Mixing
Is this theorem true?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
If $f(x)+f(y)=f(x+y)$, then:
$f(x)=a x$
where $a$ is a constant.
Is the above statement true? Is there a way of proving it?
The application of this theorem is in the last part of page 52 (second page of the chapter)
real-analysis functions functional-equations real-numbers
edited Jul 20 at 6:00
Rhys Steele
5,6101828
asked Jul 18 at 16:42
Joe
373113
add a comment |Â
up vote
2
down vote
favorite
If $f(x)+f(y)=f(x+y)$, then:
$f(x)=a x$
where $a$ is a constant.
Is the above statement true? Is there a way of proving it?
The application of this theorem is in the last part of page 52 (second page of the chapter)
real-analysis functions functional-equations real-numbers
edited Jul 20 at 6:00
Rhys Steele
5,6101828
asked Jul 18 at 16:42
Joe
373113
Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
no, it is not true in general
– Surb
Jul 18 at 16:45
1
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $f(x)+f(y)=f(x+y)$, then:
$f(x)=a x$
where $a$ is a constant.
Is the above statement true? Is there a way of proving it?
The application of this theorem is in the last part of page 52 (second page of the chapter)
real-analysis functions functional-equations real-numbers
edited Jul 20 at 6:00
Rhys Steele
5,6101828
asked Jul 18 at 16:42
Joe
373113
If $f(x)+f(y)=f(x+y)$, then:
$f(x)=a x$
where $a$ is a constant.
Is the above statement true? Is there a way of proving it?
The application of this theorem is in the last part of page 52 (second page of the chapter)
real-analysis functions functional-equations real-numbers
edited Jul 20 at 6:00
Rhys Steele
5,6101828
asked Jul 18 at 16:42
Joe
373113
edited Jul 20 at 6:00
Rhys Steele
5,6101828
edited Jul 20 at 6:00
Rhys Steele
5,6101828
edited Jul 20 at 6:00
Rhys Steele
5,6101828
5,6101828
asked Jul 18 at 16:42
Joe
373113
asked Jul 18 at 16:42
Joe
373113
asked Jul 18 at 16:42
Joe
373113
373113
Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
no, it is not true in general
– Surb
Jul 18 at 16:45
1
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47
add a comment |Â
Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
no, it is not true in general
– Surb
Jul 18 at 16:45
1
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47
Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
no, it is not true in general
– Surb
Jul 18 at 16:45
no, it is not true in general
– Surb
Jul 18 at 16:45
1
1
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If we are considering functions $mathbbR to mathbbR$, this is known as Cauchy's functional equation.
To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:
- $f$ is continuous at at least one point
- $f$ is monotonic on any interval
- $f$ is bounded on any interval
- $f$ is Lebesgue measurable
As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.
The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.
Let me know if anything needs clarification.
EDIT:
1) I haven't looked at all the proofs, but this seems to be a great resource.
2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is
$$f(x) = begincases0, text if x is irrational \1, text if x is rationalendcases$$
If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.
To prove this function is discontinous at every point, you have to prove that $lim_x to a f(x) not = f(a)$ for any $a$.
Bounded just means that the function is cut off height-wise; for example $sin x$ is bounded because it doesn't go higher/lower than $pm1$, which $x^2$ and $x^3$ are not bounded.
Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $mathbbR$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $mathbbQ$, the set of all the rational numbers on the number line, is also $0$.
Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.
answered Jul 18 at 18:09
Ovi
11.3k935105
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
 |Â
show 1 more comment
up vote
5
down vote
Hard to say anything unless you state the sets the function is defined on.
answered Jul 18 at 16:46
selflearner
10715
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If we are considering functions $mathbbR to mathbbR$, this is known as Cauchy's functional equation.
To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:
- $f$ is continuous at at least one point
- $f$ is monotonic on any interval
- $f$ is bounded on any interval
- $f$ is Lebesgue measurable
As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.
The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.
Let me know if anything needs clarification.
EDIT:
1) I haven't looked at all the proofs, but this seems to be a great resource.
2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is
$$f(x) = begincases0, text if x is irrational \1, text if x is rationalendcases$$
If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.
To prove this function is discontinous at every point, you have to prove that $lim_x to a f(x) not = f(a)$ for any $a$.
Bounded just means that the function is cut off height-wise; for example $sin x$ is bounded because it doesn't go higher/lower than $pm1$, which $x^2$ and $x^3$ are not bounded.
Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $mathbbR$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $mathbbQ$, the set of all the rational numbers on the number line, is also $0$.
Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.
answered Jul 18 at 18:09
Ovi
11.3k935105
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
 |Â
show 1 more comment
up vote
3
down vote
accepted
If we are considering functions $mathbbR to mathbbR$, this is known as Cauchy's functional equation.
To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:
- $f$ is continuous at at least one point
- $f$ is monotonic on any interval
- $f$ is bounded on any interval
- $f$ is Lebesgue measurable
As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.
The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.
Let me know if anything needs clarification.
EDIT:
1) I haven't looked at all the proofs, but this seems to be a great resource.
2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is
$$f(x) = begincases0, text if x is irrational \1, text if x is rationalendcases$$
If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.
To prove this function is discontinous at every point, you have to prove that $lim_x to a f(x) not = f(a)$ for any $a$.
Bounded just means that the function is cut off height-wise; for example $sin x$ is bounded because it doesn't go higher/lower than $pm1$, which $x^2$ and $x^3$ are not bounded.
Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $mathbbR$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $mathbbQ$, the set of all the rational numbers on the number line, is also $0$.
Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.
answered Jul 18 at 18:09
Ovi
11.3k935105
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If we are considering functions $mathbbR to mathbbR$, this is known as Cauchy's functional equation.
To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:
- $f$ is continuous at at least one point
- $f$ is monotonic on any interval
- $f$ is bounded on any interval
- $f$ is Lebesgue measurable
As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.
The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.
Let me know if anything needs clarification.
EDIT:
1) I haven't looked at all the proofs, but this seems to be a great resource.
2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is
$$f(x) = begincases0, text if x is irrational \1, text if x is rationalendcases$$
If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.
To prove this function is discontinous at every point, you have to prove that $lim_x to a f(x) not = f(a)$ for any $a$.
Bounded just means that the function is cut off height-wise; for example $sin x$ is bounded because it doesn't go higher/lower than $pm1$, which $x^2$ and $x^3$ are not bounded.
Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $mathbbR$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $mathbbQ$, the set of all the rational numbers on the number line, is also $0$.
Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.
answered Jul 18 at 18:09
Ovi
11.3k935105
If we are considering functions $mathbbR to mathbbR$, this is known as Cauchy's functional equation.
To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:
- $f$ is continuous at at least one point
- $f$ is monotonic on any interval
- $f$ is bounded on any interval
- $f$ is Lebesgue measurable
As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.
The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.
Let me know if anything needs clarification.
EDIT:
1) I haven't looked at all the proofs, but this seems to be a great resource.
2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is
$$f(x) = begincases0, text if x is irrational \1, text if x is rationalendcases$$
If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.
To prove this function is discontinous at every point, you have to prove that $lim_x to a f(x) not = f(a)$ for any $a$.
Bounded just means that the function is cut off height-wise; for example $sin x$ is bounded because it doesn't go higher/lower than $pm1$, which $x^2$ and $x^3$ are not bounded.
Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $mathbbR$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $mathbbQ$, the set of all the rational numbers on the number line, is also $0$.
Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.
answered Jul 18 at 18:09
Ovi
11.3k935105
edited Jul 19 at 17:45
answered Jul 18 at 18:09
Ovi
11.3k935105
answered Jul 18 at 18:09
Ovi
11.3k935105
answered Jul 18 at 18:09
Ovi
11.3k935105
11.3k935105
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
 |Â
show 1 more comment
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
Thank you very much for the answer... (1) Can you please give a reference to a proof of this theorem if it is not that much intricate for a physics graduate student. (2) How can a function be discontinuous at every point. It seems impossible for me to imagine such a function.
– Joe
Jul 19 at 15:09
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I haven't yet studied bounded and Lebesgue measurable functions in my course.
– Joe
Jul 19 at 15:18
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
I have read the Wikipedia article and now I understand how a function can be discontinuous at every point. Can you show me in my problem (page 52) that the function under consideration is not discontinuous at every point.
– Joe
Jul 19 at 17:45
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
@Joe I've edited the question with info about your comments. I don't know physics, so I think you may have more luck with your quetion about page $52$ on physics stackexchange or electrical engineering stackexchange physics.stackexchange.com electronics.stackexchange.com
– Ovi
Jul 19 at 17:58
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
Just one more thing to clarify. Can you explain a bit about how a function cannot be monotonic on any interval?
– Joe
Jul 20 at 13:59
 |Â
show 1 more comment
up vote
5
down vote
Hard to say anything unless you state the sets the function is defined on.
answered Jul 18 at 16:46
selflearner
10715
add a comment |Â
up vote
5
down vote
Hard to say anything unless you state the sets the function is defined on.
answered Jul 18 at 16:46
selflearner
10715
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Hard to say anything unless you state the sets the function is defined on.
answered Jul 18 at 16:46
selflearner
10715
Hard to say anything unless you state the sets the function is defined on.
answered Jul 18 at 16:46
selflearner
10715
answered Jul 18 at 16:46
selflearner
10715
answered Jul 18 at 16:46
selflearner
10715
answered Jul 18 at 16:46
selflearner
10715
10715
add a comment |Â
add a comment |Â
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Do you mind posting the complete question?
– Cornman
Jul 18 at 16:44
no, it is not true in general
– Surb
Jul 18 at 16:45
1
en.wikipedia.org/wiki/Cauchy%27s_functional_equation
– Surb
Jul 18 at 16:46
Aren't there some other conditions, say something about continuity...?
– DonAntonio
Jul 18 at 17:32
Sorry.. I don't have a back in real analysis. I came across an application of this theorem in a phycis book. Take a look at my edited question.
– Joe
Jul 18 at 17:47