Is there a sequence $(t_n)_{n\ge 0}$ such that $\lim_n t_n = \infty$, $\lim_n \Delta_n = 0$ and $\sup_n S_n < \infty$?

Question

Let $(t_n)_{n\ge 0}$ be a strictly increasing sequence of non-negative real numbers such that $t_0 =0$. We define induced sequences $(S_n)_{n\ge 1}$ and $(\Delta_n)_{n\ge 1}$ by $$ \begin{align} S_n &:= \sum_{i=0}^{n-1} \frac{t_{i+1} - t_i}{t_n - t_i}, \\ \Delta_n &:= \sup_{i \in \{ 0, \ldots,n-1\}} \frac{t_{i+1}-t_i}{t_n}. \end{align} $$

In this thread, I got that there is a sequence $(t_n)_{n\ge 0}$ such that $\lim_n t_n = \infty$ and $\sup_n S_n < \infty$. Now I would like to ask if there is a sequence $(t_n)_{n\ge 0}$ such that $\lim_n t_n = \infty$, $\lim_n \Delta_n = 0$ and $\sup_n S_n < \infty$.

Thank you so much for your elaboration!

Answer 1

The answer is indeed no.

Fix $m\in \mathbb N$

\begin{align} S_n &= \sum_{k=0}^{n-1} \frac1{1 - \frac{t_k}{t_n}}\left(\frac{t_{k+1}}{t_n} - \frac{t_{k}}{t_n}\right)\\ &\ge \sum_{k=0}^{n-1}\frac{1-\left(\frac{t_k}{t_n}\right)^m}{1 - \frac{t_k}{t_n}}\left(\frac{t_{k+1}}{t_n} - \frac{t_{k}}{t_n}\right) \end{align}

This proves that $$\sup_n S_n \ge \lim_{n\to \infty}\sum_{k=0}^{n-1}\frac{1-\left(\frac{t_k}{t_n}\right)^m}{1 - \frac{t_k}{t_n}}\left(\frac{t_{k+1}}{t_n} - \frac{t_{k}}{t_n}\right) = \int_{0}^1 \frac{1-u^m}{1-u}\mathrm d u = \sum_{\ell = 1}^{m} \frac1\ell \to \infty$$

---

What we proved here is:

If $\left(t_n\right)$ is a increasing sequence of real numbers such that: $$\lim\limits_{n\to \infty}\sup_{i\in\left\{0,\ldots,n-1\right\}} \frac{t_{i+1} - t_i}{t_n} = 0$$ then $$\sup\limits_{n}\sum_{k=0}^{n-1} \frac{t_{k+1} - t_k}{t_n - t_k} = \infty$$

Answer 2

I think the answer is no, but my proof has a little gap in it which turned out to be surprisingly difficult, and right now I'm not sure how to fill it in.

---

Suppose that such sequence $(t_n)_{n\ge 0}$ exists. Let $\tau(x)$ be a continuously differentiable strictly increasing function that interpolates $(t_n)_{n\ge 0}$, that is, $$\tau(k) = t_k\; \forall k \in \mathbb{Z}_{\ge 0}$$ Note that $$\int_{i}^{i+1} \frac{\tau'(s) ds}{t_n - \tau(s)}\le \int_{i}^{{i+1}} \frac{\tau'(s) ds}{t_n - t_{i+1}} = \frac{t_{i+1} - t_i}{t_n - t_{i+1}}$$ (the inequality follows from the fact that $\tau$ is increasing)

Integrating over intervals and summing up the inequalities, we obtain a bound $$ \int_{0}^{{n-1}} \frac{\tau'(s) ds}{t_n - \tau(s)} \le \sum_{i=0}^{n-2}\frac{t_{i+1} - t_i}{t_n - t_{i+1}} := S'_n $$ If we manage to prove that $S'_n$ are bounded, we will arrive at a contradiction, because we can compute the integral explicitly: $$ \int_{0}^{{n-1}} \frac{\tau'(s) ds}{t_n - \tau(s)} = \int_{t_0}^{t_{n-1}} \frac{d\tau}{t_n - \tau} \\ = \log (t_n - \tau)\Big|_{t_{n-1}}^{0} = \log \frac{t_n}{t_n - t_{n-1}} \ge \log \frac{1}{\Delta_n} \to \infty $$ because $\Delta_n \to 0$. So it remains to show that $S'_n$, and at the moment, I am not sure how to do this.

---

Edit: thanks to @Youem, I realized that $S'_n$ is not necessarily bounded, because in fact any of the terms $\frac{t_{i+1} - t_i}{t_n - t_{i+1}}$ can be unbounded. However, we don't need $S'_n$ to be bounded for the argument to work, we just need $\lim\inf S'_n < \infty$, and as it turns out, it suffices to show that $\lim\inf$ of the maximum of the terms is bounded.

Claim $\lim\inf \max_{i \in \{0, \ldots, n-2\}} \frac{t_{i+1} - t_i}{t_n - t_{i+1}} < 2 $

Proof:

Suppose the contrary, then from some point $N$, we can then choose $i$ such that $$ \frac{t_{i} - t_{i-1}}{t_n - t_{i}} \ge 2\; i \in \{1, \ldots, n-1\}$$ whenever $n > N$ We can then write $$t_n \le t_{n_1} + \frac{1}{2}(t_{n_1} - t_{n_1 - 1}) \le t_{n_1} + \frac{1}{2}(t_{n_1} - t_{n_2}) \le t_{n_2} + \frac{1}{2}(t_{n_2} - t_{n_2 - 1}) + \frac{1}{4}(t_{n_2} - t_{n_2 - 1}) \\ = t_{n_2} + (t_{n_2} - t_{n_2 - 1})\left(\frac{1}{2} + \frac{1}{4}\right) \le \ldots \le t_m + (t_m - t_{m-1})\left(\frac{1}{2} + \frac{1}{4} + \ldots\right) \le t_{m-1}$$ where $m \le N$. Hence $\forall n > N$ we get $t_n \le \max_{m < N} t_m$, so that $(t_n)_n$ is bounded - contradiction $\square$

Now we can prove that $\lim \inf S'_n < \infty$:

In view of the above claim, choose as large $n$ as we like, such that $$\max_{i \in \{0, \ldots, n-2\}} \frac{t_{i+1} - t_i}{t_n - t_{i+1}} < 2$$ In that case, we have $$\frac{t_{i+1} - t_i}{t_n - t_{i+1}} - \frac{t_{i+1} - t_i}{t_n - t_{i}} = \frac{t_{i+1} - t_i}{t_n - t_{i+1}}\frac{t_{i+1} - t_i}{t_n - t_{i}} \le 2 \frac{t_{i+1} - t_i}{t_n - t_{i}}$$ Therefore $$\frac{t_{i+1} - t_i}{t_n - t_{i+1}} \le 3 \frac{t_{i+1} - t_i}{t_n - t_{i}}\\ \Rightarrow S'_n \le 3 S_n$$