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Consider the Mazur's Lemma (H. Brezis - "Functional analysis, ..."):

"Assume $(x_n)$ converges weakly to $x$. Then there exists a sequence $(y_n)$ made up of convex combinations of the $x_n$'s that converges strongly to $x$.''

The Lemma says that "there exists a sequence...''.

Is it true that $\textbf{every}$ sequence $(y_n)$ made up of convex combinations of the $x_n$'s converges strongly to $x$?

In particular, if we consider $$y_n = \frac{1}{n}(x_1 + x_2 + ... +x_n),$$

is it true that $y_n$ converges strongly to $x$?

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    $\begingroup$ Well, the degenerate case $y_n = x_n$ does not work so some assumption on the convex combination is needed. $\endgroup$
    – Dirk
    Apr 23, 2015 at 7:25

2 Answers 2

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If $x_i= u_{floor(log(i)+1)}$ where $\{u_i| i=1,2,\ldots\}$ is an orthonormal basis for a Hilbert Space, then $(x_i)$ converges weakly to $0$ and $y_n = 1/n \sum_{i=1}^n x_i$ does not converge strongly to anything.

In particular, if $n = floor( e^k )$ for an integer $k\geq1$, then $||y_n - u_k|| < 2/3$ hence $||y_n||>1/3$.

Interestingly, if we define the linear operator $L_n: l_2\rightarrow R$ by

$L_n(x) = 1/n \sum_{i=1}^n x(i)$

then the $L_n$ converge arbitrarily slowly to the zero operator.

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The Banach-Saks theorem tells you that there is a subsequence $x_{n_k}$ such that $1/n \sum_{k=1}^n x_{n_k}$ is strongly convergent. Asking that the mean itself is convergent is too strong a requirement. This can happen e.g. if there is some independence between the $x_i$ as in the law of large numbers (e.g. summable correlations), and that's the idea in the proof of the Banach-Saks theorem.

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    $\begingroup$ The Banach-Saks theorem is not true for every Banach space. There exist normalized unconditional bases that converge weakly to zero and yet the norm of the average of any n terms has norm at least 1/2. $\endgroup$ Apr 22, 2015 at 20:24
  • $\begingroup$ @Johnson Sure. I don't have the Brezis at hand so I don't know what is Pomegranate's context. $\endgroup$
    – coudy
    Apr 22, 2015 at 20:59

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