Are the solutions of $f(x+h)=f(x)f(h)$of the form $a^x$ even if we consider discontinuous functions

Question

Let $$f(x):\mathbb{R}\to \mathbb{R} $$$$$$and$$f(x+h)=f(x)f(h)$$

If $f(x)$ is a continuous function then we can prove all solutions for ($f(x)$ not equal to zero at any point) are of the form $a^x$ .(Where $a^x$ is defined using sequences ) by simply using properties of $f(x)$ and continuity

But is the result still true if we also consider $f$ not continuous and ($f(x)$ not equal to zero at any point) or is there a counter-example?

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This function can be constant and the constant must be $1$ but that is the only constant value the function can achieve and since the constant function on $\mathbb{R}\to \mathbb{R}$is continuous the previous result will work .


If we can prove that such a function must be either monotonically increasing everywhere or monotonically decreasing everywhere if it is not $1$

Then we can use the theorem that a function from $\mathbb{R}\to \mathbb{R}$that is monotonically increasing must be continuous somewhere which for this function due to the functional equation would imply the function is continuous everywhere and we would have proved the question using the previous result .

Q: Can the existence of nosuch function be somehow proven by using monotonicity (or some other way )or is there a counter-example?

Answer 1

Clearly, $f(x)$ for some $x$ implies $f(x)=0$ for all $x$. Excluding this case the question can be translated to Cauchy's equation : $g(x+y)=g(x)+g(y)$ by taking logarithm. Here are some facts about $g$: If $g$ is Borel measurable (in particular if it is monotone) then $g(x)=cx$ for some constant $c$. But there exist non-measurable solution so of this equation. [Proof of the existence of such functions requires Axiom of Choice].

Hence $f(x)=a^{x}$ need not be true in general (take $f(x)=e^{g(x)})$.

Answer 2

No. There exist non-vanishing such functions which are not exponential.

First observe that, if $g(x+y)=g(x)+g(y)$, for all $x,y$, and $f(x)=\exp\big(g(x)\big)$, then $f(x+y)=f(x)f(y)$, for all $x,y$.

To obtain an additive $g$, i.e., $g(x+y)=g(x)+g(y)$, for all $x,y$, which is not of the form $g(x)=cx$, we need to use Zorn's Lemma, and in particular, the fact that $\mathbb R$ possesses a Hamel basis, as a linear space over $\mathbb Q$. In other words, there exists a $B\subset \mathbb R$, such that:

Every $x\in\mathbb R$ can be written, in a unique way as a linear combination of elements of $B$ with rational coefficients. That is, for every $x\in\mathbb R$, there exist unique $b_1,\ldots,b_n\in B,$ and $q_1,\ldots,q_n\in\mathbb Q$, such that $$ x=q_1b_1+\cdots+q_nb_n. $$

Now, for every set of $c_i$'s in $\mathbb R$, the function $$ g(x)=q_1q(b_1)+\cdots+q_ng(b_n)=q_1c_1+\cdots+q_nc_n, $$ is additive, i.e., $g(x+y)=g(x)+g(y)$, and the coresponding $f$, i.e., $f(x)=e^{g(x)}$ satisfies $$ g(x+y)=g(x)g(y). $$ If the $c_i$'s are not proportional to the $b_i$'s, then $f$ is not continuous and hence not exponential. For example, say $g(b_i)=c_i$, $i=1,2$, and $b_1c_2-b_2c_1\ne 0$. Say $c_1\ne 0$, and $\{q_n\}\in \mathbb Q$, so that $q_n\to b_1/b_2$. Then $$ f(q_nb_2)=q_nc_2\to b_1c_2/b_2 $$ while $$ q_n b_2\to b_2\cdot b_1/b_2=b_1\quad\text{and}\quad f(b_1)=c_1\ne b_1c_2/b_2. $$