If $f$ is continuous and bounded, there exists $x_n\to\infty$, with $\,f(T+x_n)-f(x_n)\to 0$

Question

Let $f$ be continuous and bounded in $(x_0, \infty)$.

Prove that for every $T>0$, there exists a sequence $\{x_n\}$, such that $x_n\to\infty\,$ and $f\,(T+x_n)-f(x_n)\to 0$, as $n\to\infty$

Attempt I tried to reason by contradiction and consider $x_{n+1}=x_n+T$. Want to prove that if we choose the starting point appropriately, such $x_n$ will converge to a limit. You do not need to continue my line of reasoning though

Answer 1

Consider the sequence $$ a_n=f(nT). $$ If $$b_n=a_n-a_{n-1}\to 0,$$ we are done. Also, if $\{b_n\}$ contains a subsequence tending to zero, we are also done.

Assume that there is no subsequence of $\{b_n\}$ tending to zero. This implies that there exists an $\varepsilon>0$ and an $n_0\in\mathbb N$, such that $$ |b_n|\ge \varepsilon, \quad \text{for all $n\ge n_0$.} $$ Clearly, this implies that there exist infinite $n$'s for which $b_n\ge \varepsilon$ and infinite $n$'s for which $b_n\le -\varepsilon$. Otherwise, we would have that $a_n\to \infty$ or $a_n\to -\infty$.

Therefore, there exist infinitely many $n$'s, such that $$ a_n-a_{n-1}=b_n\ge \varepsilon\quad\text{and}\quad a_{n+1}-a_n =b_{n+1}\le -\varepsilon. $$ or equivalently $$ f(nT)-\varepsilon\ge f\big((n-1)T),\,f\big((n+1)T) $$ Set $g(x)=f(x+T)-f(x)$. Then $$ g\big((n-1)T\big)=f(nT)-f\big((n-1)T\big)\ge\varepsilon\qquad\text{while}\qquad g(nT)=f\big((n+1)T)-f(nT)\le -\varepsilon $$ and hence IVT implies that there exists a $x_n\in\big((n-1)T,nT\big)$, such that $$ 0=g(x_n)=f(x_n+T)-f(x_n). $$

Answer 2

Hint: If false, there exists $T\in \mathbb R$ such that

$$\liminf_{x\to \infty} |f(T+x)-f(x)| = \epsilon >0.$$

This implies there exists $a>0$ such that

$$|f(T+x)-f(x)| > \epsilon/2$$

for $x>a.$ Use the continuity of $f$ to see one of the following must hold for $x>a$:

$$i) \, f(T+x)-f(x) > \epsilon/2,;\,\,\,\,ii)\, f(T+x)-f(x) < -\epsilon/2 .$$

Suppose $i)$ holds and see where that leads.