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I am preparing for an exam and I saw this exercise that our teacher left for us to prove:

Let $\{x_n \}_{n=1}^{\infty}$ be a sequence in a Hilbert space $\mathcal{H}$ such that $x_n \rightharpoonup 0$ or $\langle x_n, u \rangle \rightarrow 0$ for all $u \in \mathcal{H}.$

  1. Prove inductively that there exists a subsequence, $\{x_{n_k}\}_{k = 1}^{\infty}$ such that $\left| \langle x_{n_k}, x_{n_j} \rangle \right| \le \frac{1}{k}$ whenever $k > j.$
  2. For $N \in \mathbb{N}$, define $$y_N = \frac{1}{N}\sum_{k=1}^N x_{n_k}.$$ Prove that $y_N \rightarrow 0.$

For 1, I was thinking that there exists a convergent subsequence. For 2, I'm rather lost.

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For $(1)$:

Since $\langle x_n, x_1\rangle \xrightarrow{n\to\infty} 0$, there exists a strictly increasing sequence $(p_1(n))_{n=1}^\infty \in \mathbb{N}$ such that $\left|\langle x_{p_1(n)}, x_1\rangle\right| < \frac1{n}, \forall n \in \mathbb{N}$.

Define $x_{n_1} = x_{p_1(1)}$.

Since $\langle x_{p_1(n)}, x_{p_1(1)}\rangle \xrightarrow{n\to\infty} 0$, there exists a strictly increasing sequence $(p_2(n))_{n=1}^\infty \in \mathbb{N}$ such that $\left|\langle x_{p_2(p_1(n))}, x_{p_1(1)}\rangle\right| < \frac1{n}, \forall n \in \mathbb{N}$.

Define $x_{n_2} = x_{p_1(p_2(2))}$.

Since $\langle x_{p_2(p_1(n))}, x_{p_2(p_1(2))}\rangle \xrightarrow{n\to\infty} 0$, there exists a strictly increasing sequence $(p_3(n))_{n=1}^\infty \in \mathbb{N}$ such that $\left|\langle x_{p_3(p_2(p_1(n)))}, x_{p_2(p_1(2))}\rangle\right| < \frac1{n}, \forall n \in \mathbb{N}$.

Define $x_{n_3} = x_{p_3(p_2(p_1(3)))}$.

Continue inductively, and define $x_{n_k} = x_{p_1(p_2(\ldots p_k(k)))} = x_{(p_1 \circ \cdots \circ p_k)(k)}$ for all $k \in \mathbb{N}$.

For $k > j$ we have by construction of $p_{j}$

$$\left|\langle x_{n_k}, x_{n_j}\rangle\right| = \left|\langle x_{(p_1 \circ \cdots p_{j} \circ p_{j+1} \circ \cdots \circ p_k)(k)}, x_{(p_1 \circ \cdots \circ p_j)(j)}\rangle\right| < \frac1{(p_{j+1}\circ p_{j+2} \circ \cdots \circ p_k)(k)} < \frac1{k}$$

because all $p_i$ are strictly increasing.

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For 2:
$\Vert y_N \Vert^2 = \frac{1}{N^2} \langle \sum_{k = 1}^N y_{n_k}, \sum_{k = 1}^N y_{n_k} \rangle = \frac{1}{N^2} \sum_{k = 1}^N \sum_{j = 1}^N \langle y_{n_k}, y_{n_j} \rangle \leq \frac{2}{N^2} \sum_{k = 1}^N \sum_{j = k + 1}^N \vert \langle y_{n_k}, y_{n_j} \rangle \vert$.
At the last estimate we use that $\vert \langle y_{n_k}, y_{n_j} \rangle \vert = \vert \langle y_{n_j}, y_{n_k} \rangle \vert $.

Now we can use the properties of the sequence and further estimate
$\frac{2}{N^2} \sum_{k = 1}^N \sum_{j = k + 1}^N \vert \langle y_{n_k}, y_{n_j} \rangle \vert \leq \frac{2}{N^2} \sum_{k = 1}^N \sum_{j = k + 1}^N \frac{1}{k} = \frac{2}{N^2} \sum_{k = 1}^N \frac{N - k}{k} = \frac{2}{N}(\sum_{k = 1}^N \frac{1}{k} - 1) \to 0$.

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  • $\begingroup$ Thanks, @eddie. This proof makes a lot of sense. I think you meant $x_{n_k}$ not $y+{n_k}$. $\endgroup$ – Harry Evans May 20 '18 at 23:45
  • $\begingroup$ @HarryEvans that's exactely what i meant ;) $\endgroup$ – eddie May 21 '18 at 7:55

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