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. 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.
 A: 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 $x \in X$.
A: 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.
A: 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...
