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This is part of a homework assignment for a real analysis course taught out of "Baby Rudin." Just looking for a push in the right direction, not a full-blown solution. We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero. The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant. I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers. Is this an approach worth taking?

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    $\begingroup$ $f(0)f(0)=f^2(0)$ so $f(0)=0$ or $f(0)=1$. $f(0)\neq 0$ so $f(0)=1$. $f(n)=f^n(1)$ and... $\endgroup$
    – Mikasa
    Feb 3, 2013 at 5:27
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    $\begingroup$ $f(x)f(0)=f(x+0)\implies f(0)=1$ as $f(x)\ne0$ $\endgroup$ Feb 3, 2013 at 5:48
  • $\begingroup$ Note that by definition of derivative, f'(x)=f(x)•f'(0). So the function is differentiable either everywhere or nowhere. $\endgroup$
    – mbaitoff
    Feb 3, 2013 at 6:07
  • $\begingroup$ Similar: math.stackexchange.com/questions/22069/… $\endgroup$ Jun 7, 2021 at 9:59

3 Answers 3

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First note that $f(x) > 0$, for all $x \in \mathbb{R}$. This can be seen from the fact that $$f(x) = f\left(\dfrac{x}2 + \dfrac{x}2\right) = f \left(\dfrac{x}2\right)^2$$ Further, you can eliminate the case $f(x) = 0$, since this would mean $f \equiv 0$.

One way to go about is as follows.

$1$. Prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}^+$.

$2$. Now prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}$.

$3$. Now prove that $f(p/q) = f(1)^{p/q}$ for $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$.

$4$. Now make use of the fact that rationals are dense in $\mathbb{R}$ and hence you can find a sequence of rationals $r_n \in \mathbb{R}$ such that $r_n \to r \in \mathbb{R} \backslash \mathbb{Q}$. Now use continuity to conclude that $f(x) = f(1)^x$ for all $x \in \mathbb{R}$. You will see that you need only continuity at one point to conclude that $f(x) = f(1)^x$.

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  • $\begingroup$ I remember seeing this proof in my Stochastic Models book to show that the only continuous memoryless distribution is exponential. Its refreshing to see it again. +1 $\endgroup$
    – Inquest
    Feb 3, 2013 at 6:34
  • $\begingroup$ Thanks a lot for your help, I'll see if I'm good enough at Latex to submit my answer. $\endgroup$
    – KarlG
    Feb 9, 2013 at 23:29
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Since $f(t)=f(t+0)=f(t)f(0)$ we can conclude that $f(0)=1$. If there exists a real number $β$ such that $f(β)=0$, then for any real number $x$ we have

$f(x)=f(β+(x-β))=f(β)f(x-β)=0$

which implies that $f$ is identically zero. By the continuity of $f$ we can conclude that $f(x)>0$ for all real numbers.

For a positive integer $n$, $f(n)=f(1+1+...+1)=f(1)f(1)*...*f(1)=f(1)^n$.

Since $f(-1)=f(1+(-2))=f(1)f(-2)=f(1)f(-1)f(-1)$,

it follows that $f(-1)=f(1)^{-1}$.

For a negative integer $m$, $f(m)=f(-1+(-1)+...+(-1))=f(-1)^{-m}=(f(1)^{-1})^{-m}=f(1)^m$.

Since $f(0)=1=f(1)^0$, $f(n)=f(1)^n$ holds for all integers.

Let $q$ and $m$ be positive integers, then $f(m)=f(1/q+1/q+...+1/q)=f(1/q)^{qm}=f(1)^m$.

It follows that $f(1/q)=f(1)^{1/q}$. This will also hold for negative integers $q_1$ and $m_1$ as well.

Then, $f(s/t)=f(1/t+...+1/t)=f(1/t)^s=(f(1)^{1/t})^s=f(1)^{s/t}$, for any rational number.

Since the rationals are dense in the reals and $f$ is continuous, for any real number $x$ we can find a sequence $(r_n)$ that will converge to $x$ with $f(r_n)=f(1)^{r_n}$, so that $f(x)=f(1)^x$ for all real numbers.

Since $f(1)>0$ and is real-valued, $\log(f(1))=c$ exists. Then:

$f(x)=f(1)^x=(\exp(\log(f(1))))^x=(\exp(c))^x=(e^c)^x=e^{cx}$.

Thanks again!

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$f(x)= f(\frac{x}{2})f(\frac{x}{2}) $ and together with $f(x) \neq 0$ implies $f(x)$ is positive (actually if $f(x) $ is zero anywhere then it is zero everywhere)

Define $g(x)= log f(x)$

Then $g(x) + g(y) =g(x+y)$ which is cauchy functional equation with solution $g(x) = cx$ where $c$ is a constant

So $f(x)= e^{cx}$

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