - f(x) , vrangle = langle f'(x + theta v) cdot v , vrangle > 0 $$
and I couldn't conclude anything.
The second part is o.k.
Any thoughts?
Edit: I can't use the Mean Value Theorem here.
analysis multivariable-calculus
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
add a comment |
$begingroup$
Question:
Let $f: U to mathbb R^m$ differentiable at the convex set $U subseteq mathbb R^m$. If $$langle f'(x) cdot v , v rangle > 0 , ,,, forall,, x in U, v neq 0 in mathbb R^m $$
then $f$ is injective. If $f in C^1$ then $f$ is a diffeomorphism of $U$ over a subset of $mathbb R^m$. Give an example such that $U = mathbb R^m$, but $f$ is not surjective.
Attempt: The idea is to show $$|f(x+v) - f(x)| > 0$$
As $U$ is convex and $f$ id differentiable in $U$ then $[x,x+v] subseteq U$ for any $x, x + v in U$, by the Mean Value Theorem there exists $theta in (0,1)$ such that $$f(x + v) - f(x) = frac{partial f}{partial v}(x + theta v) = f'(x + theta v) cdot v$$
Then $$langle f(x + v) - f(x) , vrangle = langle f'(x + theta v) cdot v , vrangle > 0 $$
and I couldn't conclude anything.
The second part is o.k.
Any thoughts?
Edit: I can't use the Mean Value Theorem here.
analysis multivariable-calculus
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
add a comment |
$begingroup$
Question:
Let $f: U to mathbb R^m$ differentiable at the convex set $U subseteq mathbb R^m$. If $$langle f'(x) cdot v , v rangle > 0 , ,,, forall,, x in U, v neq 0 in mathbb R^m $$
then $f$ is injective. If $f in C^1$ then $f$ is a diffeomorphism of $U$ over a subset of $mathbb R^m$. Give an example such that $U = mathbb R^m$, but $f$ is not surjective.
Attempt: The idea is to show $$|f(x+v) - f(x)| > 0$$
As $U$ is convex and $f$ id differentiable in $U$ then $[x,x+v] subseteq U$ for any $x, x + v in U$, by the Mean Value Theorem there exists $theta in (0,1)$ such that $$f(x + v) - f(x) = frac{partial f}{partial v}(x + theta v) = f'(x + theta v) cdot v$$
Then $$langle f(x + v) - f(x) , vrangle = langle f'(x + theta v) cdot v , vrangle > 0 $$
and I couldn't conclude anything.
The second part is o.k.
Any thoughts?
Edit: I can't use the Mean Value Theorem here.
analysis multivariable-calculus
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
Question:
Let $f: U to mathbb R^m$ differentiable at the convex set $U subseteq mathbb R^m$. If $$langle f'(x) cdot v , v rangle > 0 , ,,, forall,, x in U, v neq 0 in mathbb R^m $$
then $f$ is injective. If $f in C^1$ then $f$ is a diffeomorphism of $U$ over a subset of $mathbb R^m$. Give an example such that $U = mathbb R^m$, but $f$ is not surjective.
Attempt: The idea is to show $$|f(x+v) - f(x)| > 0$$
As $U$ is convex and $f$ id differentiable in $U$ then $[x,x+v] subseteq U$ for any $x, x + v in U$, by the Mean Value Theorem there exists $theta in (0,1)$ such that $$f(x + v) - f(x) = frac{partial f}{partial v}(x + theta v) = f'(x + theta v) cdot v$$
Then $$langle f(x + v) - f(x) , vrangle = langle f'(x + theta v) cdot v , vrangle > 0 $$
and I couldn't conclude anything.
The second part is o.k.
Any thoughts?
Edit: I can't use the Mean Value Theorem here.
analysis multivariable-calculus
analysis multivariable-calculus
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
edited Jul 2 '15 at 23:14
Aaron Maroja
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
asked Jul 2 '15 at 2:39
Aaron MarojaAaron Maroja
15.5k51447
15.5k51447
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I think that you have already solved the problem. Pick $x, yin U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||cdot||y-x||>0$. As you picked $xneq y$, then $f$ must be injective. Sorry for my english.
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
$endgroup$
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
|
show 3 more comments
$begingroup$
After a while I figured it out.
Consider the function $$begin{align}psi: [0,1] &to mathbb R\ t &mapsto langle f(a + tv), v rangleend{align}$$
defined at $[0,1]$, differentiable with
$$begin{align}psi' (t) &= langle f'(a + tv)cdot v ,vrangle + langle f(a+tv),0 rangle \ &= langle f'(a + tv)cdot v rangle > 0end{align} tag {*}$$
for $v = b -a$ and for all $t in [0,1]$, then $psi$ is strictly increasing at this closed interval. Now suppose that for $a neq b$ we have $f(a) = f(b)$.
$$begin{align}psi (1) - psi (0) &= langle f(b), v rangle - langle f(a),v rangle\&= langle f(b) - f(a), v rangle\ &= 0end{align}$$
As $psi$ is continuous and $psi (1) = psi (0)$ by Rolle's Theorem there exists $c in (0,1)$ such that $psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) neq f(b)$ and it follows that $f$ is injective.
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
add a comment |
$begingroup$
What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?
answered Jan 22 at 22:39
João CostaJoão Costa
755
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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$begingroup$
I think that you have already solved the problem. Pick $x, yin U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||cdot||y-x||>0$. As you picked $xneq y$, then $f$ must be injective. Sorry for my english.
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
$endgroup$
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
|
show 3 more comments
$begingroup$
I think that you have already solved the problem. Pick $x, yin U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||cdot||y-x||>0$. As you picked $xneq y$, then $f$ must be injective. Sorry for my english.
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
$endgroup$
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
|
show 3 more comments
$begingroup$
I think that you have already solved the problem. Pick $x, yin U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||cdot||y-x||>0$. As you picked $xneq y$, then $f$ must be injective. Sorry for my english.
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
$endgroup$
I think that you have already solved the problem. Pick $x, yin U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||cdot||y-x||>0$. As you picked $xneq y$, then $f$ must be injective. Sorry for my english.
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
edited Jul 2 '15 at 15:41
Aaron Maroja
15.5k51447
15.5k51447
answered Jul 2 '15 at 2:54
MicaelMicael
18319
answered Jul 2 '15 at 2:54
MicaelMicael
18319
answered Jul 2 '15 at 2:54
MicaelMicael
18319
18319
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
|
show 3 more comments
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
Cauchy-Schwarz inequality
$endgroup$
– Micael
Jul 2 '15 at 3:02
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
I just found a counter example where I can't use the Mean Value Theorem for $f: U to mathbb R^m$, is there any way to walk around this?
$endgroup$
– Aaron Maroja
Jul 2 '15 at 15:09
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
Just for the record, this does not answer the question.
$endgroup$
– Aaron Maroja
Jul 2 '15 at 23:29
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
The point here is that, when you parametrize the segment $[x, y]subset U$ you can use the M.V.T. for functions from $R$ into $R^m$.
$endgroup$
– Micael
Jul 3 '15 at 6:11
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
$begingroup$
Could you show me the proof somewhere, a paper or a reference?
$endgroup$
– Aaron Maroja
Jul 3 '15 at 11:55
|
show 3 more comments
$begingroup$
After a while I figured it out.
Consider the function $$begin{align}psi: [0,1] &to mathbb R\ t &mapsto langle f(a + tv), v rangleend{align}$$
defined at $[0,1]$, differentiable with
$$begin{align}psi' (t) &= langle f'(a + tv)cdot v ,vrangle + langle f(a+tv),0 rangle \ &= langle f'(a + tv)cdot v rangle > 0end{align} tag {*}$$
for $v = b -a$ and for all $t in [0,1]$, then $psi$ is strictly increasing at this closed interval. Now suppose that for $a neq b$ we have $f(a) = f(b)$.
$$begin{align}psi (1) - psi (0) &= langle f(b), v rangle - langle f(a),v rangle\&= langle f(b) - f(a), v rangle\ &= 0end{align}$$
As $psi$ is continuous and $psi (1) = psi (0)$ by Rolle's Theorem there exists $c in (0,1)$ such that $psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) neq f(b)$ and it follows that $f$ is injective.
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
add a comment |
$begingroup$
After a while I figured it out.
Consider the function $$begin{align}psi: [0,1] &to mathbb R\ t &mapsto langle f(a + tv), v rangleend{align}$$
defined at $[0,1]$, differentiable with
$$begin{align}psi' (t) &= langle f'(a + tv)cdot v ,vrangle + langle f(a+tv),0 rangle \ &= langle f'(a + tv)cdot v rangle > 0end{align} tag {*}$$
for $v = b -a$ and for all $t in [0,1]$, then $psi$ is strictly increasing at this closed interval. Now suppose that for $a neq b$ we have $f(a) = f(b)$.
$$begin{align}psi (1) - psi (0) &= langle f(b), v rangle - langle f(a),v rangle\&= langle f(b) - f(a), v rangle\ &= 0end{align}$$
As $psi$ is continuous and $psi (1) = psi (0)$ by Rolle's Theorem there exists $c in (0,1)$ such that $psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) neq f(b)$ and it follows that $f$ is injective.
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
add a comment |
$begingroup$
After a while I figured it out.
Consider the function $$begin{align}psi: [0,1] &to mathbb R\ t &mapsto langle f(a + tv), v rangleend{align}$$
defined at $[0,1]$, differentiable with
$$begin{align}psi' (t) &= langle f'(a + tv)cdot v ,vrangle + langle f(a+tv),0 rangle \ &= langle f'(a + tv)cdot v rangle > 0end{align} tag {*}$$
for $v = b -a$ and for all $t in [0,1]$, then $psi$ is strictly increasing at this closed interval. Now suppose that for $a neq b$ we have $f(a) = f(b)$.
$$begin{align}psi (1) - psi (0) &= langle f(b), v rangle - langle f(a),v rangle\&= langle f(b) - f(a), v rangle\ &= 0end{align}$$
As $psi$ is continuous and $psi (1) = psi (0)$ by Rolle's Theorem there exists $c in (0,1)$ such that $psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) neq f(b)$ and it follows that $f$ is injective.
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
$endgroup$
After a while I figured it out.
Consider the function $$begin{align}psi: [0,1] &to mathbb R\ t &mapsto langle f(a + tv), v rangleend{align}$$
defined at $[0,1]$, differentiable with
$$begin{align}psi' (t) &= langle f'(a + tv)cdot v ,vrangle + langle f(a+tv),0 rangle \ &= langle f'(a + tv)cdot v rangle > 0end{align} tag {*}$$
for $v = b -a$ and for all $t in [0,1]$, then $psi$ is strictly increasing at this closed interval. Now suppose that for $a neq b$ we have $f(a) = f(b)$.
$$begin{align}psi (1) - psi (0) &= langle f(b), v rangle - langle f(a),v rangle\&= langle f(b) - f(a), v rangle\ &= 0end{align}$$
As $psi$ is continuous and $psi (1) = psi (0)$ by Rolle's Theorem there exists $c in (0,1)$ such that $psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) neq f(b)$ and it follows that $f$ is injective.
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
edited Feb 25 '16 at 13:32
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
answered Jul 2 '15 at 23:11
Aaron MarojaAaron Maroja
15.5k51447
15.5k51447
add a comment |
add a comment |
$begingroup$
What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?
answered Jan 22 at 22:39
João CostaJoão Costa
755
$endgroup$
add a comment |
$begingroup$
What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?
answered Jan 22 at 22:39
João CostaJoão Costa
755
$endgroup$
add a comment |
$begingroup$
What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?
answered Jan 22 at 22:39
João CostaJoão Costa
755
$endgroup$
What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?
answered Jan 22 at 22:39
João CostaJoão Costa
755
answered Jan 22 at 22:39
João CostaJoão Costa
755
answered Jan 22 at 22:39
João CostaJoão Costa
755
answered Jan 22 at 22:39
João CostaJoão Costa
755
755
add a comment |
add a comment |
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