Extending a Certain Result from Locally Convex Topological Vector Spaces to General Topological Vector Spaces In this Math Stack Exchange post, 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?

 A: 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 $y_n\to 0$.

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 $$  P(\lbrace |y_{2n+1}|\ge 1\rbrace) \ge P(\bigcup_{k=n}^{2n} A_k) = 1-\prod_{k=n}^{2n} (1-P(A_k)) = 1-\frac{n-1}{2n} \not\to 0$$
