Is there a relationship between $\|X\|_2 ^2$ and $(\sum_i x[i])^2$?

Question

Is there a known relationship between $\|X\|_2 ^2$ and $(\sum_i x[i])^2$, where $X$ is a vector in the real space, $\|\cdot\|_2$ is the 2-norm, and $\sum_i x[i]$ is the sum over the elements of the vector $X$.

Answer 1

Cauchy's inequality (later generalized by Buniakovski and by Schwarz, see HERE) says $$ \left(\sum_{i=1}^n x_i y_i\right)^2 \le \left(\sum_{i=1}^n |x_i|^2\right) \left(\sum_{i=1}^n |y_i|^2\right) . $$ If we take $y_i=1$ for all $i$, we obtain $$ \left(\sum_{i=1}^n x_i\right)^2 \le n\;\sum_{i=1}^n |x_i|^2 = n \|X\|_2^2 . $$

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Just for fun, let me add: Buniakovski's inequality $$ \left(\int_a^b f(t)g(t)\;dt\right)^2 \le \left(\int_a^b |f(t)|^2\;dt\right) \left(\int_a^b |g(t)|^2\;dt\right) $$ Schwarz's indquality: if $\langle \cdot,\cdot \rangle$ is an inner product, then $$ \langle\mathbf{x}, \mathbf{y}\rangle \le \|\mathbf x\|\;\|\mathbf y\| , $$ where the norm $\|\mathbf x\|$ is defined as $\langle \mathbf x, \mathbf x\rangle^{1/2}$.

Answer 2

Yes, there is a relation. If $x \in \mathbb{R}^n$ then

$ \displaystyle 0 \le \left( \sum_{i=1}^n x_i \right)^2 \le n \| x \|^2 $

To prove this, note that

$ \displaystyle \sum_{i=1}^n x_i = \mathbf{1}^T x $

where $\mathbf{1} \in \mathbb{R}^n$ and $\mathbf{1} = [1, 1, \dots, 1]^T $.

Hence,

$ \left(\sum_{i=1}^n x_i\right)^2 = x^T {\mathbf{1} \mathbf{1}}^T x $

It easy to show that the eigenvalues of the symmetric matrix ${\mathbf{1} \mathbf{1}}^T $ are $\lambda_{Min} = 0 $ and $\lambda_{Max} = n $. From the properties of quadratic forms, the inequality given above follows.

Answer 3

If you assume 2-Norm (i.e.: $\|X\|_2^2=\sum\limits_{i=1}^n X_i^2$), your vector $X \in \mathbb{R}^n$, and if by $\sum X$ you mean $\sum\limits_{i=1}^n X_i$, then the following relation seems to be the only natural one I can think of: $$ \left(\sum X\right)\cdot \left(\sum X\right) = (X_1 + X_2 + \ldots + X_n) \cdot (X_1 + X_2 + \ldots + X_n) = X_1^2 + X_2^2 + \ldots + X_n^2 + X_1 X_2 + \ldots + X_{n-1}X_n = \|X\|_2^2 + \sum\limits_{i \neq j}X_iX_j,~\text{where }i,j\in \underline{n} $$