Let $X$ be a normed space and $x_0, x_1 \in X$. Let $l \in (0,1)$ be fixed. Prove, that the sequence $x_{n+1}=l x_{n}+(1-l) x_{n-1}$ converges.

Question

I'm trying to prove the following:

Let $(X, ||.||)$ be a normed space and $x_0, x_1 \in X$. Let $l \in (0,1)$ be a fixed constant. Prove, that the sequence $x_{n+1}=l x_{n}+(1-l) x_{n-1}$ converges.

I started with noting, that because $X$ is a vector space, we know that $x_n\in X$ $\forall n\in \mathbb{N}$.

Now I need to show, that there exists $x\in X$ such that $||x_n - x||\rightarrow 0$ as $n \rightarrow \infty$.

It is clear to me, that the sequence $x_0, x_2, x_4, x_6,\cdots$ is "increasing", the elements are "moving away" from $x_0$ on the line between elements $x_0$ and $x_1$. And, the sequence $x_1, x_3, x_5, x_7,\cdots$ is "decreasing", the elements are "moving away" from $x_1$ towards $x_0$.

Also, I know that the elements of the first subsequence are all "smaller" than the elements in the second subsequence.

So the limit $x$ should be the meeting point of these two subsequences.

PS! I understand, that there is no smaller or larger, because $x_0, x_1$ can be any two elements in X. I just put them on a number line.

$x_{n+1}-x_n=-(1-l)(x_n-x_{n-1})$ The space should be complete. — Ryszard Szwarc, Feb 27 '22 at 17:24
The image of the sequence lies in a finite dimensional normed space, which is of course complete. — arnett, Feb 27 '22 at 17:27
It is not said, that the space is complete. But could you explain, aren't all the points between x0 and x1 in X anyway? — Hasarewa, Feb 27 '22 at 17:31
@BazyliZuczek: No need for completeness as the dynamics happen along a one dimensional affine space, which is homeomorphic to the real line. — Mittens, Feb 27 '22 at 18:40
@Oliver Diaz Right, I have overlooked that the problem is one dimensional. — Ryszard Szwarc, Feb 27 '22 at 18:55

score 1 · Answer 1 · answered Feb 27 '22 at 17:42

I claim that $(x_n)$ is Cauchy.

$$||x_{n+1} - x_n|| = |(l-1)|||x_n - x_{n-1}|| = (l-1)^2 ||x_{n-1} - x_{n-2} || = $$

$$ |l-1|^n||x_1 - x_0|| \to 0$$

Since $|l-1|^n \to 0$. Now, the elements in this sequence lie in a finite dimensional space (span of $x_1 - x_0$). Since every finite dimensional normed space is complete, and every Cauchy sequence in a complete normed space converges, we get than $x_n \to x$ for some $x \in X$.

Yiorgos S. Smyrlis · Accepted Answer · 2022-02-28T00:41:27.263

The recursive relation $$ x_{n+1}=\ell x_n+(1-\ell)x_{n-1} $$ implies:

(i) $x_{n+1}-x_n=(\ell-1) (x_n-x_{n-1})\quad\Longrightarrow\quad x_{n+1}-x_n=(\ell-1)^{n} (x_1-x_{0})$

(ii) $x_{n+1}-(\ell-1) x_n=x_n-(\ell-1)x_{n-1}\quad\Longrightarrow\quad x_{n+1}-(\ell-1) x_n=x_1-(\ell-1)x_{0}$

Hence $$ x_{n}=\frac{1}{2-\ell} \Big(\big(x_1-(\ell-1)x_{0}\big)-(\ell-1)^{n} (x_1-x_{0})\Big) \to \frac{1}{2-\ell}\big(x_1-(\ell-1)x_{0}\big). $$ Note that the convergence of this sequence DOES NOT require completeness of the space $X$, i.e. does not have to be a Banach space.

score 1 · Answer 3 · answered Feb 27 '22 at 18:32

The sum of a geometric sequence with common ratio $\vert 1-l \vert <1$ converges.

$$x_n = x_0 + \left(\sum_{k=0}^{n}(1-l)^k\right)(x_1-x_0) \overset{n\to\infty}{\to} x_0 + \frac{1}{l}(x_1-x_0) \in X. $$

I'm not sure this is right, because I haven't mentioned the norm, but I don't see why you have to...

Let $X$ be a normed space and $x_0, x_1 \in X$. Let $l \in (0,1)$ be fixed. Prove, that the sequence $x_{n+1}=l x_{n}+(1-l) x_{n-1}$ converges.

3 Answers3