MHB Determine all real valued differentiable functions f(x)+f(y)=f(xy)

[lfdahl]

Determine, with proof, all the real-valued differentiable functions $f$, defined
for real $x > 0$, which satisfy $f(x) + f(y) = f(xy)$ for all $x, y > 0$.

 

[HallsofIvy]

Spoiler

The first thing I would notice is that, taking y= 0, f(x)+ f(0)= f(0) so that f(x)= 0 for all x.

 

[kaliprasad]

Spoiler

because f(xy) = f(x) + f(y) and defined for x, y >0 I see a log function $f(x) =m\log\,x$ satisfies criteria where m is arbitrary constant. m= 0 gives above solution(post #2)

 

[Opalg]

Determine, with proof, all the real-valued differentiable functions $f$, defined
for real $x > 0$, which satisfy $f(x) + f(y) = f(xy)$ for all $x, y > 0$.

[sp]With $x=y=1$, $f(1) + f(1) = f(1)$, from which $f(1) = 0$.

Fix $y$, and differentiate the equation $f(x) + f(y) = f(xy)$ with respect to $x$: $f'(x) = yf'(xy)$. Therefore $xf'(x) = xyf'(xy)$.

Now let $z = xy$, so that $xf'(x) = zf'(z)$. Since that is true for all positive numbers $x$ and $z$, it follows that $xf'(x)$ is constant, say $xf'(x) = c$. Then $f'(x) = \dfrac cx$, so we can integrate to get $f(x) = c\ln x + d$ (where $d$ is another constant). But $f(1) = 0$, so that $d=0$. Thus $f(x) = c\ln x$, and those are the only solutions.[/sp]

 

[lfdahl]

Spoiler

because f(xy) = f(x) + f(y) and defined for x, y >0 I see a log function $f(x) =m\log\,x$ satisfies criteria where m is arbitrary constant. m= 0 gives above solution(post #2)

Hi, kaliprasad!
Your intuitive solution is correct indeed! Thanks for participating!

- - - Updated - - -

[sp]With $x=y=1$, $f(1) + f(1) = f(1)$, from which $f(1) = 0$.

Fix $y$, and differentiate the equation $f(x) + f(y) = f(xy)$ with respect to $x$: $f'(x) = yf'(xy)$. Therefore $xf'(x) = xyf'(xy)$.

Now let $z = xy$, so that $xf'(x) = zf'(z)$. Since that is true for all positive numbers $x$ and $z$, it follows that $xf'(x)$ is constant, say $xf'(x) = c$. Then $f'(x) = \dfrac cx$, so we can integrate to get $f(x) = c\ln x + d$ (where $d$ is another constant). But $f(1) = 0$, so that $d=0$. Thus $f(x) = c\ln x$, and those are the only solutions.[/sp]

Thanks, Opalg! for your participation. Your solution is - of course - correct!(Yes)