Rigorous Proof of a Limit, where f(x) ≡ A. Suppose that $f$(x) is a function on (0, +∞), satisfying that
$f$(2x) = $f$(x). If lim x→∞ $f$(x) = A, prove that
$f$(x) ≡ A for all x ∈ (0, +∞).
Hi, I am trying to understand this question but I have no idea how to approach this. I also don't fully understand why the 'identical to' sign is there rather than the equal sign. Any help would be appreciated!
 A: Read the condition on $f$ as $f(x/2) = f(x)$. Given $\epsilon > 0$ there exists $M$ with the property that $|f(x) - A| < \epsilon$ for all $x \in [M,\infty)$. The stated hypothesis means that $|f(x) - A| < \epsilon$ for all $x \in [M/2,\infty)$, and then $[M/4,\infty)$, and so on until you conclude that $|f(x) - A| < \epsilon$ for all $x \in (0,\infty)$. 
Now what about $\epsilon$?
A: Assume $\exists x\ge0, f(x)\neq f(1)$.
Then the sequence $(x_k)_k$ such that $x_k=2^k x$ satisfies
$$
\lim_{k\rightarrow \infty} \; f(x_k)=f(x)\neq f(1)
$$
Moreover $\lim_{k\rightarrow \infty} \; f(2^k)=f(1)$
Therefore we find two subsequences which doesn't converge to the same limit which contradicts the definition of $A$.
A: Hints:


*

*$\forall n\in\mathbb Z, f(2^n)=f(1)$

*if $f(x)\to A$ in $+\infty$ then any growing subsequence $(x_n)_n$ also has this property

*so what about $f(1)$ ?

*similarly for $x_0\in\mathbb R^{+*}, f(2^nx_0)=f(x_0)$

*conclude

*what about $f(0)$ ? can we say something about it or do we need continuity ?

