How do I prove that for continuous $f$, if $\forall x \in \mathbb{Q}, \, f(x) = f(1)^x$ then $\forall x \in \mathbb{R}, \, f(x) = f(1)^x$?

Question

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.

Answer 1

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 $g=a^x, a=f(1)$.

Answer 2

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 $Y$.

Answer 3

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 $f=g$.

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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.

Answer 4

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 $x$.