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If $f: (0,\infty)\rightarrow\mathbb{R}^{1}$ is continuous, and $\lim\limits_{n\rightarrow\infty}f(2^{n}x)=0.$ for all $x>0$. I want to prove $\lim\limits_{x\rightarrow\infty}f(x)=0$, but I have no idea.

This looks similar to the problem A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem. But when I replace $n$ with $2^{n}$, I cannot prove it. The problem is that there is no $c>0$, such that $$(c,\infty)\subset\bigcup_{n=N}^{\infty}(2^{n}a,2^{n}b).$$

I have also doubted the validity of this conclusion, and I have tried to find a counterexample, but I have not found one.I want to know if this conclusion is true, and if not, is there a counterexample? Thank you.

Sillyasker
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lkj123
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  • This answer of the problem is similar to yours, maybe it can help you. – Zhiwei Dec 13 '23 at 14:18
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    Just a tip for a general method of how to show things if you have no idea of how to start: 1. List the properties that you have. In this case continuity and the first limit property. 2. Try to disprove what you should prove. In this case try to create a continuous function with the first limit property, which also satisfies the second limit. 3. You will run into problems. Use the problems to show what you needed. (In other words, proof by contradiction). – Aksel Bergfeldt Dec 13 '23 at 14:37
  • See https://math.stackexchange.com/q/4024033/42969 – Martin R Dec 13 '23 at 15:14

2 Answers2

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Define $g(x) = f(\exp(x))$, thus $f(x) = g(\ln x)$ and $f(2^n x) = g(n + \ln x)$. $g$ is continuous, condition become $\forall x: \lim\limits_{n\to\infty} g(n + x) = 0$, and for counterexample we need $\neg\lim\limits_{x \to \infty} g(x)= 0$.

Now, it's easy to see how to build a counterexample: let $w(x)$ be bump function: $w(0) = 1$, $w(x) = 0$ if $|x| \geq 1$, $w$ is continuous. And define $g(x) = \sum_n w((x - n + 1/n) \cdot n^3)$ - narrow bumps around $n - 1/n$, that do not intersect when shifted by integer.

It's clear that $\lim\limits_{n \to \infty} g(x)$ doesn't exist. However, every $x$ can get into at most one bump when shifted, so $g$ satisfies our condition.

mihaild
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Let $g_n(x)=x^n(1-x)n$. Note that for positive integer $n$, $g(n,0)=g(n,1)=0$ and it is continuous on $[0,1]$.

Let $f(x)$ be $0$ for $x<2$, and for positive integer $n$, for $x\in [2^n,2^{n+1}]$, let $f(x)=g_n(\log_2{(x/2^n)})$.

Since $g_n(x)<x^n * n$, for fixed $x<1$, this goes to 0 as $n$ gets big. At $1$, it’s already $0$.

As $n$ gets big $g_n(1-1/n)$ goes to $1/e$, so $f$ does not approach $0$.

Eric
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  • I think it's taking the idea that "convergence everywhere doesn't lead to uniform convergence", is it? That's a good idea, and I haven't tried to think of them together before, thank you so much. – lkj123 Dec 13 '23 at 15:16