The vector space of all functions of type $f:mathbbR rightarrow mathbbR$

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I am looking to describe all functions of type $f:mathbbR rightarrow mathbbR$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as
$$g(x) = int^infty_-inftya(k)sin(kx) + b(k)cos(kx)dk$$
for $a(k), b(k) in mathbbR$, and $k, in mathbbR$. I am assuming that $sin(kx)$ is a basis for this function space.



Let's suppose we have a basis of cardinality $Card(mathbbR)$, call it $B = b_1(kx), b_2(kx), ldots , k in mathbbR$.



I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as



$$g(x) = int^infty_-inftya_1(k)b_1(kx)+ a_2(k)b_2(kx) + ldots dk$$



This seems like a continuous combination. Can someone explain this?







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  • Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
    – AHusain
    Aug 8 at 16:21














up vote
0
down vote

favorite












I am looking to describe all functions of type $f:mathbbR rightarrow mathbbR$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as
$$g(x) = int^infty_-inftya(k)sin(kx) + b(k)cos(kx)dk$$
for $a(k), b(k) in mathbbR$, and $k, in mathbbR$. I am assuming that $sin(kx)$ is a basis for this function space.



Let's suppose we have a basis of cardinality $Card(mathbbR)$, call it $B = b_1(kx), b_2(kx), ldots , k in mathbbR$.



I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as



$$g(x) = int^infty_-inftya_1(k)b_1(kx)+ a_2(k)b_2(kx) + ldots dk$$



This seems like a continuous combination. Can someone explain this?







share|cite|improve this question






















  • Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
    – AHusain
    Aug 8 at 16:21












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am looking to describe all functions of type $f:mathbbR rightarrow mathbbR$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as
$$g(x) = int^infty_-inftya(k)sin(kx) + b(k)cos(kx)dk$$
for $a(k), b(k) in mathbbR$, and $k, in mathbbR$. I am assuming that $sin(kx)$ is a basis for this function space.



Let's suppose we have a basis of cardinality $Card(mathbbR)$, call it $B = b_1(kx), b_2(kx), ldots , k in mathbbR$.



I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as



$$g(x) = int^infty_-inftya_1(k)b_1(kx)+ a_2(k)b_2(kx) + ldots dk$$



This seems like a continuous combination. Can someone explain this?







share|cite|improve this question














I am looking to describe all functions of type $f:mathbbR rightarrow mathbbR$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as
$$g(x) = int^infty_-inftya(k)sin(kx) + b(k)cos(kx)dk$$
for $a(k), b(k) in mathbbR$, and $k, in mathbbR$. I am assuming that $sin(kx)$ is a basis for this function space.



Let's suppose we have a basis of cardinality $Card(mathbbR)$, call it $B = b_1(kx), b_2(kx), ldots , k in mathbbR$.



I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as



$$g(x) = int^infty_-inftya_1(k)b_1(kx)+ a_2(k)b_2(kx) + ldots dk$$



This seems like a continuous combination. Can someone explain this?









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edited Aug 8 at 16:48

























asked Aug 8 at 16:16









Ben Sprott

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  • Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
    – AHusain
    Aug 8 at 16:21
















  • Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
    – AHusain
    Aug 8 at 16:21















Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
– AHusain
Aug 8 at 16:21




Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
– AHusain
Aug 8 at 16:21










2 Answers
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You can certainly define it as a vector space over the reals with pointwise addition as the addition of vectors, multiplication by reals as multiplication by elements of the field. You can verify all the axioms for a vector space. The additive identity is the zero function, inverses are the negative of the function, and so on.



Finding a basis is difficult. We only allow finite sums of basis vectors in a representation. The basis has to have the cardinality of the continuum.



You could also define the vector space that is spanned by the functions $sin (kx)$. That is a fine vector space but it is missing many of the functions from $Bbb R$ to $Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.






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    It depends on how you define “to describe”. Saying that it is the set of all functions from $mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $xmapstosin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      You can certainly define it as a vector space over the reals with pointwise addition as the addition of vectors, multiplication by reals as multiplication by elements of the field. You can verify all the axioms for a vector space. The additive identity is the zero function, inverses are the negative of the function, and so on.



      Finding a basis is difficult. We only allow finite sums of basis vectors in a representation. The basis has to have the cardinality of the continuum.



      You could also define the vector space that is spanned by the functions $sin (kx)$. That is a fine vector space but it is missing many of the functions from $Bbb R$ to $Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted










        You can certainly define it as a vector space over the reals with pointwise addition as the addition of vectors, multiplication by reals as multiplication by elements of the field. You can verify all the axioms for a vector space. The additive identity is the zero function, inverses are the negative of the function, and so on.



        Finding a basis is difficult. We only allow finite sums of basis vectors in a representation. The basis has to have the cardinality of the continuum.



        You could also define the vector space that is spanned by the functions $sin (kx)$. That is a fine vector space but it is missing many of the functions from $Bbb R$ to $Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.






        share|cite|improve this answer






















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You can certainly define it as a vector space over the reals with pointwise addition as the addition of vectors, multiplication by reals as multiplication by elements of the field. You can verify all the axioms for a vector space. The additive identity is the zero function, inverses are the negative of the function, and so on.



          Finding a basis is difficult. We only allow finite sums of basis vectors in a representation. The basis has to have the cardinality of the continuum.



          You could also define the vector space that is spanned by the functions $sin (kx)$. That is a fine vector space but it is missing many of the functions from $Bbb R$ to $Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.






          share|cite|improve this answer












          You can certainly define it as a vector space over the reals with pointwise addition as the addition of vectors, multiplication by reals as multiplication by elements of the field. You can verify all the axioms for a vector space. The additive identity is the zero function, inverses are the negative of the function, and so on.



          Finding a basis is difficult. We only allow finite sums of basis vectors in a representation. The basis has to have the cardinality of the continuum.



          You could also define the vector space that is spanned by the functions $sin (kx)$. That is a fine vector space but it is missing many of the functions from $Bbb R$ to $Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 16:23









          Ross Millikan

          277k21187352




          277k21187352




















              up vote
              2
              down vote













              It depends on how you define “to describe”. Saying that it is the set of all functions from $mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $xmapstosin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.






              share|cite|improve this answer
























                up vote
                2
                down vote













                It depends on how you define “to describe”. Saying that it is the set of all functions from $mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $xmapstosin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  It depends on how you define “to describe”. Saying that it is the set of all functions from $mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $xmapstosin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.






                  share|cite|improve this answer












                  It depends on how you define “to describe”. Saying that it is the set of all functions from $mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $xmapstosin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 8 at 16:20









                  José Carlos Santos

                  115k1699177




                  115k1699177






















                       

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