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Let $f(x)$ be a function increasing on $[0,+\infty)$ such that $f(0)\neq0$ and $f(x+y)=f(x)f(y)$, for any $x,y>=0$. it is known that the following statements are true:

  1. $f(0)=1$
  2. $f(kx)=f(x)^k$ for any non-negative integer $k$.
  3. $f(n)=f(1)^n$ for any non-negative integer $n$.
  4. $1/a\le f(x)/(a^x)\le a$ where $a=f(1)$

I want to prove that it is impossible that $f(X)/(a^X)>1$ and $f(Y)/(a^Y)<1$ for some non-negative real numbers $X$ and $Y$ in order to show that $f(x)/a^x=1$ for any non-negative real number $x$ but have no clue. Can anyone help?

Nathaniel Bubis
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Nick
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3 Answers3

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You know that $f(1)^n = f(n)$ for any non-negative integer $n$. Now, suppose that $f(1) = a>0$. Then, $f(\frac{n}{m}) = f(\frac{1}{m})^n$. Now, because $m * \frac{1}{m} = 1$, it follows that $f(1)=f(m*\frac{1}{m})=f(\frac{1}{m})^m$. Hence, for all rational numbers $\frac{n}{m}$, $f(\frac{n}{m}) = a^{\frac{n}{m}}$, which is well defined because $a$ is positive.

Assume continuity. Consider any real number $x$. Then, there exists a sequence of rational numbers converging to $x$, say $x_n \to x$. Now, since $f$ is densely defined, there is a unique continuous extension to the reals, by saying $f(x) = \lim_{n \to \infty} f(x_n)$. This limit is $\lim_{n\to\infty } a^{x_n} = a^x$. Hence, $f(x) = a^x$, for any $a$ positive (if a were negative, then we would have problems in square rooting, for example)

It turns out that all solutions of the functional equation(including discontinuous ones) are differing from this only at the irrationals.However, adding even Lebesgue measurability gives a unique solution $a^x$.

Sarvesh Ravichandran Iyer
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  • i think this is a valid proof. But it is required to show $f(X)/(a^X)>1$ and $f(Y)/(a^Y)<1$ before showing $f(x)=a^x$ – Nick Aug 07 '16 at 08:07
  • As I wrote, it is possible to have discontinuous solutions to the above equation. Therefore, if $X$ and $Y$ are points of discontinuity, it may very well be true that $\frac{f(X)}{a^X} > 1$ and $\frac{f(Y)}{a^Y} < 1$, but the function remains a solution. Assuming continuity though, the above is impossible. Would you like me to add that proof? – Sarvesh Ravichandran Iyer Aug 07 '16 at 08:10
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I think i have figured out the proof.
Note that for any n$\in$$\Bbb R^+$, $[f(x)/(a^x)]^n=f(nx)/(a^nx)$.
Suppose there exits some $X$$\in$$\Bbb R^+$ such that $f(X)/(a^X)$$\gt$1.
Then $f(nX)/a^{nX}$$\gt$1.
$\Rightarrow$ $\lim_{n\to \infty}$[$f(X)/(a^X)]^n$=$+\infty$.
Then for the positive number $a$, there exists N$\in$$\Bbb N$ such that n>N $\Rightarrow$$[f(x)/(a^x)]^n$$\gt$$a$.
Hence $f((N+1)X)/(a^{(N+1)X})$$\gt$$a$, which contradicts to
$1/a\le f(x)/(a^x)\le a$
Proving another part is of similar idea.

Nick
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From 4) you get $$ a^{x-1}\le f(x)\le a^{x+1} $$ Now apply that $f(x)=f(nx)^{1/n}$ to transform that initial inequality to $$ a^{x-1/n}\le f(x)\le a^{x+1/n} $$ which has to be true for all $1<n\in\Bbb N$. This is only possible for $$ f(x)=a^x. $$

Lutz Lehmann
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