Extenstion of Intermediate Value Theorem.












1












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Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?



Thank you,
JF






real-analysis vector-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?



    Thank you,
    JF






    real-analysis vector-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?



      Thank you,
      JF






      real-analysis vector-spaces







      share|cite|improve this question









      $endgroup$




      Let $f:[0,1]^{d}longrightarrow mathbb{R}^{d}$ with $dgeq 2$. $f$ is continuous and let $cin (0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $xin (0,1)^{d}$ such that $f(x)=(c,...,c)$ ?



      Thank you,
      JF






      real-analysis vector-spaces




      real-analysis vector-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jun 2 '14 at 16:29









      user154763user154763

      61




      61






















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
            $endgroup$
            – user154763
            Jun 2 '14 at 18:25












          • $begingroup$
            I see, still the concept holds, as the function can go around the point without having to go through it.
            $endgroup$
            – Ellya
            Jun 2 '14 at 18:45










          • $begingroup$
            How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
            $endgroup$
            – user154763
            Jun 2 '14 at 19:13





















          0












          $begingroup$

          I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.



          $d$ is a member of N.



          P = $mathbb{R}^{d}$



          Q is an $mathbb{R}^{d - 1}$ dimensional line in P.



          A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…




          When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:




          1. The points may become equal and unequal any number of times without becoming equal to a member of Q.


          2. The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.







          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
              $endgroup$
              – user154763
              Jun 2 '14 at 18:25












            • $begingroup$
              I see, still the concept holds, as the function can go around the point without having to go through it.
              $endgroup$
              – Ellya
              Jun 2 '14 at 18:45










            • $begingroup$
              How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
              $endgroup$
              – user154763
              Jun 2 '14 at 19:13


















            0












            $begingroup$

            I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
              $endgroup$
              – user154763
              Jun 2 '14 at 18:25












            • $begingroup$
              I see, still the concept holds, as the function can go around the point without having to go through it.
              $endgroup$
              – Ellya
              Jun 2 '14 at 18:45










            • $begingroup$
              How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
              $endgroup$
              – user154763
              Jun 2 '14 at 19:13
















            0












            0








            0





            $begingroup$

            I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.






            share|cite|improve this answer









            $endgroup$



            I believe there are many things negating this, on being how would you define $a>>b$ for vectors? and with higher dimensions, we can start of "below" a point, and then go around it to be "above" it.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 2 '14 at 16:34









            EllyaEllya

            9,57211226




            9,57211226












            • $begingroup$
              If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
              $endgroup$
              – user154763
              Jun 2 '14 at 18:25












            • $begingroup$
              I see, still the concept holds, as the function can go around the point without having to go through it.
              $endgroup$
              – Ellya
              Jun 2 '14 at 18:45










            • $begingroup$
              How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
              $endgroup$
              – user154763
              Jun 2 '14 at 19:13




















            • $begingroup$
              If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
              $endgroup$
              – user154763
              Jun 2 '14 at 18:25












            • $begingroup$
              I see, still the concept holds, as the function can go around the point without having to go through it.
              $endgroup$
              – Ellya
              Jun 2 '14 at 18:45










            • $begingroup$
              How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
              $endgroup$
              – user154763
              Jun 2 '14 at 19:13


















            $begingroup$
            If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
            $endgroup$
            – user154763
            Jun 2 '14 at 18:25






            $begingroup$
            If $x$ is $ngeq 2$ dimensional, $x>>y$ would be defined as $x_{1}>y_{1}$ and ... and $x_{n}>y_{n}$.
            $endgroup$
            – user154763
            Jun 2 '14 at 18:25














            $begingroup$
            I see, still the concept holds, as the function can go around the point without having to go through it.
            $endgroup$
            – Ellya
            Jun 2 '14 at 18:45




            $begingroup$
            I see, still the concept holds, as the function can go around the point without having to go through it.
            $endgroup$
            – Ellya
            Jun 2 '14 at 18:45












            $begingroup$
            How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
            $endgroup$
            – user154763
            Jun 2 '14 at 19:13






            $begingroup$
            How about when $frac{partial f_{i}(x_{i},x_{-i})}{partial x_{i}}>0 forall x_{-i}$? Seems like it would simplify greatly...
            $endgroup$
            – user154763
            Jun 2 '14 at 19:13













            0












            $begingroup$

            I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.



            $d$ is a member of N.



            P = $mathbb{R}^{d}$



            Q is an $mathbb{R}^{d - 1}$ dimensional line in P.



            A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…




            When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:




            1. The points may become equal and unequal any number of times without becoming equal to a member of Q.


            2. The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.







            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.



              $d$ is a member of N.



              P = $mathbb{R}^{d}$



              Q is an $mathbb{R}^{d - 1}$ dimensional line in P.



              A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…




              When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:




              1. The points may become equal and unequal any number of times without becoming equal to a member of Q.


              2. The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.







              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.



                $d$ is a member of N.



                P = $mathbb{R}^{d}$



                Q is an $mathbb{R}^{d - 1}$ dimensional line in P.



                A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…




                When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:




                1. The points may become equal and unequal any number of times without becoming equal to a member of Q.


                2. The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.







                share|cite|improve this answer











                $endgroup$



                I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem.



                $d$ is a member of N.



                P = $mathbb{R}^{d}$



                Q is an $mathbb{R}^{d - 1}$ dimensional line in P.



                A 0 dimensional line is a point, a 2 dimensional line is a plane, etc…




                When moving a pair of values $(v, w)$ continuously in P, that start a finite real distance from each other, exactly one of the following two statements are true:




                1. The points may become equal and unequal any number of times without becoming equal to a member of Q.


                2. The points may not become equal without first becoming equal to a member of Q. In this case a $d - 1$ dimensional line segment from $v$ to $w$ must contain a point that is a member of Q.








                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 13:56

























                answered Dec 4 '18 at 13:48









                alan2herealan2here

                504219




                504219






























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