Dgdrxrt

Inappropriate reference requests from Journal reviewers

Does it take more energy to get to Venus or to Mars?

What connection does MS Office have to Netscape Navigator?

How do I go from 300 unfinished/half written blog posts, to published posts?

Elegant way to replace substring in a regex with optional groups in Python?

Written every which way

Would a completely good Muggle be able to use a wand?

Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?

How do scammers retract money, while you can’t?

sp_blitzCache results Memory grants

What does "Its cash flow is deeply negative" mean?

Which kind of appliances can one connect to electric sockets located in an airplane's toilet?

Complex fractions

Can we say or write : "No, it'sn't"?

MessageLevel in QGIS3

How did the Bene Gesserit know how to make a Kwisatz Haderach?

Why didn't Khan get resurrected in the Genesis Explosion?

What benefits would be gained by using human laborers instead of drones in deep sea mining?

If a black hole is created from light, can this black hole then move at speed of light?

Why has the US not been more assertive in confronting Russia in recent years?

Is micro rebar a better way to reinforce concrete than rebar?

Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

How to start emacs in "nothing" mode (`fundamental-mode`)

Is it possible to have $f(x)f(y) = g(x)+g(y)$?

1

1

$begingroup$

Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy

$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$

By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).

functional-equations

share|cite|improve this question

asked 2 days ago

flawrflawr

11.7k32546

$endgroup$

add a comment | 

1

1

$begingroup$

Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy

$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$

By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).

functional-equations

share|cite|improve this question

asked 2 days ago

flawrflawr

11.7k32546

$endgroup$

add a comment | 

$begingroup$

Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $Omega$ of) $mathbb R$ or $mathbb C$ that satisfy

$$f(x)f(y) = g(x) + g(y) :forall x,y in Omega$$

By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = fracc^22$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).

functional-equations

share|cite|improve this question

asked 2 days ago

flawrflawr

11.7k32546

$endgroup$

share|cite|improve this question

asked 2 days ago

flawrflawr

11.7k32546

share|cite|improve this question

asked 2 days ago

flawrflawr

11.7k32546

asked 2 days ago

flawrflawr

11.7k32546

asked 2 days ago

flawrflawr

11.7k32546

2 Answers
2

active

oldest

votes

1

$begingroup$

Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have

$$cf(x) = g(x) + d$$

for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.

Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation

$$c = frac2dc implies c^2 = 2d.$$

---

Looking at the diagonal case $x = y$, we have

$$f(x)^2 = 2g(x)$$

which reduces us to

$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$

There aren't so many solutions to this.

To be even more explicit, we have a quadratic equation

$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$

which implies that $f$ is constant.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

T. BongersT. Bongers

23.5k54762

$endgroup$

add a comment | 

1

$begingroup$

Notice that deriving both side ($dx$):

$$f'(x)f(y)=g'(x)$$

And now both sides $dy$:

$$f'(x)f'(y)=0$$

So $f'(x)=0 Rightarrow f(x)=K $

And:

$$K^2=g(x)+g(y)$$

That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").

share|cite|improve this answer

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164724%2fis-it-possible-to-have-fxfy-gxgy%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have

$$cf(x) = g(x) + d$$

for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.

Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation

$$c = frac2dc implies c^2 = 2d.$$

---

Looking at the diagonal case $x = y$, we have

$$f(x)^2 = 2g(x)$$

which reduces us to

$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$

There aren't so many solutions to this.

To be even more explicit, we have a quadratic equation

$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$

which implies that $f$ is constant.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

T. BongersT. Bongers

23.5k54762

$endgroup$

add a comment | 

1

$begingroup$

Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have

$$cf(x) = g(x) + d$$

for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.

Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation

$$c = frac2dc implies c^2 = 2d.$$

---

Looking at the diagonal case $x = y$, we have

$$f(x)^2 = 2g(x)$$

which reduces us to

$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$

There aren't so many solutions to this.

To be even more explicit, we have a quadratic equation

$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$

which implies that $f$ is constant.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

T. BongersT. Bongers

23.5k54762

$endgroup$

add a comment | 

$begingroup$

Suppose that $0 in Omega$ and that $f(0) = c, g(0) = d$. We then have

$$cf(x) = g(x) + d$$

for all $x in Omega$. If $c = 0$, then $g(x)$ is also constant $0$.

Otherwise, we have $$f(x) = fracg(x) + dc.$$
Note that $x = 0$ gives us the equation

$$c = frac2dc implies c^2 = 2d.$$

---

Looking at the diagonal case $x = y$, we have

$$f(x)^2 = 2g(x)$$

which reduces us to

$$f(x) = fracfrac 1 2 f(x)^2 + fracf(0)^22f(0) = frac 1 2f(0) (f(x)^2 + f(0)^2).$$

There aren't so many solutions to this.

To be even more explicit, we have a quadratic equation

$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$

which implies that $f$ is constant.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

T. BongersT. Bongers

23.5k54762

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

T. BongersT. Bongers

23.5k54762

answered 2 days ago

T. BongersT. Bongers

23.5k54762

answered 2 days ago

T. BongersT. Bongers

23.5k54762

1

$begingroup$

Notice that deriving both side ($dx$):

$$f'(x)f(y)=g'(x)$$

And now both sides $dy$:

$$f'(x)f'(y)=0$$

So $f'(x)=0 Rightarrow f(x)=K $

And:

$$K^2=g(x)+g(y)$$

That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").

share|cite|improve this answer

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

Notice that deriving both side ($dx$):

$$f'(x)f(y)=g'(x)$$

And now both sides $dy$:

$$f'(x)f'(y)=0$$

So $f'(x)=0 Rightarrow f(x)=K $

And:

$$K^2=g(x)+g(y)$$

That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").

share|cite|improve this answer

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Notice that deriving both side ($dx$):

$$f'(x)f(y)=g'(x)$$

And now both sides $dy$:

$$f'(x)f'(y)=0$$

So $f'(x)=0 Rightarrow f(x)=K $

And:

$$K^2=g(x)+g(y)$$

That has trivially as only solution $g(x)=fracK^22$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").

share|cite|improve this answer

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

EurekaEureka

423112

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

EurekaEureka

423112