Given the differentiable/continuous real-valued function $f(x)f(y) = f(x + y)$ I got so far as to show that $\forall x \in \mathbb{Q}, \, f(x) = f(1)^x$.

I am trying to show that because $f$ is continuous, and because rationals are dense in the reals, then $\forall x \in \mathbb{R}, \, f(x) = f(1)^x$ too.

Continuity here being $\forall a \in \mathbb{R}, \, \lim_{x \to a} f(x) = f(a)$ I believe.

Rationals dense in the reals: $\forall r, \epsilon \in \mathbb{R}, \exists q \in \mathbb{Q} : \, (|q - r| < \epsilon)$

Vaguely I understand that I need to show the following: If, for any real there exists a rational arbitrarily close to it, we can formulate a sequence of rationals approaching some real, and I think this behaves like a limit. And due to continuity, since $f(x)$ exists at each rational, the definition says it also exists for $f(a)$ where $a$ is real?

I'm not 100% sure if that's the right idea or if that's what's being demonstrated (what technically tells us that $f(a)$ exists for all $a \in \mathbb{R}$?), but I would appreciate help with the formal representation of how to show this.

  • 1
    $\begingroup$ Just take a sequence of rationals $x_n \to x$. By continuity of $f$, we have $$1=\lim_{n \to \infty} \frac{f(x_n)}{f(1)^{x_n}} = \frac{f(x)}{f(1)^x}.$$ $\endgroup$ – Dzoooks Jun 13 '19 at 2:03

You need to get a few things straight here:

  • There should be a working definition of symbol $a^b$ where $a>0,b$ are real. Moreover under this definition you should have established that the function $g$ defined by $g(x) =a^x$ is continuous on $\mathbb{R} $. This part is non-trivial.
  • $f(1)>0$.

Using these two facts one can show that $f(x) =f(1)^ x$ for all $x\in\mathbb {R} $. And as per your question one only needs to consider the case when $x$ is irrational.

Let $x_n$ be any sequence of rationals such that $x_n\to x$. Then we have via continuity of $f$ $$f(x) =\lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} f(1)^{x_n}=f(1)^x$$ where the last equality is a consequence of continuity of $g=a^x, a=f(1)$.

  • $\begingroup$ I am not quite sure what you mean by that first bullet point. Second bullet point, I was able to show that $f(x) > 0$ I believe (my proof attempt is here math.stackexchange.com/questions/3259229/…) $\endgroup$ – user681336 Jun 13 '19 at 5:37
  • $\begingroup$ @user681336: you need to know the meaning of $f(1)^x$ for irrational $x$ and only then you can think of proving $f(x) = f(1)^x$. The meaning of an irrational exponent is not available with algebra and has to be provided using careful analysis. There are many ways to do so some of which are described in my blog posts: paramanands.blogspot.com/2014/05/… $\endgroup$ – Paramanand Singh Jun 13 '19 at 11:15

Just a share of thought: Don't take it as an answer. I am writing it here because the comment section is a small area.

Given that $f(x)=f(1)^x$ for all $x\in \Bbb Q$, so $f(1)$ exists finitely as $1\in \Bbb Q$. Since $\Bbb Q$ is dense in $\Bbb R$ so, for any $r\in \Bbb R$ there is a sequence $\{x_n : n \in \Bbb N\}$ in $\Bbb Q$ such that $x_n\to r$ as $n\to \infty$. Given that $f$ is continuous so, $f(x_n)\to f(r)$ as $n\to\infty$ but $f(x_n)=f(1)^{x_n}\to f(1)^r$. Due to uniqueness of limits (as $\Bbb R$ is Hausdroff) $f(r)=f(1)^r$, and this is true for all $r\in\Bbb R$.

Now the answer to your question "what technically tells that $f(r)$ exists for all $r\in\Bbb R$" is the Dense-ness of $\Bbb Q$ in $\Bbb R$. It can be shown that, if $(X,\tau)$ and $(Y,\sigma)$ are $T_2$ spaces and $D\subset X$ is dense in $X$, $E\subset Y$ is dense in $Y$ then, any continuous map from $D$ to $E$ can be extended continuously to a map from $X$ to $Y$.


There is a more general theorem at play here (not too abstract though). (By the way first note that in your definition of denseness, you have to specify that $\epsilon > 0$.)

Theorem: (I remember doing this as an exercise in Spivak's Calculus)

Let $A$ be a dense subset of $\Bbb{R}$. Let $f,g: \Bbb{R} \to \Bbb{R}$ be given continuous functions. If for every $x\in A$ we have $f(x) = g(x)$, then it follows that $f=g$ (on all of $\Bbb{R}$).

So, in words this says that if two continuous functions agree on a dense set, then they agree everywhere

To prove this, consider instead the difference $h = f-g$. Our hypothesis now says that $h|_A = 0$, so to prove it is zero everywhere, pick any $a \in \Bbb{R}\setminus A$, and let $\epsilon > 0$ be arbitrary. Since by assumption $h$ is continuous at $a$, there is a $\delta > 0$ such that for all $x \in (a-\delta, a+ \delta)$, we have $|h(a) -h(x)| < \epsilon$.

Since $A$ is dense in $R$, we can find a $\xi \in A$ such that $\xi \in (a- \delta, a+\delta)$. For this particular $\xi$, we have that \begin{align} |h(\xi) - h(a)| = |0-h(a)| = |h(a)| < \epsilon \end{align} Since $\epsilon > 0$ was arbitrary, it follows that $h(a) = 0$. Since $a$ was arbitrary, it follows that $h=0$, and thus $f=g$.

Now, if you assume that the function $g: \Bbb{R} \to \Bbb{R}$ defined by $g(x) = [f(1)]^x$ is continuous (from what I remember continuity isnt easy to show, because you have to first define properties of exponential function etc), then the function $f$ satisfying the properties listed in your question and the function $g$ agree on the dense set $\Bbb{Q}$. Hence, by the theorem above, they agree on $\Bbb{R}$, which is what you wanted to show.


Observe that $f$ is never negative because $f(x) = f\left( \dfrac x2 \right)^2$ for all $x$.

If $f(x_0) = 0$ for some $x_0$, then $0 = f(x_0)f(x-x_0) = f(x)$ so that $f(x) = 0$ for all $x$.

Unless $f$ is identically zero you can assume that $f(x) > 0$ for all $x$. Now take the logarithm: if $x$ is rational then

$$\log f(x) = x \log f(1).$$

Define $g(x) = e^{x \log f(1)}$ for arbitrary real $x$. Then since $f$ is continuous, $g$ is continuous, and $f(x) = g(x)$ for all rational $x$ you obtain $f(x) = g(x) = f(1)^x$ for all real $x$.


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