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Is there any closed form solution for a function $f$ on $\mathbb R$ satisfying $f(t+s)=f(t)f(s), t, s \in \mathbb R$? I've figured out at least either $f(t) = 0$ or $f(t) = e^{-\lambda t}$ can be a solution. I don't know whether there exist some other solutions. Besides, I've no idea how to solve it for a closed form.

Thanks in advance.

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    $\begingroup$ en.wikipedia.org/wiki/Cauchy%27s_functional_equation $\endgroup$ – user175968 Oct 28 '15 at 21:39
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    $\begingroup$ What do you mean by closed form? There are infinitely many possible solutions. You named an infinite many solutions since the second expression is true for all $\lambda$. $\endgroup$ – Mark Viola Oct 28 '15 at 21:42
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    $\begingroup$ If you have $g(x)=ln(f(x))$ then $g(t+s)=g(t)+g(s)$ $\endgroup$ – user175968 Oct 28 '15 at 21:44
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    $\begingroup$ @Dr.MV: It is, if you take the log (provided you show $f > 0$). $\endgroup$ – Alex M. Oct 28 '15 at 21:44
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    $\begingroup$ Taking in account that $f(0)=1$ on the class of differentible function you get $(f(t+s)-f(t))/s=(f(s)-f(0)/s f(t)$, In the limit you have $f'=f'(0) f$ which implies the exponent. $\endgroup$ – Alexander Vigodner Oct 28 '15 at 21:54
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So we want to find all functions $f:\mathbf{R}\to\mathbf{R}$ that are continuous, and satisfy $$f(x+y)=f(x)f(y).\tag*{$\forall(x,y)\in\mathbf{R}^2$}$$

First, note that $f(x)=f(x/2)^2$, hence $f(x)\geqslant0$. Now we would like to know what happens if the meant function equals $0$ at a certain point. So assume there's an $\alpha$ for which $f(\alpha)=0$. Then $$f(x)=f(x-\alpha+\alpha)=f(x-\alpha)f(\alpha)=0,$$ which means that the function would be zero everywhere. We can also understand from that that if any solution to that functional equation didn't vanish identically, then they cannot vanish at any point.

So we will exclusively consider solutions that do not vanish, hence $f(x)\gt0$ for all $x$. Thus its natural logarithm will be well-defined. So we can write $g=\ln\circ f$, which would yield $$g(x)+g(y)=g(x+y),\tag*{$\forall(x,y)\in\mathbf{R}^2$}$$ and this is exactly Cauchy's functional equation, which has the unique continuous solution $g:x\mapsto cx$. From this it can be easily seen that $f:x\mapsto b^x$ is the only continuous solution, along with $x\mapsto0$ and $x\mapsto1$, since putting $x=y=0$ yields $f(0)^2=f(0)$.

Another solution can be proposed by imitating this proof of Cauchy's functional equation, as follows: (We'll deal with only non vanishing solutions) One can easily see that by putting $x=y=0$ one gets: $$f(0)=f(0)^2\iff f(0)=1,$$ since $f(0)\neq0$. One can show using induction that for any $n\in\mathbf{N}$ with $n\neq0$, $$f(x_1+\cdots+x_n)=f(x_1)\cdots f(x_n).$$ If we set $x_i=x$ for all $i$, then we get $$f(nx)=f(x)^n.$$ If $x=pz/q$, where $(p,q)\in\mathbf{N}^2$ with $q\neq0$, then $$f(pz)=f\left(q\left(\dfrac p q z\right)\right)=f\left(\dfrac{pz}{q}\right)^q.$$ And since $f(pz)=f(z)^p$, one gets: $$f(z)^p=f\left(\dfrac pq z\right)^q \implies f\left(\dfrac pq z\right)=f(z)^{p/q}.$$ By putting $x=-y$ in the functional equation one finds: $$f(0)=f(x)f(-x)\implies f(-x)=\dfrac{1}{f(x)}.$$ Hence from the two above relations we conclude that $$f\left(-\dfrac pq z\right)=f(z)^{-p/q}.$$ That is, for all reals $z$, and for all rationals $r$, $$f(rz)=f(z)^r.$$ When $z=1$ we get $f(r)=\lambda^r$, by defining $\lambda:=f(1)$. We wish to extend this solution to all real numbers. So let $\phi\in\mathbf{R}$, we know that there is a sequence $\{r_n\}$ such that $$\lim_{n\to\infty}r_n=\phi.$$ For all the terms of the sequence we have $f(r_n)=\lambda^{r_n}$, and since it is continuous $$f(\phi)=f\left(\displaystyle\lim_{n\to\infty}r_n\right)=\lim_{n\to\infty}f(r_n)=\lim_{n\to\infty}\lambda^{r_n}=\lambda^\phi.$$

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The only continuous solutions are the constant functions $0$ and $1$, along with exponential functions. There are, however, some other very complicated solutions. This problem can be converted into the Cauchy functional equation by taking logarithms of both sides. Explicitly, $$\log f(x+y)=\log f(x)+\log f(y),$$ so if $\log f$ is continuous, then $\log f$ is a linear function.

We still need to justify taking logarithms, though. It is not too hard to check that if $f$ is continuous and not identically $0$, then it is always positive. If not, then $f$ must be $0$ somewhere by the Intermediate Value Theorem. The functional equation immediately yields $f(x)=0$ for all $x$. Hence, either $f(x)=0$ or $\log f(x)$ is linear, assuming $f$ is continuous.

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  • $\begingroup$ How did you know for continuous solutions, only three answers, 0, 1 and exponential functions? $\endgroup$ – Bear and bunny Oct 28 '15 at 21:51
  • $\begingroup$ @Bearandbunny because the linear functions ($f(x) = cx$) are the only continuous solutions to the Cauchy equation (follow the link). $\endgroup$ – Henno Brandsma Oct 28 '15 at 21:55
  • $\begingroup$ @HennoBrandsma: Thanks. $\endgroup$ – Bear and bunny Oct 28 '15 at 22:00
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    $\begingroup$ The constant $1$ is also an exponential function: $1 = \exp(0x)$. $\endgroup$ – celtschk Oct 28 '15 at 22:29
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Assume that $f$ is differentiable. Differentiate with respect to $t$. Then $f'(t+s)= f'(t)f(s)$. Set $t=0$. Then $f'(s)=f'(0)f(s)$.

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    $\begingroup$ don't use $f'$ since it confusing in the context of two variable - use $\frac{\partial}{\partial t}$ instead $\endgroup$ – Michael Medvinsky Oct 28 '15 at 21:46
  • $\begingroup$ @MichaelMedvinsky : But $f$ is a function of a single variable, and $\frac{\partial}{\partial t}f(t+s)$ seems to me like a sloppy notation for $\frac{\partial}{\partial t}g (s,t)$ where $g(s,t) = f(s+t)$. $\endgroup$ – Joel Cohen Oct 28 '15 at 22:07
  • $\begingroup$ $f(x)$ is a function of a single variable, but $f(s+t)$ is a function of 2 variables, i.e. it is a $g(s,t)=f(s+t)$. Furthermore, the expression $f'(t+s)=f'(t)f(s)$ is very confusing, you in general differentiate two sides of the equation but the two sides aren't of one variable, therefore the notations of partial derivative would be more appropriate. $\endgroup$ – Michael Medvinsky Oct 28 '15 at 22:16
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    $\begingroup$ $f$ is a function of one variable, so I see no confusion. $f'(s+t)$ is the derivative of the single variable function $f$ evaluated at $s+t$ $\endgroup$ – Dunham Oct 29 '15 at 1:05
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To add a cultural note to other people's answer :

If $g : \mathbb{R} \to \mathbb{C}$ is a solution to Cauchy's equation

$$g(x+y) = g(x) + g(y)$$

then $f = e^g$ is a solution of your equation. We can show that solution of Cauchy's equation are $\mathbb{Q}$-linear (meaning that $g(qx) = q \,g(x)$ for $q \in \mathbb{Q}$). So continuous solution are just linear functions $g(x) = \lambda x$ (with $\lambda \in \mathbb{C}$), while there exist many other discontinuous solutions (called Hamel functions).

Conversely, if $f$ is real-valued, then we can show $f > 0$ (apart from the solution $f = 0$) and define $g = \ln f$, which satisfies Cauchy's equation.

However, if $f$ is complex-valued, we can find many other solutions to this equation by considering solutions $g : \mathbb{R} \to \mathbb{C}/2i\pi \mathbb{Z}$ to the equation

$$g(x+y) \equiv g(x) + g(y) \mod 2i \pi$$

and taking $f = e^g$ (conversely, such a function $g$ always exists). It turns out there are many more solutions because this equation does not imply that $g$ is $\mathbb{Q}$-linear (meaning that $g(qx) \equiv q \,g(x) \mod 2i\pi$ for $q \in \mathbb{Q}$) as was the case for Cauchy's equation (unless $g$ is continuous). It's possible to give a full description of solutions of this equation on $\mathbb{Q}$ but it's quite involved and uses $p$-adic numbers (see for example this document), suffice it to say it's much bigger than just linear functions ! As for solutions on $\mathbb{R}$, you can obtain them by choosing a Hamel basis and gluing together solutions on $\mathbb{Q}$ as in the classical Cauchy's equation.

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    $\begingroup$ Thanks for your extension. I will upvote as well. $\endgroup$ – Bear and bunny Oct 29 '15 at 3:03

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