Mixing
Solving functional equation $f(x)f(y) = f(x+y)$
$begingroup$
I'm having some trouble solving the following equation for $f: A rightarrow B$ where $A subseteq mathbb{R}$ and $B subseteq
mathbb{C}$ such as:
$$f(x)f(y) = f(x+y) quad forall x,y in A$$
The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.
Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think
calculus functional-equations
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
$endgroup$
|
show 4 more comments
$begingroup$
I'm having some trouble solving the following equation for $f: A rightarrow B$ where $A subseteq mathbb{R}$ and $B subseteq
mathbb{C}$ such as:
$$f(x)f(y) = f(x+y) quad forall x,y in A$$
The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.
Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think
calculus functional-equations
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
$endgroup$
$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
1
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
2
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09
|
show 4 more comments
$begingroup$
I'm having some trouble solving the following equation for $f: A rightarrow B$ where $A subseteq mathbb{R}$ and $B subseteq
mathbb{C}$ such as:
$$f(x)f(y) = f(x+y) quad forall x,y in A$$
The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.
Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think
calculus functional-equations
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
$endgroup$
I'm having some trouble solving the following equation for $f: A rightarrow B$ where $A subseteq mathbb{R}$ and $B subseteq
mathbb{C}$ such as:
$$f(x)f(y) = f(x+y) quad forall x,y in A$$
The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.
Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think
calculus functional-equations
calculus functional-equations
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
edited Jan 29 '15 at 23:10
Samuel
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
asked Jan 29 '15 at 21:55
SamuelSamuel
1569
1569
$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
1
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
2
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09
|
show 4 more comments
$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
1
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
2
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09
$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
1
1
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
2
2
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The general solution of the given functional equation depends on conditions imposed to $f$.
We have:
The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $exp(ax+bbar x)$, where $bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.
If we request that $f$ be differentiable then the general solution is $f(x)=exp (ax)$, as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that $f(x+y)=f(x)f(y) Rightarrow f(x)=f(x+0)=f(x)f(0) Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:
$$
f'(y)=f'(0)f(y) Rightarrow f(y)=kexpleft(f'(0)yright)
$$
and, given $f(0)=1$ we must have $k=1$.
The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
add a comment |
$begingroup$
If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $Bbb R^2toBbb R^2$ defined by
$$(x,y)mapsto (f(x)f(y), f(x+y)).$$
The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:
$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$
from which you can easily argue that
$$frac d{dx}log f(x) = frac{f'(x)}{f(x)} = frac{f'(0)}{f(0)} = constant$$
so that the result follows (in the differentiable case).
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
$endgroup$
add a comment |
$begingroup$
Let $y=1$ and $f(1)=e^k$ for some $kinmathbb{C}$, then
$$
f(x+1)=e^kf(x)\
f(x)=e^kf(x-1)\
f(x)=e^{2k}f(x-2)\
f(x)=e^{3k}f(x-3)\
vdots\
f(x)=e^{(x-1)k}f(1)\
f(x)=e^{kx}\
$$
$endgroup$
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The general solution of the given functional equation depends on conditions imposed to $f$.
We have:
The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $exp(ax+bbar x)$, where $bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.
If we request that $f$ be differentiable then the general solution is $f(x)=exp (ax)$, as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that $f(x+y)=f(x)f(y) Rightarrow f(x)=f(x+0)=f(x)f(0) Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:
$$
f'(y)=f'(0)f(y) Rightarrow f(y)=kexpleft(f'(0)yright)
$$
and, given $f(0)=1$ we must have $k=1$.
The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
add a comment |
$begingroup$
The general solution of the given functional equation depends on conditions imposed to $f$.
We have:
The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $exp(ax+bbar x)$, where $bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.
If we request that $f$ be differentiable then the general solution is $f(x)=exp (ax)$, as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that $f(x+y)=f(x)f(y) Rightarrow f(x)=f(x+0)=f(x)f(0) Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:
$$
f'(y)=f'(0)f(y) Rightarrow f(y)=kexpleft(f'(0)yright)
$$
and, given $f(0)=1$ we must have $k=1$.
The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
add a comment |
$begingroup$
The general solution of the given functional equation depends on conditions imposed to $f$.
We have:
The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $exp(ax+bbar x)$, where $bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.
If we request that $f$ be differentiable then the general solution is $f(x)=exp (ax)$, as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that $f(x+y)=f(x)f(y) Rightarrow f(x)=f(x+0)=f(x)f(0) Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:
$$
f'(y)=f'(0)f(y) Rightarrow f(y)=kexpleft(f'(0)yright)
$$
and, given $f(0)=1$ we must have $k=1$.
The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
The general solution of the given functional equation depends on conditions imposed to $f$.
We have:
The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $exp(ax+bbar x)$, where $bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.
If we request that $f$ be differentiable then the general solution is $f(x)=exp (ax)$, as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that $f(x+y)=f(x)f(y) Rightarrow f(x)=f(x+0)=f(x)f(0) Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:
$$
f'(y)=f'(0)f(y) Rightarrow f(y)=kexpleft(f'(0)yright)
$$
and, given $f(0)=1$ we must have $k=1$.
The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
edited Jan 30 '15 at 13:20
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
answered Jan 30 '15 at 12:37
Emilio NovatiEmilio Novati
52.2k43574
52.2k43574
add a comment |
add a comment |
$begingroup$
If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $Bbb R^2toBbb R^2$ defined by
$$(x,y)mapsto (f(x)f(y), f(x+y)).$$
The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:
$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$
from which you can easily argue that
$$frac d{dx}log f(x) = frac{f'(x)}{f(x)} = frac{f'(0)}{f(0)} = constant$$
so that the result follows (in the differentiable case).
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
$endgroup$
add a comment |
$begingroup$
If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $Bbb R^2toBbb R^2$ defined by
$$(x,y)mapsto (f(x)f(y), f(x+y)).$$
The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:
$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$
from which you can easily argue that
$$frac d{dx}log f(x) = frac{f'(x)}{f(x)} = frac{f'(0)}{f(0)} = constant$$
so that the result follows (in the differentiable case).
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
$endgroup$
add a comment |
$begingroup$
If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $Bbb R^2toBbb R^2$ defined by
$$(x,y)mapsto (f(x)f(y), f(x+y)).$$
The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:
$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$
from which you can easily argue that
$$frac d{dx}log f(x) = frac{f'(x)}{f(x)} = frac{f'(0)}{f(0)} = constant$$
so that the result follows (in the differentiable case).
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
$endgroup$
If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $Bbb R^2toBbb R^2$ defined by
$$(x,y)mapsto (f(x)f(y), f(x+y)).$$
The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:
$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$
from which you can easily argue that
$$frac d{dx}log f(x) = frac{f'(x)}{f(x)} = frac{f'(0)}{f(0)} = constant$$
so that the result follows (in the differentiable case).
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
answered Jan 29 '15 at 22:53
doetoedoetoe
1,348811
1,348811
add a comment |
add a comment |
$begingroup$
Let $y=1$ and $f(1)=e^k$ for some $kinmathbb{C}$, then
$$
f(x+1)=e^kf(x)\
f(x)=e^kf(x-1)\
f(x)=e^{2k}f(x-2)\
f(x)=e^{3k}f(x-3)\
vdots\
f(x)=e^{(x-1)k}f(1)\
f(x)=e^{kx}\
$$
$endgroup$
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
add a comment |
$begingroup$
Let $y=1$ and $f(1)=e^k$ for some $kinmathbb{C}$, then
$$
f(x+1)=e^kf(x)\
f(x)=e^kf(x-1)\
f(x)=e^{2k}f(x-2)\
f(x)=e^{3k}f(x-3)\
vdots\
f(x)=e^{(x-1)k}f(1)\
f(x)=e^{kx}\
$$
$endgroup$
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
add a comment |
$begingroup$
Let $y=1$ and $f(1)=e^k$ for some $kinmathbb{C}$, then
$$
f(x+1)=e^kf(x)\
f(x)=e^kf(x-1)\
f(x)=e^{2k}f(x-2)\
f(x)=e^{3k}f(x-3)\
vdots\
f(x)=e^{(x-1)k}f(1)\
f(x)=e^{kx}\
$$
$endgroup$
Let $y=1$ and $f(1)=e^k$ for some $kinmathbb{C}$, then
$$
f(x+1)=e^kf(x)\
f(x)=e^kf(x-1)\
f(x)=e^{2k}f(x-2)\
f(x)=e^{3k}f(x-3)\
vdots\
f(x)=e^{(x-1)k}f(1)\
f(x)=e^{kx}\
$$
edited Jan 30 '15 at 9:21
answered Jan 29 '15 at 23:00
user164524
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This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
add a comment |
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
This derivation holds only for integer $x$.
$endgroup$
– Winther
Jan 30 '15 at 8:09
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
@Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $ainmathbb{R},f(a)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:18
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
I'm referring to the $f(x-3)cdots f(1)$ step which only works for $x$ integer.
$endgroup$
– Winther
Jan 30 '15 at 9:33
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
$begingroup$
@Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $yto0^+$ and $f(y)=e^k$.
$endgroup$
– user164524
Jan 30 '15 at 9:50
2
2
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
$begingroup$
That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here
$endgroup$
– Winther
Jan 30 '15 at 9:53
add a comment |
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$begingroup$
Which set is $A$?
$endgroup$
– ajotatxe
Jan 29 '15 at 21:58
$begingroup$
For my case, it's $U(1) = lbrace e^{ivarphi} mid varphi in left[ 0, 2pi right] rbrace$ which can be represented by $ varphi in left[ 0, 2pi right]$ so I guess that A is $left[ 0, 2pi right]$
$endgroup$
– Samuel
Jan 29 '15 at 22:02
1
$begingroup$
@Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2pi]$. In particular, $0$ and $2pi$ have the same image under $varphimapsto e^{ivarphi}$.
$endgroup$
– Hayden
Jan 29 '15 at 22:04
$begingroup$
So $left[ 0, 2pi right[$ ?
$endgroup$
– Samuel
Jan 29 '15 at 22:05
2
$begingroup$
Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$)
$endgroup$
– Winther
Jan 29 '15 at 22:09