# If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t

Let $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$.

I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as $g(xy) = g(x) + g(y)$

I looked up the hint and it says let $g(x) = \log f(a^x)$

The wikipedia page for functional equations only states the form of the solutions without proof.

Attempt Using the hint (which was like pulling a rabbit out of the hat)

Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$ so that we can define the real function $g(x) = \log f(a^x)$ and we have $$g(x+y) = g(x)+ g(y)$$

i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).

Letting $\log_a f(a) = p$ we get $f(a^x) =a^p$. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited

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@Qia: I don't think this is a dupe. This is $g(xy) = g(x) + g(y)$. Not $g(x + y) = g(x) g(y)$ –  Aryabhata Jun 7 '11 at 22:16
Sure, but the method is identical. –  Qiaochu Yuan Jun 7 '11 at 22:26
perhaps this is true but i think it is by no means evident at least for an unexperienced person as me –  miracle173 Jun 7 '11 at 23:10
Using your hint: $g(x)=f(a^x)\implies g(x+y)=f(a^{x+y})=f(a^xa^y)=f(a^x)f(a^y)=g(x)g(y).$ This transforms it into the problem that other users were referring to. –  Nick Strehlke Jun 8 '11 at 0:05

So, we assume $f$ is continuous. Letting $g(x) = \ln(f(a^x))$, we get \begin{align*} g(x+y) &= \ln(f(a^{x+y})) = \ln(f(a^xa^y)) = \ln(f(a^x)f(a^y))\\ &= \log(f(a^x)) + \ln(f(a^y))\\ &= g(x)+g(y). \end{align*} So $g$ satisfies the Cauchy functional equation; if you assume $f$ is continuous, then so is $g$, hence $g(x) = xg(1)$ for all $x\gt 0$.

Since $g(1) = \ln(f(a))$, we have $$f(a^x) = e^{g(x)} = e^{g(1)x} = (e^{x})^{g(1)}.$$ Given $r\in \mathbb{R}$, $r\gt 0$, we have $r = a^{\log_a(r)}$, hence \begin{align*} f(r) &= f\left(a^{\log_a(r)}\right)\\ &= \left(e^{\log_a(r)}\right)^{g(1)}\\ &= \left(e^{\ln(r)/\ln(a)}\right)^{g(1)}\\ &= \left(e^{\ln(r)}\right)^{g(1)/\ln(a)}\\ &= r^{g(1)/\ln(a)}, \end{align*} where we have used the change-of-base formula for the logarithm, $$\log_a(r) = \frac{\ln r}{\ln a}.$$ Finally, since $g(1) = \ln(f(a))$, we have $$f(r) = r^{\ln(f(a))/\ln(a)}.$$ As this works for any positive $a$, $a\neq 1$, taking $a=e$ we get $$f(r) = r^{\ln(f(e))}.$$

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Please use the Align command in the second line so that answer doesn't cross the border. –  user9413 Jun 8 '11 at 4:07
@Chandru: The formula does not "cross the border" on my screen, so I was unaware of the problem. –  Arturo Magidin Jun 8 '11 at 4:07
By border i meant, where the related questions are written. –  user9413 Jun 8 '11 at 4:09
@Chandru: I know what you meant; it just didn't happen in my display. And I've changed it. –  Arturo Magidin Jun 8 '11 at 4:10

There are at least three interesting features of the problem as currently stated. First of all, it specifies that $x, y \ge 0$. Secondly, apart from $x,y \ge 0$, it specifies neither the domain nor the codomain. And finally, it does not ask for continuity.

We will have to narrow down the question in order to answer it, but we will narrow it down a tiny bit less than in the very thorough answer by Arturo Magidin.

Let's decide that the domain consists of the non-negative reals. In order for the answer $x^p$ to make sense, we cannot have the codomain be $(0,\infty)$, we must at least insist on $[0,\infty)$. (After all, we are told that the functional equation holds for $x,y \ge 0$.)

Thus the functional equation admits the solution $f$ identically equal to $0$.

Assume that the codomain of $f$ is the reals. By putting $x=y=\sqrt{t}$, we can see that $f(t)=(f(\sqrt{t}))^2$, so $f(t)$ is always $\ge 0$.

Suppose that there is a non-zero $a$ such that $f(a)=0$. Then from the functional equation, we find that $f(x)=f(a)f(x/a)=0$, and therefore $f$ is identically $0$.

Thus from now on we can confine attention to functions $f$ such that $f(x)>0$ when $x>0$. It follows that for $x>0$, we can take logarithms freely and proceed along the lines suggested in the hint.

When looking at functional equations, particularly in a contest setting, it is useful to identify $f(0)$. From the fact that $f(0)=(f(0))^2$, we see that $f(0)=0$ or $f(0)=1$. But if $f(0)=1$, then from $f(0 \cdot x)=f(0)f(x)$ we conclude that $f(x)=1$. Thus apart from the case $f$ identically equal to $1$, which is presumably covered by the case $p=0$ of the answer, we can take $f(0)=0$.

Note however that if we forget about continuity, the answer $f(0)=0$, $f(x)=1$ if $x \ne 0$ is perfectly fine. And forgetting about continuity, at least at $0$, has certain advantages. For instance, in the solution by Arturo Magidin, the question of whether $\ln(f(e))$ can be negative is not directly addressed. The answer is that if we specify continuity for $x>0$, but not necessarily at $0$, then $\ln(f(e))$ can be negative.

The conclusion is that the full list of functions that are continuous for $x>0$, but not necessarily at $0$, and that satisfy the functional equation, consists of the following functions.

$1$. The identically $0$ function and the identically $1$ function.

$2$. The function which is $0$ at $0$ and $1$ elsewhere.

$3$. The functions $f(x)=x^p$, where $p$ is positive.

$4$. The functions $f(0)=0$, $f(x)=x^p$ for $x>0$, where $p$ is negative.

Comment: The only Lebesgue measurable functions that satisfy the Cauchy functional equation are the obvious ones, so the above list gives a list of all measurable functions that satisfy the functional equation of the problem. And if, for example, we are willing to jettison the Axiom of Choice in favour of the Axiom of Determinacy, then all functions from $\mathbb{R}$ to $\mathbb{R}$ are Lebesgue measurable, and then the above is a complete list.

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Both the answers above are very good and thorough, but given an assumption that the function is differentiable, the DE approach strikes me as the easiest.

$\frac{\partial}{\partial y} f(x y) = x f'(xy) = f(x)f'(y)$

Evaluating y at 1 gives:

$xf'(x) = f(x)f'(1)$

The above is a separable DE:

Let $p = f'(1)$ and $z = f(x)$

$x\frac{dz}{dx} = pz \implies \int \frac{dz}{z} = p\int \frac{dx}{x}$

$\therefore \ln|z| = p\ln|x| + C$

Let $A = e^C$. $\implies C = \ln(A)$

$x > 0 \implies |x| = x$

$\therefore \ln|z| = p\ln(x) + \ln(A) = \ln(x^p) + \ln(A) = \ln(Ax^p)$

Hence $|z| = Ax^p$; $z = \pm Ax^p = f(x)$

Let $B = \pm A$ and now $f(x) = Bx^p$

Now using the initial property:

$f(x)f(y) = Bx^p By^p = B^2 (xy)^p = f(xy) = B (xy)^p$

$B^2 = B \implies B$ is $0$ or $1$.

If B is zero, that provides the constant function $f(x) = 0$, otherwise the solution is $f(x) = x^p$.

As can be seen from the other answers, this does not capture all possible solutions, but sometimes that's the price of simplicity.

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