Mixing
Some problems in additive functions.
$begingroup$
Let f be a function that satisfies $forall xin R: f(x)+f(y)=f(x+y)$ .
I'm thinking about equivalnce
$$f(x_0)=0 Leftrightarrow x_0=0$$
Where f(x) $neq 0$. It's true of course if $f$ is a continous function (because then $f(x)=f(1)x$), but I do not know if it is going to work if $f$ is not continous.
I managed to prove the following equivalence:
$f(x_0)=0 Longleftrightarrow x_0=0 Longleftrightarrow f$ is an injective function(for f(x) $neq 0$), but it does not help me so much. It is also hard to even draw not continous additive functions, because they are not continous in any point.
real-analysis calculus
asked Feb 1 at 14:06
LucianLucian
396
$endgroup$
add a comment |
$begingroup$
Let f be a function that satisfies $forall xin R: f(x)+f(y)=f(x+y)$ .
I'm thinking about equivalnce
$$f(x_0)=0 Leftrightarrow x_0=0$$
Where f(x) $neq 0$. It's true of course if $f$ is a continous function (because then $f(x)=f(1)x$), but I do not know if it is going to work if $f$ is not continous.
I managed to prove the following equivalence:
$f(x_0)=0 Longleftrightarrow x_0=0 Longleftrightarrow f$ is an injective function(for f(x) $neq 0$), but it does not help me so much. It is also hard to even draw not continous additive functions, because they are not continous in any point.
real-analysis calculus
asked Feb 1 at 14:06
LucianLucian
396
$endgroup$
$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
1
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
1
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39
add a comment |
$begingroup$
Let f be a function that satisfies $forall xin R: f(x)+f(y)=f(x+y)$ .
I'm thinking about equivalnce
$$f(x_0)=0 Leftrightarrow x_0=0$$
Where f(x) $neq 0$. It's true of course if $f$ is a continous function (because then $f(x)=f(1)x$), but I do not know if it is going to work if $f$ is not continous.
I managed to prove the following equivalence:
$f(x_0)=0 Longleftrightarrow x_0=0 Longleftrightarrow f$ is an injective function(for f(x) $neq 0$), but it does not help me so much. It is also hard to even draw not continous additive functions, because they are not continous in any point.
real-analysis calculus
asked Feb 1 at 14:06
LucianLucian
396
$endgroup$
Let f be a function that satisfies $forall xin R: f(x)+f(y)=f(x+y)$ .
I'm thinking about equivalnce
$$f(x_0)=0 Leftrightarrow x_0=0$$
Where f(x) $neq 0$. It's true of course if $f$ is a continous function (because then $f(x)=f(1)x$), but I do not know if it is going to work if $f$ is not continous.
I managed to prove the following equivalence:
$f(x_0)=0 Longleftrightarrow x_0=0 Longleftrightarrow f$ is an injective function(for f(x) $neq 0$), but it does not help me so much. It is also hard to even draw not continous additive functions, because they are not continous in any point.
real-analysis calculus
real-analysis calculus
asked Feb 1 at 14:06
LucianLucian
396
asked Feb 1 at 14:06
LucianLucian
396
edited Feb 1 at 14:57
Lucian
asked Feb 1 at 14:06
LucianLucian
396
asked Feb 1 at 14:06
LucianLucian
396
asked Feb 1 at 14:06
LucianLucian
396
396
$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
1
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
1
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39
add a comment |
$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
1
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
1
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39
$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
1
1
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
1
1
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39
add a comment |
1 Answer
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$begingroup$
$$f(x)=0$$ is a continuous and additive function.
But
$$f(x_0)=0iff x_0=0$$ does not hold.
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
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$begingroup$
$$f(x)=0$$ is a continuous and additive function.
But
$$f(x_0)=0iff x_0=0$$ does not hold.
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
$begingroup$
$$f(x)=0$$ is a continuous and additive function.
But
$$f(x_0)=0iff x_0=0$$ does not hold.
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
$begingroup$
$$f(x)=0$$ is a continuous and additive function.
But
$$f(x_0)=0iff x_0=0$$ does not hold.
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
$endgroup$
$$f(x)=0$$ is a continuous and additive function.
But
$$f(x_0)=0iff x_0=0$$ does not hold.
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:46
Yves DaoustYves Daoust
133k676231
133k676231
add a comment |
add a comment |
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$begingroup$
$f(x)=0$ is continuous and it get the value zero even when $x_0not = 0$. You can easily prove that $f(0)=0$ because $f(0)+f(0)=f(0+0)=f(0)$.
$endgroup$
– Yanko
Feb 1 at 14:09
1
$begingroup$
en.wikipedia.org/wiki/Cauchy%27s_functional_equation The non-continuous functions are basically a collection of many different linear functions, one for each scaled copy of $Bbb Q$ (I believe they require the axiom of choice to construct).
$endgroup$
– Arthur
Feb 1 at 14:10
1
$begingroup$
@Arthur They do require more than ZF. If the functions are measurable, then they are linear, and there exist models for ZF in which all functions are measurable.
$endgroup$
– Theo Bendit
Feb 1 at 14:17
$begingroup$
@TheoBendit Obviously they don't require the full strength of the AoC. But among the axioms to add to ZF which guarantees that nonlinear solutions exist, AoC is definitely the most commonly used one.
$endgroup$
– Arthur
Feb 1 at 14:39