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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?

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Presumably you want your topological vector spaces to be Hausdorff? –  Qiaochu Yuan Dec 16 '13 at 4:32
1  
@QiaochuYuan: For the sake of simplicity, we can assume that $ X $ is Hausdorff, although the proof above shows that $ \displaystyle \lim_{n \to \infty} x_{n} = x $ without caring if $ x $ is the unique limit of $ (x_{n})_{n \in \mathbb{N}} $ or not. –  user36116 Dec 16 '13 at 4:38

1 Answer 1

up vote 4 down vote accepted

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

share|improve this answer
    
Thanks, Jochen. I need to spend some time thinking about your answer. In particular, I need to think about how to construct an appropriate sequence $ (A_{n})_{n \in \mathbb{N}} $ of measurable subsets of $ [0,1] $ that would result in $ y_{n} \not\to 0 $. –  user36116 Dec 17 '13 at 3:14

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