Schwarz inequality
$
\def \EN {\quad \mbox{and} \quad}
\def \OF {\quad \mbox{or} \quad}
\def \hieruit {\quad \Longrightarrow \quad}
$
Inclining towards Linear Algebra, but with the theory of point clouds in mind,
we will redefine the inner product of two vectors as:
$$
(\vec{a} \cdot \vec{b}) := \sum_k w_k a_k b_k
$$
It can be shown easily that such an inner product obeys all of the usual rules:
\begin{eqnarray*}
(\vec{a} \cdot \vec{b}) = (\vec{b} \cdot \vec{a}) \\
(\vec{a} \cdot ( \vec{b} + \vec{c}) ) =
(\vec{a} \cdot \vec{b} ) + (\vec{a} \cdot \vec{c} ) \\
(\vec{a} \cdot \lambda \vec{b} ) = \lambda (\vec{a} \cdot \vec{b} ) \\
(\vec{a} \cdot \vec{a}) \ge 0
\end{eqnarray*}
For the last rule to be obeyed, it is necessary that masses $w_k$ be positive
(or zero). Schwartz inequality can now be conjectured for this inner product:
$$
(\vec{a} \cdot \vec{b})^2 \le (\vec{a} \cdot \vec{a}) (\vec{b} \cdot \vec{b})
$$
Proof:
$$
(\lambda \vec{a} - \vec{b} \cdot \lambda \vec{a} - \vec{b} )^2 \ge 0 \hieruit
$$ $$
\lambda^2 (\vec{a} \cdot \vec{a} ) - 2 \lambda ( \vec{a} \cdot \vec{b} )
+ ( \vec{b} \cdot \vec{b} ) \ge 0
$$
This is a quadratic inequality in $\lambda$. In order for this inequality to
hold, its discriminant must be negative or zero:
$$
4 ( \vec{a} \cdot \vec{b} )^2 -
4 ( \vec{a} \cdot \vec{a} ) ( \vec{b} \cdot \vec{b} ) \le 0 \hieruit
$$ $$
(\vec{a} \cdot \vec{b})^2 \le (\vec{a} \cdot \vec{a}) (\vec{b} \cdot \vec{b})
$$
Which completes the proof. Returning now to our original problem, define for
example:
$$
\vec{a} = (x_1 , x_2 , ... , x_k , ... , x_N) \EN
\vec{b} = (1 , 1 , ... , 1 , ... , 1)
$$
Then:
$$
(\vec{a} \cdot \vec{b}) = \sum_k w_k x_k \EN
(\vec{a} \cdot \vec{a}) = \sum_k w_k x_k^2 \EN
(\vec{b} \cdot \vec{b}) = \sum_k w_k = 1
$$
Therefore:
$$
\left( \sum_k w_k x_k \right)^2 \le
\left( \sum_k w_k x_k^2 \right) \hieruit
\sum_k w_k x_k^2 - \left( \sum_k w_k x_k \right)^2 \ge 0
$$
This result is equivalent with our previous finding that:
$$
\sum_k w_k (x_k - \overline{x})^2 \ge 0
$$
New results are obtained when applying Schwarz inequality to the quantity known
formerly as cross correllation moment. For the sake of simplicity, the midpoint
is set as the origin $(0,0)$ of the coordinate system. Now define:
$$
\vec{a} = (x_1 , x_2 , ... , x_k , ... , x_N) \EN
\vec{b} = (y_1 , y_2 , ... , y_k , ... , y_N)
$$
Then:
$$
(\vec{a} \cdot \vec{b}) = \sum_k w_k x_k y_k = \sigma_{xy}
$$ $$
(\vec{a} \cdot \vec{a}) = \sum_k w_k x_k^2 = \sigma_{xx} \EN
(\vec{b} \cdot \vec{b}) = \sum_k w_k y_k^2 = \sigma_{yy}
$$
Therefore, according to Schwarz inequality, the following relationship must
hold:
$$
\sigma_{xy}^2 \le \sigma_{xx} \sigma_{yy} \OF
\sigma_{xx} \sigma_{yy} - \sigma_{xy}^2 \ge 0
$$
As a consequence, the following quantity is bound like a (co)sine-function:
$$
\rho := \frac{\sigma_{xy}}{\sqrt{\sigma_{xx} \sigma_{yy}}}
\qquad \mbox{where} \qquad 0 \le \left| \rho \right| \le 1
$$
The thus defined quantity $\rho$ has already been mentioned as
the cross-correlation, in a narrower sense.