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Let $a=\{a_i\}$ and $b=\{b_i\}$ be real sequences such that $\lim\limits_{N \to \infty} \frac{1}{N}\sum\limits_{i=1}^{N}a_i^2$ and $\lim\limits_{N \to \infty} \frac{1}{N}\sum\limits_{i=1}^{N}b_i^2$ exist and are finite.

The first question is: does the limit $\lim\limits_{N \to \infty} \frac{1}{N}\sum\limits_{i=1}^{N}a_i b_i$ always exist?

If it exists, then we can define a semi-inner product $\langle a,b\rangle := \lim\limits_{N \to \infty} \frac{1}{N}\sum\limits_{i=1}^{N}a_i b_i$ on a linear space $Y:=\left\{a\in\mathbb{R}^{\mathbb{N}}: \lim\limits_{N \to \infty} \frac{1}{N}\sum\limits_{i=1}^{N}a_i^2 \text{ exists and is finite } \right\} $.

(Note that $\langle ,\rangle$ is not positive definite, as pointed out by Nate (see Comment 1 below).)

Setting $W:=\{a \in Y: \langle a, a\rangle =0 \}$ and using the semi-inner product $\langle ,\rangle$, we construct an inner product $(,)$ on the quotient space $Y/ W$ through setting $(a+W, b+W):=\langle a, b\rangle$.

Another question: does $\left(Y/W, (,)\right)$ form a Hilbert space?

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    Now $Y$ is not a vector space. Take $a = (1,1,1,\dots)$ and $b = (2,1,1,\dots)$. Then $a,b \in Y$ but $a-b \notin Y$. Perhaps you want to take a quotient instead?2011-10-05

1 Answers 1

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No, the limit does not always exist. Consider the case where all $a_i = 1$ while $b_i = (-1)^k$ if $2^k < i \le 2^{k+1}$.

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    Thank you. I completely understand. The Cesàro mean of $a_ib_i$ oscillates.2011-10-09