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$$\|X_m-X_n\|^2_2.$$ To be clear, I am not asking about the $X_m - X_n$ part. I am asking what the $\|\ \|^2_2$ thing means.

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closed as unclear what you're asking by Shailesh, Saad, Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Rebellos Dec 2 '18 at 9:29

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  • $\begingroup$ Welcome to Math.SE. Please use MathJax. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 2 '18 at 7:23
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    $\begingroup$ It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced). $\endgroup$ – Jyrki Lahtonen Dec 2 '18 at 9:07
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$\|x\|_2$ means the $2$-norm, which is $$\sqrt{\sum_{i=1}^n x_i^2}$$

while the $2$ on top, is its square $$\|x\|^2=\sum_{i=1}^n x_i^2.$$

Overall, it means the sum of square of the components.

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  • $\begingroup$ So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything? $\endgroup$ – chie Dec 2 '18 at 10:30
  • $\begingroup$ First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $\|x_m-x_n\|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up. $\endgroup$ – Siong Thye Goh Dec 2 '18 at 10:34
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In general, $|| \cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(\Omega,{\cal A}, \mu)$ to $\Bbb R$ defined by $f \mapsto (\int |f|^2 \,{\rm d}\mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in .

In particular, when $\Omega = \Bbb{N}$, $\mu$ is the counting measure, we have $||f||_2 = (\int |f|^2 \,{\rm d}\mu)^{1/2} = (\sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean $$||x_m-x_n||_2$$ a norm of vector.

Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $\rm \LaTeX$ in all levels of math writing.

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