Prove or disprove $\lim\limits_{n \to \infty}\Delta x_n=0.$

Question

For a sequence $\{x_n\}_{n=1}^{\infty}$, define $$\Delta x_n:=x_{n+1}-x_n,~\Delta^2 x_n:=\Delta x_{n+1}-\Delta x_n,~(n=1,2,\ldots)$$ which are named 1-order and 2-order difference, respectively.

The problem is stated as follows:

Let $\{x_n\}_{n=1}^{\infty}$ be bounded , and satisfy $\lim\limits_{n \to \infty}\Delta^2 x_n=0$. Prove or disprove $\lim\limits_{n \to \infty}\Delta x_n=0.$

By intuiton, the conclusion is likely to be true. According to $\lim\limits_{n \to \infty}\Delta^2 x_n=0,$ we can estimate $\Delta x_n$ almost equal with an increasing $n$. Thus, $\{x_n\}$ looks like an arithmetic sequence. If $\lim\limits_{n \to \infty}\Delta x_n \neq 0$, then $\{x_n\}$ can not be bounded.

But how to prove it rigidly?

Answer 1

Yes: If $(x_n)$ is bounded and $\lim_{n \to \infty}\Delta^2 x_n = 0$ then $\lim_{n \to \infty}\Delta x_n = 0$. That is a consequence of the following general estimate:

If $(x_n)$ is a sequence with $|x_n| \le M$ and $|\Delta^2 x_n| \le K$ for all $n$ then $$ \tag{*} |\Delta x_n|^2 \le 4MK \, . $$ for all $n$.

In our case $\lim_{n \to \infty}\Delta^2 x_n=0$, so that the above can be applied to tail sequences $(x_n)_{n \ge n_0}$ with $K$ arbitrarily small, and $\lim_{n \to \infty}\Delta x_n=0$ follows.

Proof of the claim. It suffices to prove $(*)$ for $n=0$. Without loss of generality assume that $\Delta x_0 \ge 0$. We have $$ x_n = x_0 + \sum_{j=0}^{n-1} \Delta x_j = x_0 + \sum_{j=0}^{n-1} \left( \Delta x_0 + \sum_{k=0}^{j-1} \Delta^2 x_k \right) \\ = x_0 + n \Delta x_0 + \sum_{j=0}^{n-1}\sum_{k=0}^{j-1} \Delta^2 x_k \, . $$ Using the given bounds $-M \le x_n \le M$ and $\Delta^2 x_n \ge -K$ it follows that $$ M \ge -M + n \Delta x_0 - \frac{(n-1)n}{2}K \\ \implies 0 \le \frac{(n-1)n}{2}K - n \Delta x_0 + 2M $$

If $K=0$ then $0 \le \Delta x_0 \le 2M/n$ implies $\Delta x_0 = 0$, and we are done. Otherwise the quadratic inequality can be rearranged (by “completing the square”) to $$ 0 \le \left(n - \left(\frac{\Delta x_0}{K} + \frac 12 \right) \right)^2 + \frac{4M}{K} - \left(\frac{\Delta x_0}{K} + \frac 12 \right)^2 \, . $$

Now choose the non-negative integer $n$ such $\left| n - \left(\frac{\Delta x_0}{K} + \frac 12 \right) \right| \le \frac 12$. Then $$ 0 \le \frac 14 + \frac{4M}{K} - \left(\frac{\Delta x_0}{K} + \frac 12 \right)^2 = \frac{4M}{K} - \left(\frac{\Delta x_0}{K} \right)^2 - \frac{\Delta x_0}{K} \\ \le \frac{4M}{K} - \left(\frac{\Delta x_0}{K} \right)^2 $$ and the desired conclusion $(*)$ follows.

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Remarks: There is a “similar” inequality for differentiable functions:

Let $f: \Bbb R \to \Bbb R$ be twice differentiable. Then $$ \tag{**}\sup_{x \in \Bbb R} \left| f'\left( x\right) \right| ^{2}\le 4\sup_{x \in \Bbb R} \left| f\left( x\right) \right| \sup_{x \in \Bbb R} \left| f''\left( x\right) \right|$$

which goes back to Edmund Landau. See

• Landau–Kolmogorov inequality,
• Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| $,
• Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?.

The proofs resemble each other: We have $$ f(t) = f(0) + t f'(0) + \int_{u=0}^t \int_{v=0}^u f''(v) \, dv $$ which implies $$ 0 \le \frac{t^2}2 \sup_{x \in \Bbb R} \left| f''\left( x\right) \right| - t f'(0) + 2 \sup_{x \in \Bbb R} \left| f\left( x\right) \right| \, . $$ Then $t$ is chosen such that the right-hand side is minimal. The same is done in above prove for sequences, only that $n$ is restricted to integers and cannot be chosen arbitrarily.

Landau also proved that the factor $4$ in $(**)$ is best possible. It would be interesting to know if $4$ is also the best possible factor for sequences in $(*)$.

Answer 2

Suppose that $\Delta x_n \not\to 0$. Then there is - without loss of generality, replace $x_n$ with $-x_n$ if necessary - a $c > 0$ such that $\Delta x_n > 2c$ for infinitely many $n$. Now for every $\varepsilon > 0$ there exists an $N_{\varepsilon}$ such that $\lvert \Delta^2 x_n\rvert < \varepsilon$ for all $n \geqslant N_{\varepsilon}$. Pick an $n_1 \geqslant N_{\varepsilon}$ with $\Delta x_{n_1} > 2c$. Then $$\Delta x_{n_1 + k} \geqslant \Delta x_{n_1} - k\varepsilon > c$$ for $0 \leqslant k \leqslant c/\varepsilon$. It follows that $$x_{n_1+k+1} - x_{n_1} = \sum_{\kappa = 0}^k \Delta x_{n_1 + \kappa} > (k+1)\cdot c$$ for $0 \leqslant k \leqslant c/\varepsilon$, and thus

$$\sup_n x_n - \inf_n x_n \geqslant \frac{c^2}{\varepsilon}$$ for every $\varepsilon > 0$, which says that $x_n$ is unbounded.

Answer 3

Let $\{x_n\}$ be bounded by $X$ (i.e. $|x_n|< X$ for all $n$) and $\lim_{n\to\infty} \Delta^2x_n = 0$. Take an arbitrary $\varepsilon>0$. I will show that there is an $N\in \Bbb N$ such that $|\Delta x_n|<\varepsilon$ for all $n>N$, thus showing that $\lim_{n\to\infty}\Delta x_n = 0$.

First fix some natural number $m\geq \frac{4X}{\varepsilon} + 1$ (but also make sure that $m\geq 3$). Next, fix some $N$ such that $|\Delta^2x_n|<\frac{\varepsilon}{m-2} = Y$ for all $n>N$. Now, assume for contradiction, that there is some $n>N$ such that $\Delta x_n\geq\varepsilon$. Then we have $$ \begin{align} x_{m+n} - x_n &= \Delta x_n + \Delta x_{n+1} + \Delta x_{n+1} + \cdots + \Delta x_{n+m-1}\\ &\geq \Delta x_n + (\Delta x_{n} - Y) + (\Delta x_n - 2Y) + \cdots + (\Delta x_{n} - (m-2)Y)\\ &= (m-1)\Delta x_n - \frac{(m-1)(m-2)}{2}Y\\ &\geq (m-1)\varepsilon - \frac{(m-1)(m-2)}{2}Y\\ &= (m-1)\left(\varepsilon - \frac{m-2}2 \cdot \frac{\varepsilon}{m-2}\right)\\ &\geq \left(\frac{4X}\varepsilon + 1 - 1\right)\cdot \frac\varepsilon2\\ &= 2X \end{align} $$ But $|x_n|<X$ and $|x_{n+m}|<X$, so we can't have $x_{n+m} - x_n\geq 2X$. Thus we have a contradiction. So we must have $\Delta x_n < \varepsilon$ for all $n>N$. A very similar contradiction argument shows that $\Delta x_n > -\varepsilon$. It follows that $\Delta x_n\to 0$.

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It may look like $\frac{4X}\varepsilon + 1$ and $\frac{\varepsilon}{m-2}$ are pulled out of thin air, and that it just magically works out in the end. This is not the case. They are derived the following way:

We want some natural number $m$ to indicate how many $\Delta x_n$ terms we are adding together, and we want some $Y$ to bound $\Delta^2x_n$. With those bounds named, but without knowing what they are, we can actually do most of the working-out above. We get $$ \begin{align} x_{m+n} - x_n &\geq\Delta x_n + (\Delta x_{n} - Y) + (\Delta x_n - 2Y) + \cdots + (\Delta x_{n} - (m-2)Y)\\ &\geq \varepsilon + (\varepsilon - Y) + (\varepsilon - 2Y) + \cdots + (\varepsilon - (m-2)Y)\\ & \geq (m-1)\left(\varepsilon - \frac{m-2}{2}Y\right) \end{align} $$ We want $\varepsilon - (m-2)Y\geq 0$. (We don't want to add enough terms that we allow $\Delta x_{n+m}$ to become negative again. That's wasting terms.) And we want the final $(m-1)\left(\varepsilon - \frac{m-2}{2}Y\right)$ to be at least $2X$. Solving these two inequalities give $m\geq \frac{4X}{\varepsilon} + 1$ and $Y\leq \frac{\varepsilon}{m-2}$, which is what I used in the proof above.

Note that we don't really need $m\geq 3$. If $\frac{4X}\varepsilon \leq 1$, and we happen to pick $m = 2$, then we can choose whichever value we want for $Y$, and the argument works out in the end. However, because the general expressions require division by $m-2$, I added the $m\geq 3$ requirement for simplicity.

Answer 4

I wonder if this prove is correct.

Assume $\Delta x_n$ does not converge to 0, than there will be infinitely many $n$ with $\Delta x_n>c$(or $\Delta x_n<c$). as $\Delta^2x_n \to 0$, there is $N>0$ for all $\epsilon$ such that $|\Delta x_{n+1}-\Delta x_n|<\epsilon$. Since there are infinitely many $n$ that satisfy $\Delta x_n>c$, for that $n$ and $n>N$ we can write $|\Delta x_{n+1}-c|<\epsilon$. and for sufficiently large m, we can say that $\sup_{m}x_m-\inf_{m}x_m=2\epsilon M$ when M is the number of m's which mathches two condition. and this makes contradiction.

Answer 5

Since $x_n$ is bounded, choose $M$ so that $|x_n|\le M$.

Suppose that $\lim\limits_{n\to\infty}\Delta x_n\ne0$. Then $\exists\epsilon\gt0:\forall n_0,\exists n\ge n_0:|\Delta x_n|\ge\epsilon$.

Let $\delta=\frac{\epsilon^2}{6M}$. Since $\lim\limits_{n\to\infty}\Delta^2x_n=0$, choose $n_0$ so that if $n\ge n_0$, we have $\left|\Delta^2x_n\right|\le\delta$.

Choose $n\ge n_0$ so that $|\Delta x_n|\ge\epsilon$.

Let $k=\left\lceil\frac{6M}\epsilon\right\rceil$. Note that $(k-1)\delta\lt\epsilon$ and $k\epsilon\ge6M$.

By the choice of $n_0$ and $n$, $$ |\Delta x_{n+j}|\ge\epsilon-j\delta $$ Therefore, $$ \begin{align} |x_{n+k}-x_n| &\ge\sum_{j=0}^{k-1}(\epsilon-j\delta)\\ &=k\epsilon-\frac{k(k-1)}2\delta\\[3pt] &\gt\frac{k\epsilon}2\\[9pt] &\ge3M \end{align} $$ which contradicts the choice of $M$. Thus, $$ \lim_{n\to\infty}\Delta x_n=0 $$