Mixing
Logarithmic functions in terms of vector space theory
$begingroup$
We can consider $mathbb{R}^+$ with two operations:
+: $mathbb{R}^+ times mathbb{R}^+to mathbb{R}^+$
that maps $(a,b)$ to $a+b:=ab$ and
$ cdot : mathbb{R} times mathbb{R}^+ to mathbb{R}^+$
that maps $(lambda,a)$ to $a^lambda$.
With respect to these operations we have that $mathbb{R}^+$ is a vector space.
We want to find an example of linear application from $mathbb{R}$ to $mathbb{R}^+$.
We can fix $ain mathbb{R}^+$ and we can define
$f_a:mathbb{R}to mathbb{R}^+$ that maps every $lambda$ to $ f_a(lambda):=a^lambda$.
From the rules of powers, the map $f_a$ is a linear map for every $ain mathbb{R}^+$.
A natural question can be if all linear maps from $mathbb{R}$ to $mathbb{R}^+$ are exponential maps.
The answer is positive because for each linear map $F: mathbb{R}to mathbb{R}^+$ we have that
$F(lambda)=F(lambda cdot 1)=(F(1))^lambda=f_{F(1)}(lambda)$
so
$F=f_{F(1)}.$
We can also observe that the map $f_a$ is bijective and its inverse is the logarithmic function $log_a mathbb{R}^+to mathbb{R}$ that it is linear because of logarithmic rules or because it is the inverse of a linear map.
So the logarithmic maps are elements of the dual of $mathbb{R}^+$.
A natural question can be if all non-zero functional of $mathbb{R}^+$ must be a logarithmic function.
The answer is positive because $Gin (mathbb{R}^+)^*/{0}$ is a linear map between two vector space of dimension 1 so it is bijective but $G^{-1}: mathbb{R}to mathbb{R}^+$ is a linear map so $G^{-1}=f_{G^{-1}(1)}$.
Hence $G=(f_{G^{-1}(1)})^{-1}=log_{G^{-1}(1)}$.
So the dual of $mathbb{R}^+$ is
$ (mathbb{R}^+)^*={log_a : ain mathbb{R}^+}$
Is this correct?
real-analysis calculus linear-algebra vector-spaces
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
add a comment |
$begingroup$
We can consider $mathbb{R}^+$ with two operations:
+: $mathbb{R}^+ times mathbb{R}^+to mathbb{R}^+$
that maps $(a,b)$ to $a+b:=ab$ and
$ cdot : mathbb{R} times mathbb{R}^+ to mathbb{R}^+$
that maps $(lambda,a)$ to $a^lambda$.
With respect to these operations we have that $mathbb{R}^+$ is a vector space.
We want to find an example of linear application from $mathbb{R}$ to $mathbb{R}^+$.
We can fix $ain mathbb{R}^+$ and we can define
$f_a:mathbb{R}to mathbb{R}^+$ that maps every $lambda$ to $ f_a(lambda):=a^lambda$.
From the rules of powers, the map $f_a$ is a linear map for every $ain mathbb{R}^+$.
A natural question can be if all linear maps from $mathbb{R}$ to $mathbb{R}^+$ are exponential maps.
The answer is positive because for each linear map $F: mathbb{R}to mathbb{R}^+$ we have that
$F(lambda)=F(lambda cdot 1)=(F(1))^lambda=f_{F(1)}(lambda)$
so
$F=f_{F(1)}.$
We can also observe that the map $f_a$ is bijective and its inverse is the logarithmic function $log_a mathbb{R}^+to mathbb{R}$ that it is linear because of logarithmic rules or because it is the inverse of a linear map.
So the logarithmic maps are elements of the dual of $mathbb{R}^+$.
A natural question can be if all non-zero functional of $mathbb{R}^+$ must be a logarithmic function.
The answer is positive because $Gin (mathbb{R}^+)^*/{0}$ is a linear map between two vector space of dimension 1 so it is bijective but $G^{-1}: mathbb{R}to mathbb{R}^+$ is a linear map so $G^{-1}=f_{G^{-1}(1)}$.
Hence $G=(f_{G^{-1}(1)})^{-1}=log_{G^{-1}(1)}$.
So the dual of $mathbb{R}^+$ is
$ (mathbb{R}^+)^*={log_a : ain mathbb{R}^+}$
Is this correct?
real-analysis calculus linear-algebra vector-spaces
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
1
$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32
add a comment |
$begingroup$
We can consider $mathbb{R}^+$ with two operations:
+: $mathbb{R}^+ times mathbb{R}^+to mathbb{R}^+$
that maps $(a,b)$ to $a+b:=ab$ and
$ cdot : mathbb{R} times mathbb{R}^+ to mathbb{R}^+$
that maps $(lambda,a)$ to $a^lambda$.
With respect to these operations we have that $mathbb{R}^+$ is a vector space.
We want to find an example of linear application from $mathbb{R}$ to $mathbb{R}^+$.
We can fix $ain mathbb{R}^+$ and we can define
$f_a:mathbb{R}to mathbb{R}^+$ that maps every $lambda$ to $ f_a(lambda):=a^lambda$.
From the rules of powers, the map $f_a$ is a linear map for every $ain mathbb{R}^+$.
A natural question can be if all linear maps from $mathbb{R}$ to $mathbb{R}^+$ are exponential maps.
The answer is positive because for each linear map $F: mathbb{R}to mathbb{R}^+$ we have that
$F(lambda)=F(lambda cdot 1)=(F(1))^lambda=f_{F(1)}(lambda)$
so
$F=f_{F(1)}.$
We can also observe that the map $f_a$ is bijective and its inverse is the logarithmic function $log_a mathbb{R}^+to mathbb{R}$ that it is linear because of logarithmic rules or because it is the inverse of a linear map.
So the logarithmic maps are elements of the dual of $mathbb{R}^+$.
A natural question can be if all non-zero functional of $mathbb{R}^+$ must be a logarithmic function.
The answer is positive because $Gin (mathbb{R}^+)^*/{0}$ is a linear map between two vector space of dimension 1 so it is bijective but $G^{-1}: mathbb{R}to mathbb{R}^+$ is a linear map so $G^{-1}=f_{G^{-1}(1)}$.
Hence $G=(f_{G^{-1}(1)})^{-1}=log_{G^{-1}(1)}$.
So the dual of $mathbb{R}^+$ is
$ (mathbb{R}^+)^*={log_a : ain mathbb{R}^+}$
Is this correct?
real-analysis calculus linear-algebra vector-spaces
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
We can consider $mathbb{R}^+$ with two operations:
+: $mathbb{R}^+ times mathbb{R}^+to mathbb{R}^+$
that maps $(a,b)$ to $a+b:=ab$ and
$ cdot : mathbb{R} times mathbb{R}^+ to mathbb{R}^+$
that maps $(lambda,a)$ to $a^lambda$.
With respect to these operations we have that $mathbb{R}^+$ is a vector space.
We want to find an example of linear application from $mathbb{R}$ to $mathbb{R}^+$.
We can fix $ain mathbb{R}^+$ and we can define
$f_a:mathbb{R}to mathbb{R}^+$ that maps every $lambda$ to $ f_a(lambda):=a^lambda$.
From the rules of powers, the map $f_a$ is a linear map for every $ain mathbb{R}^+$.
A natural question can be if all linear maps from $mathbb{R}$ to $mathbb{R}^+$ are exponential maps.
The answer is positive because for each linear map $F: mathbb{R}to mathbb{R}^+$ we have that
$F(lambda)=F(lambda cdot 1)=(F(1))^lambda=f_{F(1)}(lambda)$
so
$F=f_{F(1)}.$
We can also observe that the map $f_a$ is bijective and its inverse is the logarithmic function $log_a mathbb{R}^+to mathbb{R}$ that it is linear because of logarithmic rules or because it is the inverse of a linear map.
So the logarithmic maps are elements of the dual of $mathbb{R}^+$.
A natural question can be if all non-zero functional of $mathbb{R}^+$ must be a logarithmic function.
The answer is positive because $Gin (mathbb{R}^+)^*/{0}$ is a linear map between two vector space of dimension 1 so it is bijective but $G^{-1}: mathbb{R}to mathbb{R}^+$ is a linear map so $G^{-1}=f_{G^{-1}(1)}$.
Hence $G=(f_{G^{-1}(1)})^{-1}=log_{G^{-1}(1)}$.
So the dual of $mathbb{R}^+$ is
$ (mathbb{R}^+)^*={log_a : ain mathbb{R}^+}$
Is this correct?
real-analysis calculus linear-algebra vector-spaces
real-analysis calculus linear-algebra vector-spaces
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
edited Jan 29 at 23:21
J. W. Tanner
4,2361320
4,2361320
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
asked Jan 29 at 20:08
Federico FalluccaFederico Fallucca
2,270210
2,270210
1
$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32
add a comment |
1
$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32
1
1
$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32
$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32
add a comment |
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$begingroup$
Seems all correct to me.
$endgroup$
– J.F
Jan 29 at 20:32