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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

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asked Mar 27 at 16:30

flawrflawr

11.9k32546

$endgroup$

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 Mar 27 at 16:30

flawrflawr

11.9k32546

$endgroup$

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 Mar 27 at 16:30

flawrflawr

11.9k32546

$endgroup$

share|cite|improve this question

asked Mar 27 at 16:30

flawrflawr

11.9k32546

share|cite|improve this question

asked Mar 27 at 16:30

flawrflawr

11.9k32546

asked Mar 27 at 16:30

flawrflawr

11.9k32546

asked Mar 27 at 16:30

flawrflawr

11.9k32546

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 Mar 27 at 16:51

answered Mar 27 at 16:40

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 Mar 27 at 16:44

EurekaEureka

907115

$endgroup$

add a comment | 

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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 Mar 27 at 16:51

answered Mar 27 at 16:40

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 Mar 27 at 16:51

answered Mar 27 at 16:40

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 Mar 27 at 16:51

answered Mar 27 at 16:40

T. BongersT. Bongers

23.5k54762

$endgroup$

share|cite|improve this answer

edited Mar 27 at 16:51

answered Mar 27 at 16:40

T. BongersT. Bongers

23.5k54762

answered Mar 27 at 16:40

T. BongersT. Bongers

23.5k54762

answered Mar 27 at 16:40

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 Mar 27 at 16:44

EurekaEureka

907115

$endgroup$

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 Mar 27 at 16:44

EurekaEureka

907115

$endgroup$

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 Mar 27 at 16:44

EurekaEureka

907115

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:44

EurekaEureka

907115

answered Mar 27 at 16:44

EurekaEureka

907115

answered Mar 27 at 16:44

EurekaEureka

907115