$f(x+y)=f(x) \times f(y)$, $f$ taking values in $[0,1]$

Question

Is it true that the unique functions that are right-continuous, $0\leq f(x)\leq 1$ (taking values in [0,1]), $\forall x \geq 0$ and $$ f(x+y)=f(x) \times f(y), \ \forall x,y \geq 0$$

are $f(x)=0$,$f(x)=1$ and $f(x)=e^{\lambda x}$, $\forall x \geq 0, \lambda \geq 0$ ?

I´m trying to find all the functions that satisfies these properties, right-continuous and transforming sums into products are the exponential and the zero functions.

Is it correct?

Could someone help me to prove this or give me some hints, pls. Thanks for your time and help.

Answer 1

I assume that the domain of $f$ is $[0,\infty)$.

Let $a=f(1)$. Let $n\in\mathbb{N}$. Note that $a=f(1)=f(\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n})=\left[f(\frac{1}{n})\right]^{n}$ ($\frac{1}{n}$ appears $n$ times), so $f\left(\frac{1}{n}\right)=a^{\frac{1}{n}}$. For any $r\in\mathbb{Q}\cap(0,\infty)$, choose $m,n\in\mathbb{N}$ such that $r=\frac{m}{n}$, then $f(r)=f(\frac{1}{n}+\cdots+\frac{1}{n})=\left[f(\frac{1}{n})\right]^{m}=a^{\frac{m}{n}}=a^{r}$ ($\frac{1}{n}$ appears $m$ times). Lastly, by right continuity of $f$ and the fact that $\mathbb{Q}\cap(0,\infty)$ is dense in $[0,\infty)$ with respect to the right-hand topology (i.e., for each $x\in[0,\infty)$, there exists a sequence $(r_{n})$ in $\mathbb{Q}\cap(0,\infty)$ such that $r_{1}\geq r_{2}\geq\ldots$ and $r_{n}\rightarrow x$) we have $f(x)=a^{x}$ for any $x\in[0,\infty)$. (Here, we need to distinguish the cases $a=0$ and $a\in(0,1].$ If $a=0$, then $f(r)=0$ for all $r\in\mathbb{Q}\cap(0,\infty)$ and hence $f(x)=0$ for all $x\in[0,\infty)$. If $a\in(0,1]$, then $f(x)=a^{x}$. We make such a distinction because $0^{0}$ is undefined. By writing $f(x)=a^x$, it is understood that $f(x)=0$ for all $x\in[0,\infty)$ if $a=0$, by convention. )

Answer 2

The following problem may be helpful for you.

If $f:R\to R$ is continuous an satisfies the condition $f(x+y)=f(x)f(y)$ then $f$ is the zero function or $f$ is of the form $f(x)=[f(1)]^x$. or equivalently $f(x)=e^{\lambda x}$, where $\lambda=Ln[f(1)]$. Thus I think that $\lambda$ must be non-positive, since $0<f(1)<=1$.

The foregoin problem contains a stronger hipothesis, namely, the continuity (right and left). So, try to show that if $f$ is right-continuous then $f$ is left-continous.

To solve the given problem you can show first that $f$ agrees with $f[1]^x$ for all rational number $x$. Then, use the fact that If two continous functions agrees at a dense subset then they are igual on that set.