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?
functional-analysis functions
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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?
functional-analysis functions
Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
â AHusain
Aug 8 at 16:21
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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?
functional-analysis functions
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?
functional-analysis functions
edited Aug 8 at 16:48
asked Aug 8 at 16:16
Ben Sprott
363312
363312
Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
â AHusain
Aug 8 at 16:21
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Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
â 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?
â 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?
â AHusain
Aug 8 at 16:21
add a comment |Â
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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up vote
2
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It depends on how you define âto describeâ. 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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Aug 8 at 16:23
Ross Millikan
277k21187352
277k21187352
add a comment |Â
add a comment |Â
up vote
2
down vote
It depends on how you define âto describeâ. 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.
add a comment |Â
up vote
2
down vote
It depends on how you define âto describeâ. 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.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It depends on how you define âto describeâ. 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.
It depends on how you define âto describeâ. 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.
answered Aug 8 at 16:20
José Carlos Santos
115k1699177
115k1699177
add a comment |Â
add a comment |Â
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Calling that integral $f(x)$, what is $f(-x)$? Can you write $cos x$ as some integral of that type?
â AHusain
Aug 8 at 16:21