In this [Math Stack Exchange post](https://math.stackexchange.com/questions/606172/if-lim-n-to-infty-2-x-n-1-x-n-x-then-is-it-true-that-li/608550#608550), I proved the following result.



> 
> **Theorem:** Let $ X $ be a locally convex topological vector space. Let $ x \in X $ and suppose that $ (x\_{n})\_{n \in \mathbb{N}} $ is a sequence in $ X $ satisfying $ \displaystyle \lim\_{n \to \infty} (2 x\_{n + 1} - x\_{n}) = x $. Then $ \displaystyle \lim\_{n \to \infty} x\_{n} = x $.
> 
> 
> 


At the end of the proof, I asked if the conclusion of the theorem would still hold if $ X $ was a topological vector space in general. I suspect that the answer is ‘no’, but I am unable to provide a counterexample.


In order to explain my suspicion, let me first recast my original proof, which uses the language of semi-norms, in terms of good ol’ fashioned open neighborhoods.


*Proof*


Define a sequence $ (y\_{n})\_{n \in \mathbb{N}} $ in $ X $ by
$$
\forall n \in \mathbb{N}: \quad
y\_{n} \stackrel{\text{def}}{=} x\_{n} - x.
$$
As
$$
\forall n \in \mathbb{N}: \quad
2 x\_{n + 1} - x\_{n} - x = 2 y\_{n + 1} - y\_{n},
$$
we obtain $ \displaystyle \lim\_{n \to \infty} (2 y\_{n + 1} - y\_{n}) = 0\_{X} $. Next, let $ V \subseteq X $ be an arbitrary open neighborhood of $ 0\_{X} $. As $ X $ is locally convex, we can find an open neighborhood $ U $ of $ 0\_{X} $ such that (i) $ U $ is convex and balanced and (ii) $ U + U \subseteq V $.


Now, there exists an $ N \in \mathbb{N} $ sufficiently large so that
$$
\forall k \in \mathbb{N}: \quad
2 y\_{N + k} - y\_{N + k - 1} \in U.
$$
In particular,
$$
\forall k \in \mathbb{N}: \quad
2^{k} y\_{N + k} - 2^{k - 1} y\_{N + k - 1} \in 2^{k - 1} U.
$$
It follows readily that
\begin{align\*}
(\spadesuit) \quad
\forall m \in \mathbb{N}: \quad
 2^{m} y\_{N + m} - y\_{N}
& = \sum\_{k = 1}^{m} (2^{k} y\_{N + k} - 2^{k - 1} y\_{N + k - 1}) \\
& \in \sum\_{k = 1}^{m} 2^{k - 1} U \\
& = (2^{m} - 1) U. \quad (\text{As $ U $ is convex.})
\end{align\*}
Hence,
$$
\forall m \in \mathbb{N}: \quad
y\_{N + m} - \frac{1}{2^{m}} y\_{N} \in \left( 1 - \frac{1}{2^{m}} \right) U,
$$
or equivalently,
$$
\forall m \in \mathbb{N}: \quad
y\_{N + m} \in \frac{1}{2^{m}} y\_{N} + \left( 1 - \frac{1}{2^{m}} \right) U.
$$
By the continuity of scalar multiplication, we can find an $ m \in \mathbb{N} $ such that
$$
\forall n \in \mathbb{N}\_{\geq m}: \quad
\frac{1}{2^{n}} y\_{N} \in U.
$$
This implies that
\begin{align\*}
\forall n \in \mathbb{N}\_{\geq m}: \quad
 y\_{N + n}
& \in U + \left( 1 - \frac{1}{2^{n}} \right) U \\
& \subseteq U + U \quad (\text{As $ U $ is balanced.}) \\
& \subseteq V,
\end{align\*}
or equivalently,
$$
\forall n \in \mathbb{N}\_{\geq N + m}: \quad
y\_{n} \in V.
$$
As $ V $ is arbitrary, we conclude that $ \displaystyle \lim\_{n \to \infty} y\_{n} = 0\_{X} $, thus yielding $ \displaystyle \lim\_{n \to \infty} x\_{n} = x $. $ \quad \blacksquare $


As you can see in the step indicated by ($ \spadesuit $), the convexity of $ U $ is required for it to work. For a non-locally convex topological vector space, one does not have a neighborhood base of $ 0\_{X} $ consisting of convex sets, so the existence of $ U $ is not guaranteed. As I am unable to find an alternative argument, local convexity is a condition that I am unable to avoid. This leads me to the following question.



> 
> **Question:** Can one do away with the requirement of local convexity and still obtain the conclusion of the theorem?
> 
> 
> 



Extending a Certain Result from Locally Convex Topological Vector Spaces to General Topological Vector Spaces
I think that you can construct a counterexample if $X$ is the space of measurable functions on a nice probability space (like the unit interval with the Lebesgue measure) endowed with stochastic convergence where $y\_n \to 0$ if and only if $P(|y\_n| >\varepsilon) \to 0$ for all $\varepsilon >0$. Take a suitable sequence $A\_n$ of measurable sets and (large) constants $c\_n$ and define $r\_n=c\_n I\_{A\_n}$ (the characteristic function). Then $r\_n \to 0$ whenever $P(A\_n)\to 0$. Next define $y\_n$ so that $r\_n= 2y\_{n+1}-y\_n$ (namely $y\_{n+1}=\sum\limits\_{k=1}^n 2^{-(n-k+1)}r\_k$ or something alike). Then you will see how to choose $A\_n$ in order to avoid $y\_n\to 0$.




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Some more details: For $c\_n= 2^{n+1}$ you get $$y\_{2n+1} \ge \sum\_{k=n}^{2n}c\_k 2^{-(2n-k+1)}I\_{A\_n} \ge \sum\_{k=n}^{2n} I\_{A\_k}.$$ Therefore $\bigcup\_{k=n}^{2n} A\_k$ is contained in $\lbrace |y\_{2n+1}|\ge 1\rbrace$. If $A\_n$ are independent with $P(A\_n)=1/n$ you get $$ P(\lbrace |y\_{2n+1}|\ge 1\rbrace) \ge P(\bigcup\_{k=n}^{2n} A\_k) = 1-\prod\_{k=n}^{2n} (1-P(A\_k)) = 1-\frac{n-1}{2n} \not\to 0$$


