"Change of variables" in $\limsup_{x\rightarrow\infty}f(x+t)$

Question

Let $f:(0,\infty)\rightarrow \mathbb{R}$ be a real-valued function, such that for some $t,C>0$, \begin{equation} \limsup_{x\rightarrow\infty} f(x+t)\leq C \end{equation} Is it also true that $\limsup_{x\rightarrow\infty} f(x)\leq C$? Particularly, is it true that $\limsup_{x\rightarrow\infty} f(x+t)=\limsup_{x\rightarrow\infty} f(x)$?

Answer 1

We recall that

$$ \limsup_{x \to \infty} f(x) = L $$

if and only if the following two conditions hold:

1. For every sequence $x_n \to \infty$, we have $\limsup_{n \to \infty} f(x_n) \leq L$.
2. There exists a sequence $x_n \to \infty$ such that $\lim_{n \to \infty} f(x_n) = \limsup_{n \to \infty} f(x_n)= L$.

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We first show that $\limsup_{x \to \infty} f(x) \leq \limsup_{x \to \infty} f(x+t)$.

Fix a sequence $x_n \to \infty$ satisfying condition (2) such that $\lim_{n \to \infty} f(x_n) = \limsup_{x \to \infty} f(x)$, and let $y_n = x_n - t$. Then $\lim_{n \to \infty} f(y_n + t) = \lim_{n \to \infty} f(x_n)$, so by condition (1), the inequality $$ \limsup_{x \to \infty} f(x) \leq \limsup_{x \to \infty} f(x+t) \tag{a} $$ holds.

We next show that $\limsup_{x \to \infty} f(x+t) \leq \limsup_{x \to \infty} f(x)$.

Fix a sequence $x_n \to \infty$ satisfying condition (2) such that $\lim_{n \to \infty} f(x_n + t) = \limsup_{x \to \infty} f(x+t)$, and let $y_n = x_n + t$. Then $\lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} f(x_n+t)$, so by condition (1), the inequality $$ \limsup_{x \to \infty} f(x+t) \leq \limsup_{x \to \infty} f(x) \tag{b} $$ holds.

Combining (a) and (b) gives the desired equality.

Answer 2

This is easy to prove straight from the definition of $\limsup$. By definition, $$\limsup_{x \to \infty}f(x + t) = \lim_{x \to \infty}\sup_{y > x}f(y + t).$$ Thus $$\limsup_{x \to \infty}f(x + t) = \lim_{x \to \infty}\sup_{y > x + t}f(y).$$ Now by a standard $\varepsilon$ argument for $\lim$ (probably what you call "change of variables"), we get $$\lim_{x \to \infty}\sup_{y > x + t}f(y) = \lim_{z \to \infty}\sup_{y > z}f(y),$$ and this is exactly $\limsup_{z \to \infty}f(z)$.