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Let $(e_n)_{n=1}^\infty$ be the canonical vectors in $\ell^\infty$. Consider the sequence $$(e_n + e_{n+1} +\cdots + e_{2n})_{n=1}^\infty$$

Does it converge weakly in $\ell^\infty$?


Obviously $e_n + e_{n+1} + \cdots + e_{2n} \xrightarrow{n\to\infty} 0$ coordinate-wise, so the only possible candidate for the weak limit is $0$.

I believe it indeed weakly converges to $0$.

Let $f \in (\ell^\infty)^*$ be a bounded linear functional. Assume $f(e_1 + e_{n+1} \cdots +e_{2n}) \not\to 0$. Therefore, there exists $\varepsilon > 0$ and a subsequence $$(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})_{n=1}^\infty$$ such that $|f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})| \ge \varepsilon$ for all $n \in \mathbb{N}$.

We have $$f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)}) = |f(e_{p(n)} + e_{p(n)+1} +\cdots + e_{2p(n)})|e^{i\phi_n}$$

for some $\phi_n \in \mathbb{R}$.

Inductively we construct a further subsequence $(e_{q(p(n))} + e_{q(p(n))+1} +\cdots + e_{2q(p(n))})_{n=1}^\infty$ such that the supports of $e_{q(p(n))} + e_{q(p(n))+1} +\cdots + e_{2q(p(n))}$ and $e_{q(p(n+1))} + e_{q(p(n+1))+1} +\cdots + e_{2q(p(n+1))}$ are disjoint:

Set $q(p(1)) = p(1)$. Let $p(k)$ be such that $p(k) \ge 2p(1)$. Set $q(p(2)) = p(k)$. Le $p(k')$ be such that $p(k') \ge 2p(k) = 2q(p(2))$. Set $q(p(3)) = p(k')$ and so on.

Now consider $$x_n = \sum_{k=1}^n e^{-i\phi_{q(k)}}(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})$$ Since the summands have disjoint supports and $|e^{-i\phi_{q(k)}}| = 1$ it follows $\|x_n\|_\infty = 1$ for all $n \in \mathbb{N}$.

However, $$|f(x_n)| = \left|\sum_{k=1}^n e^{-i\phi_{q(k)}} f(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})\right| = \sum_{k=1}^n |f(e_{q(p(k))} + e_{q(p(k))+1} +\cdots + e_{2q(p(k))})| \ge n\varepsilon \xrightarrow{n\to\infty} +\infty$$

so $f$ is unbounded. A contradiction.

Therefore $f(e_n + e_{n+1} + \cdots + e_{2n}) \xrightarrow{n\to\infty} 0$ for all $f \in (\ell^\infty)*$ so $e_n + e_{n+1} + \cdots + e_{2n} \rightharpoonup 0$.

Is my proof correct? Is there an easier (and still elementary) way to conclude that it converges weakly to $0$?

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Edit: I lied - there is another proof that might well be regarded as much simpler. Inspired by the OP's comment on what he was really trying to do:

Trivial Lemma Suppose $X$ is a Banach space and $Y$ is a closed subspace. If $y_n\to y$ weakly in $Y$ then $y_n\to y$ weakly in $X$.

Proof: Suppose $f\in X^*$. Let $g=f|_Y$. Then $g\in Y^*$, so $$f(y_n)=g(y_n)\to g(y)=f(y).$$

So to answer the original question it's enough to show that $e_n+\dots+e_{2n}\to0$ weakly in $c_0$. This is clear, since $c_0^*=\ell_1$.

Original: Looks right. Is there an easier proof? I doubt it - what you did is already very simple, just applying a few definitions. If it doesn't look simple that's just because of the notation.

I'd write the very same proof as follows - possibly looks simpler, possibly easier to read (in fact I confess I didn't read what you wrote very carefully, just sort of skimmed it and saw that the idea clearly worked):

Say $x_n=e_n+\dots+e_{2n}$. Suppose $x_n$ does not tend to $0$ weakly. Then there exists $f\in\ell_\infty^*$ so $f(x_n)\not\to0$.

So there exist a subsequence $(x_{n_j})$, $\epsilon>0$, and complex numbers $\alpha_j$ with $|\alpha_j|=1$, so that $$f(y_j)\ge\epsilon$$if $$y_j=\alpha_jx_{n_j}.$$We may assume that $n_{j+1}>2n_j$, so that the $y_j$ have disjoint support. Since the $y_j$ have disjoint support we have $$||y_1+\dots+y_n||=1,$$although$$f(y_1+\dots+y_n)\ge n\epsilon,$$contradicting the fact that $f$ is bounded.

That looks simpler to me - I think it's much easier to read, since the displayed formulas are less intricate. But it's exactly the same proof.

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  • $\begingroup$ Ok, thanks for the feedback. Not really related to the question, but is there an example of a sequence in $c_0$ which converges weakly in $c_0$ but not in $\ell^\infty$? The sequence $(e_n + \cdots + e_{2n})_{n=1}^\infty$ was my attempt to construct such an example, but we have established that it converges weakly to $0$ in both $c_0$ and $\ell^\infty$. $\endgroup$ Jan 7 '18 at 12:49
  • $\begingroup$ @mechanodroid See edit. $\endgroup$ Jan 7 '18 at 13:07
  • $\begingroup$ Great, exactly the kind of thing I was hoping for! $\endgroup$ Jan 7 '18 at 13:12

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