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?

  • 1
    $\begingroup$ Your inner product is not positive definite. If $a \in \ell^2$, you will have $<a,a>=0$. $\endgroup$ Commented Oct 5, 2011 at 14:30
  • $\begingroup$ Thank you for your comment. I have modified the definition of $Y$. $\endgroup$
    – Yamamoto
    Commented Oct 5, 2011 at 14:53
  • 4
    $\begingroup$ 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? $\endgroup$ Commented Oct 5, 2011 at 17:51

1 Answer 1


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

  • $\begingroup$ Thank you for your help. But I'm not clear about the definition of $b_i$. What is $k$? And What values does $b_i$ take if $i \leq2^k$ or $i>2^{k+1}$ ? $\endgroup$
    – Yamamoto
    Commented Oct 7, 2011 at 12:16
  • $\begingroup$ What I mean is, for each positive integer $i$ take $k$ to be the unique integer such that $2^k < i \le 2^{k+1}$. Thus $b_1 = (-1)^{-1} = -1$ because $2^{-1} < 1 \le 2^{0}$, $b_2 = 1$, $b_3 = b_4 = -1$, $b_5 = \ldots = b_8 = 1$, etc. $\endgroup$ Commented Oct 7, 2011 at 19:41
  • $\begingroup$ Thank you. I completely understand. The Cesàro mean of $a_ib_i$ oscillates. $\endgroup$
    – Yamamoto
    Commented Oct 9, 2011 at 1:36

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