Mixing
Functional Equation simple problem
$begingroup$
How do I show that if there are functions $f,g$ such that$$
f(g(x)+g(y))=bx+cy
$$holds for all $b,cinmathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?
Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$
y^2+z^2=f(x,g(y-x)+g(z-x)).
$$
Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
functions functional-equations
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
$endgroup$
add a comment |
$begingroup$
How do I show that if there are functions $f,g$ such that$$
f(g(x)+g(y))=bx+cy
$$holds for all $b,cinmathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?
Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$
y^2+z^2=f(x,g(y-x)+g(z-x)).
$$
Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
functions functional-equations
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
$endgroup$
add a comment |
$begingroup$
How do I show that if there are functions $f,g$ such that$$
f(g(x)+g(y))=bx+cy
$$holds for all $b,cinmathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?
Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$
y^2+z^2=f(x,g(y-x)+g(z-x)).
$$
Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
functions functional-equations
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
$endgroup$
How do I show that if there are functions $f,g$ such that$$
f(g(x)+g(y))=bx+cy
$$holds for all $b,cinmathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?
Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$
y^2+z^2=f(x,g(y-x)+g(z-x)).
$$
Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
functions functional-equations
functions functional-equations
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
edited Jan 10 at 15:24
sam wolfe
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
asked Jan 10 at 12:57
sam wolfesam wolfe
1,008525
1,008525
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.
Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.
Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $mathbb{R}$ into a $mathbb{Q}$-basis of $mathbb{R}$, which we denote as $mathcal{B}=(g(x))_x$.
Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).
Take now $f(x,y)=0$ if $y notin mathcal{B} + mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.
answered Jan 10 at 13:00
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068616%2ffunctional-equation-simple-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.
Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.
Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $mathbb{R}$ into a $mathbb{Q}$-basis of $mathbb{R}$, which we denote as $mathcal{B}=(g(x))_x$.
Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).
Take now $f(x,y)=0$ if $y notin mathcal{B} + mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.
answered Jan 10 at 13:00
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
|
show 5 more comments
$begingroup$
For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.
Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.
Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $mathbb{R}$ into a $mathbb{Q}$-basis of $mathbb{R}$, which we denote as $mathcal{B}=(g(x))_x$.
Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).
Take now $f(x,y)=0$ if $y notin mathcal{B} + mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.
answered Jan 10 at 13:00
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
|
show 5 more comments
$begingroup$
For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.
Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.
Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $mathbb{R}$ into a $mathbb{Q}$-basis of $mathbb{R}$, which we denote as $mathcal{B}=(g(x))_x$.
Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).
Take now $f(x,y)=0$ if $y notin mathcal{B} + mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.
answered Jan 10 at 13:00
MindlackMindlack
3,53717
$endgroup$
For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.
Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.
Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $mathbb{R}$ into a $mathbb{Q}$-basis of $mathbb{R}$, which we denote as $mathcal{B}=(g(x))_x$.
Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).
Take now $f(x,y)=0$ if $y notin mathcal{B} + mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.
answered Jan 10 at 13:00
MindlackMindlack
3,53717
edited Jan 10 at 17:08
answered Jan 10 at 13:00
MindlackMindlack
3,53717
answered Jan 10 at 13:00
MindlackMindlack
3,53717
answered Jan 10 at 13:00
MindlackMindlack
3,53717
3,53717
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
|
show 5 more comments
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know!
$endgroup$
– sam wolfe
Jan 10 at 15:25
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
So $f$ is a two-variable function?
$endgroup$
– Mindlack
Jan 10 at 16:35
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
In the extra case, yes.
$endgroup$
– sam wolfe
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
$begingroup$
With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction.
$endgroup$
– Mindlack
Jan 10 at 16:47
1
1
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
$begingroup$
What I did prove is that there is no contradiction (at least if the axiom of choice is granted).
$endgroup$
– Mindlack
Jan 17 at 7:35
|
show 5 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068616%2ffunctional-equation-simple-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
