Suppose $f$ is a function satisfying
$$
f(x)f(y)=f(x+y) \label{*}\tag{*} \\
$$
for all $x$ and $y$. If we require that $f$ be continuous, then it can be shown that $f(x)=a^x$ for some constant $a$, or $f=0$. In Michael Spivak's Calculus, it is mentioned that there in fact infinitely many discontinuous functions satisfying $\eqref{*}$. What are the most notable examples?
Discontinuous function satisfying $f(x)f(y)=f(x+y)$
calculuscontinuityexponential functionreal-analysis
Related Solutions
Given a homeomorphism $\varphi$ of $\Bbb R_+=[0,\infty)$ (which is necessarily increasing and $\varphi(0)=0$), for each $t\in [0,\infty)$ put $\underline{\varphi}(t)=\inf_{s>0} \varphi(st)/ \varphi(s)$ and $\overline{\varphi}(t)=\sup_{s>0} \varphi(st)/ \varphi(s)$. In particular, for each $t\in\Bbb R_+$ we have $$\underline{\varphi}(t)\le \frac{{\varphi}(t)}{\varphi(1)}\le \overline{\varphi}(t).$$ Remark that it is possible that $\overline{\varphi}(t)=\infty$ for some $t>0$. Since the map $\varphi$ is increasing and $\varphi(0)=0$, the maps $\underline{\varphi}$ and $\overline{\varphi}$ are non-decreasing and $\underline{\varphi}(0)= \overline{\varphi}(0)=0$. Also we have $\underline{\varphi}(1)= \overline{\varphi}(1)=1$. Moreover, it is easy to show the following Multiplicative Inequality: for each $t,t’>0$ we have $\underline{\varphi}(tt’)\ge \underline{\varphi}(t) \underline{\varphi}(t’)$ and $\overline{\varphi}(tt’)\le \overline{\varphi}(t) \overline{\varphi}(t’)$.
If $\psi_1$ and $\psi_2$ are functions from $\Bbb R_+$ to $\Bbb R_+$ satisfying F1 then $\psi_1(t)\le \underline{\varphi}(t)$ and $\psi_2(t)\ge \overline{\varphi}(t)$ for each $t>0$.
So if such a function $\psi_2$ exists then the function $\overline{\varphi}$ is finite, that is $\overline{\varphi}(t)<\infty$ for each $t>0$. Multiplicative Inequality implies that the function $\overline{\varphi}$ is finite iff there exists $\overline{t}_+>1$ such that $\overline{\varphi}(\overline{t}_+)$ is bounded.
Nevertheless, if $\varphi(s)=\ln (1+s)$ for each $s\in\Bbb R_+$ then there is no homeomorphism $\psi_2$ of $\Bbb R_+$, satisfying F1, because $\overline{\varphi}(t)=\max\{1,t\}$ for each $t>0$. Indeed, $\varphi(st)/ \varphi(s)=\log_{1+s} (1+st)$ for each $s>0$.
If $t\ge 1$ then $(1+s)^t\ge 1+st$ by Bernoulli's inequality, so $\log_{1+s} (1+st)\le t$. On the other hand, by L'Hôpital's rule, $$\lim_{s\to +0} \log_{1+s} (1+st)= \lim_{s\to +0} \frac {\ln (1+st)}{\ln(1+s)}= \lim_{s\to +0} \frac {\frac{t}{1+st}}{\frac 1{1+s}}=t .$$
If $0<t\le 1$ then $1+s\ge 1+st$, so $\log_{1+s} (1+st)\le 1$. On the other hand, $$\frac {\ln (1+st)}{\ln(1+s)}\ge \frac {\ln (t+st)}{\ln(1+s)}= \frac {\ln t+\ln s}{\ln(1+s)},$$ and the last value tends to $1$ when $s$ tends to infinity.
So in order to provide that there exists a homeomorphism $\psi_2$ of $\Bbb R_+$, satisfying F1, we need an addtional condition. A necessarily condition is there exists $0<\overline{t}_-<1$ such that $\overline{\varphi}(\overline{t}_-)<1$.
We claim that if the points $\overline{t}_-$ and $\overline{t}_+$ exist then there exists such a homeomorphism $\psi_2$. Indeed, by Multiplicative Inequality for each non-negative integer $n$ we have $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-)^n$ if $t\le \overline{t}_-^n$ and $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_+)^n$, if $t\le \overline{t}_+^n$. In particular, $\overline{\varphi}(0)=0$. So we can put $\psi_2(0)=0$ and for each $t>0,$
$$\psi_2(t)=\max\left\{\overline{\varphi}(\overline{t}_+)^{1+\tfrac{\ln t}{\ln \overline{t}_+}},\overline{\varphi}(\overline{t}_-)^{\tfrac {\ln t}{\ln \overline{t}_-}-1}\right\}.$$
Indeed, it is easy to check that $\psi_2(0)=0$, $\lim_{t\to\infty} \psi_2(t)=+\infty$ and a function $\psi_2$ is continuous and increasing. So it is open that is it is a homeomorphism of $\Bbb R_+$, that is there exists a continuous inverse bijection $\psi_2^{-1}:\Bbb R_+\to \Bbb R_+$.
Moreover,
$$\psi_2(1)=\max\left\{\overline{\varphi}(\overline{t}_+),\overline{\varphi}(\overline{t}_-)^{-1}\right\}\ge 1=\overline{\varphi}(1).$$
If $t>1$ then let $n$ be a positive integer such that $\overline{t}_+^{n-1}<t\le \overline{t}_+^n$. Then $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_+)^n\le \psi_2(t)$, because $1+\log_{\overline{t}_+} t\ge n$.
If $t<1$ then let $n$ be a positive integer such that $\overline{t}_-^{n-1}>t\ge \overline{t}_-^n$. Then $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-)^{n-1}\le \psi_2(t)$, because $\log_{\overline{t}_-} t\le n$.
Also the following properties of the function $\overline{\varphi}$ can be useful. Since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality for each natural number $n$ we have
$$\overline{\varphi}(0)\le \lim_{t\to 0+} \overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-^n)\le \overline{\varphi}(\overline{t}_-)^n\to 0,$$ and
$$\lim_{t\to\infty} \overline{\varphi}(t)\ge \overline{\varphi}(1)/\overline{\varphi}(\overline{t}_-^n) \to+\infty.$$
So $\overline{\varphi}(0)=0$ and $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
We claim that the function $\overline{\varphi}$ is increasing. Indeed, suppose to the contrary that there exist real numbers $t<t’$ such that $\overline{\varphi}(t)=\overline{\varphi}(t’)$.
If $t=0$ then for each $t^*>0$ there exists a natural number $n$ such that $t^*\le t\overline{t}_+^n$ so, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality $$\overline{\varphi}(t^*)\le \overline{\varphi}(t) \overline{\varphi}(\overline{t}_+^n)=0,$$ a contradiction with $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
If $t>0$ then by the above $\overline{\varphi}(t)>0$ and, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality
$$1=\overline{\varphi}(1)\le \overline{\varphi}(t’/t)\le \overline{\varphi}(t’)/ \overline{\varphi}(t)=1.$$
So $\overline{\varphi}(t’/t)=1$. Then for each $t^*>0$ there exists a natural number $n$ such that $t^*\le (t’/t)^n$ so, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality $$\overline{\varphi}(t^*)\le \overline{\varphi}(1) \overline{\varphi}((t’/t)^n)=1,$$ a contradiction with $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
The $\psi_1$ case is considered similarly. In order to provide that there exists a homeomorphism $\psi_1$ of $\Bbb R_+$, satisfying F1, we need that $\lim_{t\to+\infty} \underline{\varphi}(t)=+\infty$. Multiplicative Inequality implies that this condition holds iff there exists $\underline{t}_+>1$ such that $\underline{\varphi}(\underline{t}_+)>1$. This condition is not automatic. Indeed, if again $\varphi(s)= ln(1+s)$ for each $s\in\Bbb R_+$ then $\varphi(st)/ \varphi(s)=\log_{1+s} (1+st)$ for each $s>0$. If $s\ge t$ then $(1+s)^2=1+2s+s^2>1+s^2\ge 1+st$, so $\underline{\varphi}(t)\le 2$ for each $t$.
An other necessary condition to provide that there exists a homeomorphism $\psi_1$ of $\Bbb R_+$, satisfying F1, is: there exists $0<\underline{t}_-<1$ such that $\underline{\varphi}(\underline{t}_-)>0$. This condition is not automatic too, as shows the following example. Let $\varphi(s)=e^s-1$ for each $s\in\Bbb R_+$. Then for each $s,t>0$, $\varphi(st)/ \varphi(s)=\tfrac{e^{st}-1}{e^s-1}$. If $t<1$ then $\underline{\varphi}(t)=0$ because $$\lim_{s\to +\infty}\frac{e^{st}-1}{e^s-1}=\lim_{s\to +\infty} \frac{e^{s{t-1}}-e^{-s}}{1-e^{-s}}=0.$$
Now assume that $t\ge 1$ and $s>0$. Put $x=e^s-1$. By Bernoulli's inequality, $(1+x)^t\ge 1+xt$. Thus $$\frac{e^{st}-1}{e^s-1}= \frac{(1+x)^t-1}{x}\ge t.$$ Then $\underline{\varphi}(t)=t$ because by L'Hôpital's rule $$\lim_{s\to +0} \frac{e^{st}-1}{e^s-1}= \lim_{s\to +0} \frac{te^{s}}{e^s}=t.$$
We claim that if the points $\underline{t}_-$ and $\underline{t}_+$ exist then there exists such a homeomorphism $\psi_1$. Indeed, pick arbitrary numbers $\alpha_+,\alpha_-, C>0$ such that $C\underline{t}_+\le 1$, $\underline{t}_+^{2\alpha_+}\le\underline{\varphi}(\underline{t}_+)$, $C\le \underline{\varphi}(\underline{t}_-)$, and $\underline{t}_-^{\alpha_-}\le \underline{\varphi}(\underline{t}_-)$. It is easy to check that $C\underline{t}_+^{\alpha_+(n+1)}\le \underline{\varphi}(\underline{t}_+)^n$ and $C\underline{t}_-^{\alpha_- n}\le \underline{\varphi}(\underline{t}_-)^{n+1}$ for each non-negative integer $n$.
Put $\psi_1(0)=0$, and $\psi_1(t)=Ct^{\alpha_-}$ for any $0\le t\le 1$, and $\psi_1(t)=Ct^{\alpha_+}$ for any $t\ge 1$. Iit is easy to check that $\psi_1(0)=0$, $\lim_{t\to\infty} \psi_1(t)=+\infty$ and a function $\psi_1$ is continuous and increasing. So it is open that is it is a homeomorphism of $\Bbb R_+$, that is there exists a continuous inverse bijection $\psi_1^{-1}:\Bbb R_+\to \Bbb R_+$.
If $t\ge 1$ then let $n$ be the largest integer such that $t\ge \underline{t}_+^n$. Then $t<\underline{t}_+^{n+1}$ so
$$\psi_1(t)\le \psi_1(\underline{t}_+^{n+1})=C\underline{t}_+^{\alpha_+(n+1)}\le \underline{\varphi}(\underline{t}_+)^n\le \underline{\varphi}(\underline{t}_+^n)\le \underline{\varphi}(t).$$
If $t\le 1$ then let $n$ be the largest integer such that $t\le \underline{t}_-^n$. Then $t>\underline{t}_-^{n+1}$ so
$$\psi_1(t)\le \psi_1(\underline{t}_-^n)=C\underline{t}_-^{\alpha_- n}\le \underline{\varphi}(\underline{t}_-)^{n+1}\le \underline{\varphi}(\underline{t}_-^{n+1})\le \underline{\varphi}(t).$$
Best Answer
Since there exists a discontinuous function satisfying the Cauchy functional equation $g(x)$ ,$e^{g(x)}$ will satisfy $f(x+y)=f(x)f(y)$ and $k(x)=e^{g(x)}$ must be discontinuous since composition of continuous functions are always continuous but $\ln(k(x))=g(x)$ which is discontinuous. Since there are infinitely many discontinuous functions satisfying the Cauchy functional equation there are infinitely many discontinuous functions satisfying your equation .