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.

  • $\begingroup$ How about $f(x) = 1$? $\endgroup$ – Jair Taylor Apr 3 '18 at 20:10
  • $\begingroup$ o sure I forgot that hehehe but How can I prove that these are the unique functions? $\endgroup$ – Rachel Apr 3 '18 at 20:15
  • $\begingroup$ Well you didn't forget, it is the case lambda=0 $\endgroup$ – Tal-Botvinnik Apr 3 '18 at 20:32
  • $\begingroup$ Moreover, please remove that tag "real-analysis". This question is simple and does not fall into the category of "real-analysis". It should belong to "elementary analysis". (For example, no measure theory, no technique of functional analysis, and no general topology is involved.) $\endgroup$ – Danny Pak-Keung Chan Apr 3 '18 at 21:46
  • 1
    $\begingroup$ $\lambda$ should be non-positive. If $\lambda>0$, then $\exp(\lambda x)>1$ for all $x\in(0,\infty]$. $\endgroup$ – Danny Pak-Keung Chan Apr 3 '18 at 22:06

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. )

  • $\begingroup$ Short Answer: $f(x)=\exp(\lambda x)$ for some $\lambda\in(-\infty,0]$ for $f(x)=0$. $\endgroup$ – Danny Pak-Keung Chan Apr 3 '18 at 22:09
  • $\begingroup$ typo, "for" should be replaced by "or" $\endgroup$ – Danny Pak-Keung Chan Apr 4 '18 at 1:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.