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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!

3 Answers 3

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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 $\epsilon$?

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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 $A$.

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    Very nice answer. +1 Though I'd say that there are two sequences $\;x_n,\,y_n\to\infty\;$ s.t. $\;\lim f(x_n)\neq\lim f(y_n)\implies\;$ contradiction to the fact that $\;\lim\limits_{x\to\infty}f(x)\;$ **exists** – 2017-02-08
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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 ?