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Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function. I have two questions as follows.

  1. Suppose that $f(x+n)\to L$ for all $x\in \mathbb{R}$ when $n\to \infty$. Is it guaranteed that $f(x) \to L$ when $x\to +\infty$?

2. Suppose that $f(x+n)$ converges for all $x\in \mathbb{R}$ when $n \to \infty$. Is it guaranteed that $f(x)$ converges when $x \to +\infty$? (As pointed out by @Ninad Munshi, the answer is obviously negative by considering a simple example with $f(x) = \sin(2\pi x)$. Thank @Ninad Munshi.)

Remarks:

  1. Here we may require that the limit of any convergence must be a finite number. However, what if we consider a generalized situation that the limit is allowed to be infinity?

  2. The question is an additive analogs of $f(nx)\to 0$ as $n\to+\infty$ , If $\,\lim_{n\to\infty}f(nx)\,$ exists, for all $x\in\mathbb R$, then so does $\,\lim_{x\to\infty}f(x)\,$ , and If $\lim_{n \to +\infty} \ f(\alpha n)$ exists , does this imply $\lim_{x \to +\infty} f(x)$ exists? , which can be solved using Baire's category theorem.

Any comments or criticism will be appreciated. Thank you.

Nuno
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1 Answers1

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Let $f_n(x)$ denote the function whose graph is an isosceles triangle with base $[0,1/n]$ and height $1$. Specifically: $$f_n(x) = \begin{cases} 2nx & \text{if }0 \leq x \leq \frac{1}{2n} \\ 2 - 2nx & \text{if }\frac{1}{2n} \leq x \leq \frac{1}{n} \\ 0 & \text{otherwise} \end{cases}$$ and let $$f(x) = \sum_{n=1}^{\infty}f_n(x-n)$$ So, the graph of $f$ consists of a series of increasingly narrow triangles, where the $n$'th triangle has base $[n, n+1/n]$ and height $1$. It's straightforward to verify that $$\lim_{n \to \infty}f(x+n) = 0$$ for all $x$, but $\lim_{x \to \infty}f(x)$ does not exist.