If $langle f'(x) cdot v , v rangle > 0$ then $f$ is injective












4












$begingroup$


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







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 2 '15 at 23:14







      Aaron Maroja

















      asked Jul 2 '15 at 2:39









      Aaron MarojaAaron Maroja

      15.5k51447




      15.5k51447






















          3 Answers
          3






          active

          oldest

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          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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





















          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              active

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              active

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              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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


















              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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
















              1












              1








              1





              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$



              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.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jul 2 '15 at 15:41









              Aaron Maroja

              15.5k51447




              15.5k51447










              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




















              • $begingroup$
                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













              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$



                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 25 '16 at 13:32

























                  answered Jul 2 '15 at 23:11









                  Aaron MarojaAaron Maroja

                  15.5k51447




                  15.5k51447























                      0












                      $begingroup$

                      What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?






                          share|cite|improve this answer









                          $endgroup$



                          What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 22 at 22:39









                          João CostaJoão Costa

                          755




                          755






























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                              The Binding of Isaac: Rebirth/Afterbirth

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                              Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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                              What does “Dominus providebit” mean?

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                              5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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