Proving a function is constant in $\mathbb{R}$ if $f(x)=f(2x)$ and f is continuous at zero

Question

$f$ is defined as a function on $\mathbb{R}$ which is continuous in 0 and $\forall x \in \mathbb{R}:f(x) = f(2x)$. I do have to proof that $f$ is constant in $\mathbb{R}$.

I figured that this translates to:

$\forall x:f(x) = f(2x) => \forall x \forall y : f(x) = f(y)$

I now want to proof this by contradiction (A = $\neg \neg A$)

$\neg(\forall x:f(x) = f(2x) => \forall x \forall y : f(x) = f(y))$

$\neg (\neg \forall x:f(x) = f(2x) \vee \forall x \forall y : f(x) = f(y)) $

$\forall x : f(x) = f(2x) \wedge \exists x \exists y : f(x) \neq f(y)$

But this does not seem to be sufficient as there could still be a function fulfilling those properties (even though I cannot name one). What am I missing, as I know that such a function cannot exist, but I cannot find any proof within those terms.

Answer 1

Consider any $x \in \mathbb{R}$. Consider the sequence $x_n = \dfrac{x}{2^n}$.

Clearly, we have $\displaystyle \lim_{n \rightarrow \infty} x_n = 0$.

Now, since we have $f(a) = f(2a)$, $\forall a \in \mathbb{R}$, using this, it is easy to prove by induction that $$f(x) = f\left(\dfrac{x}{2^n} \right)$$ $\forall x \in \mathbb{R}$ and $\forall n \in \mathbb{N}$

Now recall that every continuous function is sequentially continuous i.e. $$\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f \left(\lim_{n \rightarrow \infty} x_n \right) $$

Using the above arguments, we get that $\forall x \in \mathbb{R}$, $$f(x) = \displaystyle \lim_{n \rightarrow \infty} f(x) = \displaystyle \lim_{n \rightarrow \infty} f\left(\dfrac{x}{2^n} \right) = f\left(\displaystyle \lim_{n \rightarrow \infty} \dfrac{x}{2^n} \right) = f(0)$$

Hence, $f(x) = f(0)$, $\forall x \in \mathbb{R}$. Hence, $f(x)$ is a constant.

Answer 2

Put $f(0) = a$, and let $\epsilon > 0$. Choose $\delta > 0$ so that $|t|<\delta \implies $|f(t) -a| < \epsilon$.

Since $f(2t) = f(t)$ we have $f(2^n t) = f(t)$ for all $n \ge 0$. Choose $x > 0$ and $n$ so $\delta> x/2^n$. Then by our periodicity condition, $|f(x) - a| < \epsilon$.
We can do the same thing if $x < 0$ so $|f(x) -a | < \epsilon$ for all $x$. Since $\epsilon$ was chosen arbitrarily, $f(x) = a$ on the entire line.

Answer 3

Let $f$ be such a function. Pick a real number $x \in \mathbb R$. Consider the sequence $x_n=x/2^{n}$ then $f(x)=f(x_n)$ by assumption. Furthermore $x_n \rightarrow 0$. Since $f$ is continuous at $0$ we may conclude $\lim_{n\rightarrow \infty} f(x_n)=f(0)$. In particular we have that $f(x)=f(0)$ for every $x$ so $f$ is constant.