# Problem #13: Function

• Jul 6th 2014, 10:13 AM
topsquark
Problem #13: Function
Determine all real valued differentiable functions f, such that f(x) + f(y) = f(xy).

(Source: Purdue University Math Department POTW.)

-Dan
• Jul 6th 2014, 11:51 AM
romsek
Re: Problem #13: Function
Spoiler:
Let $g(x)$ be linear, i.e. $g(x+y)=g(x)+g(y)$

Let $f(z)=g(\ln(z))$

Then

$f(xy)=g(\ln(xy))=g(\ln(x)+\ln(y))=g(\ln(x))+g(\ln (y))=f(x)+f(y)$

Is this all of them? No idea. But this form works.
• Jul 6th 2014, 03:20 PM
topsquark
Re: Problem #13: Function
Quote:

Originally Posted by romsek
Spoiler:
Let $g(x)$ be linear, i.e. $g(x+y)=g(x)+g(y)$

Let $f(z)=g(\ln(z))$

Then

$f(xy)=g(\ln(xy))=g(\ln(x)+\ln(y))=g(\ln(x))+g(\ln (y))=f(x)+f(y)$

Is this all of them? No idea. But this form works.

That depends. My solution does not appear to be as general as the solution you have, but I wonder if it isn't the same. Could you give me an example of a linear map g(z)? I agree with your solution, but doesn't the linear map g(z) you mention have to be of the form $\displaystyle a~ln(x)$, where a is a constant? (This is the function that I derived.)

Also, you have a correct answer, but how did you derive it? Or was it an educated guess?

-Dan
• Jul 6th 2014, 04:08 PM
Deveno
Re: Problem #13: Function
The only (continuous) $\Bbb R$-additive functions $g:\Bbb R \to \Bbb R$ are of the form: $g(x) = ax$ where $a \in \Bbb R$ is a constant. To see this, note that $g$ is completely determined by $g(1)$ since:

$\displaystyle g(n) = g\left(\sum_{i = 1}^n 1\right) = \sum_{i = 1}^n g(1) = n\cdot g(1)$, for any natural number $n > 0$.

Since $g(0) = g(0 + 0) = g(0) + g(0)$, it follows that $g(0) = 0$.

Finally, $g(-n) + g(n) = g(-n + n) = g(0) = 0$ so that: $g(-n) = -g(n)$.

Next we consider $g\left(\dfrac{1}{k}\right)$. By a similar argument as above, we see that for $k \in \Bbb N$:

$k\cdot g\left(\dfrac{1}{k}\right) = g\left(\dfrac{k}{k}\right) = g(1)$, so that $g\left(\dfrac{1}{k}\right) = \dfrac{1}{k}\cdot g(1)$.

Hence for any RATIONAL number $r$, we have $g(r) = r\cdot g(1)$.

Now if $\{r_j\}$ is a Cauchy rational sequence converging to the real number $x$, we obtain a Cauchy real sequence $\{g(r_j)\}$ converging to $g(1)\cdot x$.

Taking $a = g(1)$, we have the desired conclusion (note we need $g$ to be a CONTINUOUS additive function-there are bizarre NON-continuous counter-examples).

None of the above is necessary if $g$ is (by dint of being $\Bbb R$-linear) stiipulated to respect scalar multiplication, in which case all we need to know is that the matrix with respect to any two bases of $\Bbb R$ is just a real number (1x1 matrix), and since $\{1\}$ is a basis for $\Bbb R$, we have $g$ is completely determined by $g(1)$.

Spoiler:
Given, then, that $f(x) = a\ln(x)$, if $f$ is not the 0-function, taking $b = \dfrac{1}{a}$, we have $f(x) = \dfrac{\ln(x)}{b}$, and since $\ln$ is surjective, we may write $b = \ln(c)$ so that:

$f(x) = \dfrac{\ln(x)}{\ln(c)} = \log_c(x)$, that is: $f$ is a logarithm function for "some base" $c$
• Jul 6th 2014, 05:22 PM
JeffM
Re: Problem #13: Function
I have no idea how to solve this kind of question (nor did I look at Deveno's answer before coming up with my own partial solution). The function that I found is admittedly trivial, but it definitely shows that romsek's solution is not unique. It also applies to all real x and y, which is how I read the problem. Again, I admit it is trivial.

Spoiler:
$x,\ y \in \mathbb R\ and\ f(x) = 0 \implies f(x) + f(y) = 0 + 0 = 0 = f(xy).$

• Jul 6th 2014, 10:49 PM
Deveno
Re: Problem #13: Function
It appears there has been some mis-understanding about my prior post: it does not "derive" the answer topsquark is looking for. Let's correct this.

The additive condition in my previous post is called the Cauchy equation. It's really the "hard part" of the problem as we will see.

Spoiler:

Suppose that $f$ is not the 0-function. Thus we may choose $c > 0$ such that $f(c) \neq 0$. Let $n$ be any natural number. Then there are integers $k,m$ such that:

$f(c^k) = kf(c) > n$ and $f(c^m) = mf(c) < -n$. We conclude if $f$ is not the 0-function, $f$ is surjective (by the intermediate value theorem).

In particular, there exists integers $k,m$ such that $f(c^k) > 0$ and $f(c^m) < 2$, which means that there is $a > 0$ such that $f(a) = 1$ (since $a$ is in the interval with $c^k,c^m$ as endpoints).

Now define $g(x) = f(a^x)$. Note that $g$ is continuous, being the composition of two continuous functions ($f$, being differentiable, is a fortiori, continuous).

Now $g(r+s) = f(a^{r+s}) = f(a^ra^s) = f(a^r) + f(a^s) = g(r) + g(s)$.

Thus, by my previous post, $g(x) = f(a)x$ (it satisfies the Cauchy equation).

Since $f(a) = 1$, we have $g(x) = x$, that is: $f(a^x) = x$, so that $f$ is the inverse function to $a^x$, that is: $f = \log_a$.
• Jul 9th 2014, 08:32 AM
johng
Re: Problem #13: Function
Dan,
The problem makes no mention of the domain of the function, so I made up my own domain specification. Furthermore, my derivation makes no use of the differentiability of f, and so I stated the problem only for continuous f.

http://i61.tinypic.com/17s66x.png
http://i61.tinypic.com/2ikacds.png
• Jul 9th 2014, 06:55 PM
topsquark
Re: Problem #13: Function
Quote:

Originally Posted by johng
Dan,
The problem makes no mention of the domain of the function, so I made up my own domain specification. Furthermore, my derivation makes no use of the differentiability of f, and so I stated the problem only for continuous f.

Sorry about the domain thing. Yes, $\displaystyle D = \left ( - \infty, 0 \right ) \cup \left ( 0, \infty \right )$.

-Dan
• Jul 16th 2014, 05:24 PM
topsquark
Re: Problem #13: Function
Interesting. No one used Calculus beyond continuity. This one does:

We have that $\displaystyle f(x) + f(y) = f(xy)$ and that f() is differentiable.

Thus
$\displaystyle \frac{df(x)}{dx} + \frac{df(y)}{dx} = \frac{df(xy)}{dx}$

$\displaystyle \frac{df(x)}{dx} = y \cdot \frac{df(xy)}{d(xy)}$

$\displaystyle \frac{df(xy)}{d(xy)} = \frac{1}{y} \cdot \frac{df(x)}{dx}$

Similarly we have that
$\displaystyle \frac{df(xy)}{d(xy)} = \frac{1}{x} \cdot \frac{df(y)}{dy}$

Equating these two lines gives
$\displaystyle \frac{df(xy)}{d(xy)} = \frac{1}{y} \cdot \frac{df(x)}{dx} = \frac{1}{x} \cdot \frac{df(y)}{dy}$

Thus we have
$\displaystyle x\cdot \frac{df}{dx} = y\cdot \frac{df}{dy}$

And now the usual "the LHS only depends on x and the RHS depends only on y" so we get
$\displaystyle x\cdot \frac{df}{dx} = a$

where a is a constant.

Solving for f gives
$\displaystyle f(x) = a \ln|x| + C$

Now, we know from the original that $\displaystyle f(1) + f(1) = f(1 \cdot 1) = f(1) \implies f(1) = 0$, giving C = 0. Thus f(x) = a ln|x| is the function desired.

Thanks to all that participated!

-Dan