Is the set of all exponential functions a subspace of the vector space of all continuous functions?












0












$begingroup$


Given the vector space, $ C(-infty,infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U={a^xmid a ge 1 }$, a subspace of the given vector space?



As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).



My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-infty,infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
    $endgroup$
    – StackTD
    Jul 15 '17 at 14:55










  • $begingroup$
    I did and I've corrected that, thank you!
    $endgroup$
    – jeanquilt
    Jul 15 '17 at 14:56










  • $begingroup$
    The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
    $endgroup$
    – user463383
    Jul 15 '17 at 14:58






  • 3




    $begingroup$
    What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
    $endgroup$
    – Hagen von Eitzen
    Jul 15 '17 at 14:58


















0












$begingroup$


Given the vector space, $ C(-infty,infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U={a^xmid a ge 1 }$, a subspace of the given vector space?



As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).



My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-infty,infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
    $endgroup$
    – StackTD
    Jul 15 '17 at 14:55










  • $begingroup$
    I did and I've corrected that, thank you!
    $endgroup$
    – jeanquilt
    Jul 15 '17 at 14:56










  • $begingroup$
    The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
    $endgroup$
    – user463383
    Jul 15 '17 at 14:58






  • 3




    $begingroup$
    What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
    $endgroup$
    – Hagen von Eitzen
    Jul 15 '17 at 14:58
















0












0








0


1



$begingroup$


Given the vector space, $ C(-infty,infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U={a^xmid a ge 1 }$, a subspace of the given vector space?



As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).



My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-infty,infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




Given the vector space, $ C(-infty,infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U={a^xmid a ge 1 }$, a subspace of the given vector space?



As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).



My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-infty,infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 10:49









José Carlos Santos

157k22126227




157k22126227










asked Jul 15 '17 at 14:54









jeanquiltjeanquilt

289212




289212












  • $begingroup$
    If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
    $endgroup$
    – StackTD
    Jul 15 '17 at 14:55










  • $begingroup$
    I did and I've corrected that, thank you!
    $endgroup$
    – jeanquilt
    Jul 15 '17 at 14:56










  • $begingroup$
    The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
    $endgroup$
    – user463383
    Jul 15 '17 at 14:58






  • 3




    $begingroup$
    What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
    $endgroup$
    – Hagen von Eitzen
    Jul 15 '17 at 14:58




















  • $begingroup$
    If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
    $endgroup$
    – StackTD
    Jul 15 '17 at 14:55










  • $begingroup$
    I did and I've corrected that, thank you!
    $endgroup$
    – jeanquilt
    Jul 15 '17 at 14:56










  • $begingroup$
    The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
    $endgroup$
    – user463383
    Jul 15 '17 at 14:58






  • 3




    $begingroup$
    What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
    $endgroup$
    – Hagen von Eitzen
    Jul 15 '17 at 14:58


















$begingroup$
If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
$endgroup$
– StackTD
Jul 15 '17 at 14:55




$begingroup$
If you say exponential functions, did you mean to write $a^x$ instead of $x^a$?
$endgroup$
– StackTD
Jul 15 '17 at 14:55












$begingroup$
I did and I've corrected that, thank you!
$endgroup$
– jeanquilt
Jul 15 '17 at 14:56




$begingroup$
I did and I've corrected that, thank you!
$endgroup$
– jeanquilt
Jul 15 '17 at 14:56












$begingroup$
The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
$endgroup$
– user463383
Jul 15 '17 at 14:58




$begingroup$
The zero in $C(-infty,infty)$ is the constant function the always returns zero. That function is not in $U$.
$endgroup$
– user463383
Jul 15 '17 at 14:58




3




3




$begingroup$
What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
$endgroup$
– Hagen von Eitzen
Jul 15 '17 at 14:58






$begingroup$
What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$?
$endgroup$
– Hagen von Eitzen
Jul 15 '17 at 14:58












3 Answers
3






active

oldest

votes


















0












$begingroup$

As originally written, no, since $0notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.



Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.



Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
    $endgroup$
    – jeanquilt
    Jul 15 '17 at 17:17










  • $begingroup$
    You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
    $endgroup$
    – Cameron Buie
    Jul 15 '17 at 17:33



















1












$begingroup$

Not, it is not a subspace. For instance the null function does not belong to $ U$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    (...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).




    Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?



    Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      As originally written, no, since $0notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.



      Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.



      Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
        $endgroup$
        – jeanquilt
        Jul 15 '17 at 17:17










      • $begingroup$
        You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
        $endgroup$
        – Cameron Buie
        Jul 15 '17 at 17:33
















      0












      $begingroup$

      As originally written, no, since $0notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.



      Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.



      Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
        $endgroup$
        – jeanquilt
        Jul 15 '17 at 17:17










      • $begingroup$
        You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
        $endgroup$
        – Cameron Buie
        Jul 15 '17 at 17:33














      0












      0








      0





      $begingroup$

      As originally written, no, since $0notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.



      Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.



      Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$






      share|cite|improve this answer











      $endgroup$



      As originally written, no, since $0notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.



      Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.



      Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jul 15 '17 at 16:05

























      answered Jul 15 '17 at 15:00









      Cameron BuieCameron Buie

      85.2k771155




      85.2k771155












      • $begingroup$
        Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
        $endgroup$
        – jeanquilt
        Jul 15 '17 at 17:17










      • $begingroup$
        You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
        $endgroup$
        – Cameron Buie
        Jul 15 '17 at 17:33


















      • $begingroup$
        Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
        $endgroup$
        – jeanquilt
        Jul 15 '17 at 17:17










      • $begingroup$
        You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
        $endgroup$
        – Cameron Buie
        Jul 15 '17 at 17:33
















      $begingroup$
      Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
      $endgroup$
      – jeanquilt
      Jul 15 '17 at 17:17




      $begingroup$
      Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation!
      $endgroup$
      – jeanquilt
      Jul 15 '17 at 17:17












      $begingroup$
      You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
      $endgroup$
      – Cameron Buie
      Jul 15 '17 at 17:33




      $begingroup$
      You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $bigl{(x,y)inBbb R^2:x>0bigr}$ is closed under addition, but not scalar multiplication; the set $bigl{(x,y)inBbb R^2:x=0text{ or }y=0bigr}$ is closed under scalar multiplication, but not addition.
      $endgroup$
      – Cameron Buie
      Jul 15 '17 at 17:33











      1












      $begingroup$

      Not, it is not a subspace. For instance the null function does not belong to $ U$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Not, it is not a subspace. For instance the null function does not belong to $ U$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Not, it is not a subspace. For instance the null function does not belong to $ U$.






          share|cite|improve this answer









          $endgroup$



          Not, it is not a subspace. For instance the null function does not belong to $ U$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 15 '17 at 14:58









          José Carlos SantosJosé Carlos Santos

          157k22126227




          157k22126227























              0












              $begingroup$


              (...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).




              Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?



              Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$


                (...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).




                Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?



                Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  (...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).




                  Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?



                  Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?






                  share|cite|improve this answer









                  $endgroup$




                  (...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).




                  Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?



                  Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 15 '17 at 15:00









                  StackTDStackTD

                  22.6k2049




                  22.6k2049






























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