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Let $ z = ( z_1, z_2, \dots, z_m) \in \mathbb{C}^m$ and $z_j>0$ for all $ 1 \leq j \leq m.$ Then $\| \sum_{j} z_j \|_2 = \sum_j \| z_j \|_2$ if and only if $ z_j = \alpha_j z_1$ for some $ \alpha_j>0.$

This side is clear: If $ z_j = \alpha_j z_1$ for $ \alpha_j>0,$ then $ \| \sum_{j} z_j \|_2 = \| z_1 + \alpha_2 z_1 + \dots + \alpha_m z_1 \|_2 = \| ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m )z_1\|_2 = ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m ) \|z_1\|_2 = \|z_1\|_2 + \|z_2\|_2 + \dots + \| z_m\|_2 = \sum_{j} \|z_j\|_2.$

For the other side of the proof, I know that I need to use the case of equality in the Cauchy-Bunyakovskii-Schwarz(CBS) Inequality, which says that: For all $x,y \in \mathbb{C}^{m \times 1 }$ we have \begin{align} |x^ * y| = \|x\| \|y\| \text{, where $x^*$ is the conjugate transpose of x,} \end{align} if and only if $ y = \alpha x$ for $ \alpha = \dfrac{x^ * y}{x^ * x}.$

Any directions or help with the proof?

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If the $z_j$ are real numbers as you says in your first line, then $\| \sum_{j} z_j \|_2$ neither $\|z_j\|_2$ makes sense. I think you mean something like $\|(z_1,z_2,\ldots,z_n)\|_2=\sum_j |z_j|\iff z_j=\alpha_j z_1$ for some $\alpha_j\gt 0$ –  leo Apr 1 '12 at 2:10
    
So what is the difference, is not $\|z_j\|_2 = |z_j|$? –  Zizo Apr 1 '12 at 2:18
    
I think you want the condition to be $\alpha_j\geq0$, because otherwise the assertion is not true. Also, you probably want to say that each $z_j\in\mathbb{R}^n$; otherwise, the proof doesn't need CBS. Finally, the statement of CBS is wrong, as the condition you mention is the one that characterizes equality (the inequality holds no matter which $x,y$ you choose). –  Martin Argerami Apr 1 '12 at 2:35
    
Given $z=(z_1,z_2,\ldots,z_n)\in\mathbb{R}^n$, $\|(z_1,z_2,\ldots,z_n)\|_2=\sqrt{\sum_{j=1}^n z_j^2}$. Usually $|x|$ is just the absolute value of $x\in\mathbb{R}$. –  leo Apr 1 '12 at 2:44
    
@leo: I thought if we define a norm of $z_j \in \mathbb{R}$ by $\|z_j\|_p$ then as $\| \cdot\|_p \geq 0,$ we have $\|z_j\|_p = \left( z_j^p \right)^{ \frac{1}{p} } = |z_j|.$ –  Zizo Apr 1 '12 at 3:25

2 Answers 2

up vote 2 down vote accepted

Here is an answer for the newly formulated question. I will actually assume that $z_j\in\mathbb{C}^k$, as this does not change the proof.

Let us start with $m=2$. Then we have $\|z_1+z_2\|_2=\|z_1\|_2+\|z_2\|_2$; using that $\|x\|_2=x^*x$ and squaring we get $$ \|z_1\|_2+\|z_2\|_2+2\text{Re }z_1^*z_2=\|z_1\|_2+\|z_2\|_2+2\|z_1\|_2\|z_2\|_2. $$ So $$ \text{Re }z_1^*z_2=\|z_1\|_2\|z_2\|_2. $$ But then $$ \|z_1\|_2\|z_2\|_2=\text{Re }z_1^*z_2\leq|z_1^*z_2|\leq\|z_1\|_2\|z_2\|_2, $$ which implies that $$ z_1^*z_2=\|z_1\|_2\|z_2\|_2. $$ This is equality in the CBS inequality, so $z_2=\alpha_2\,z_1$, with $\alpha_2=z_1^*z_2/\|z_1\|_2^2\geq0$. Note that if $z_1,z_2$ are complex numbers, then $\alpha_2>0$.

Now we proceed by induction. If we have the equality for $h+1$ vectors, then $$ \sum_1^{h+1}\|z_j\|_2=\|\sum_1^{h+1}z_j\|_2\leq\|\sum_1^{h}z_j\|_2+\|z_{h+1}\|_2\leq\sum_1^{h+1}\|z_j\|_2. $$ Then we get equality for the first $h$ vectors and we can apply the inductive hypothesis to get $z_j=\alpha_jz_1$ for $j=1,\cdots,h$. Now $\sum_1^hz_j=(\sum_1^h\alpha_j)z_1$ and we can apply the $m=2$ case to get the final result.

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We assume $$\left|\sum_{k=1}^mz_k\right|=\sum_{k=1}^m|z_k|. $$ We consider two cases. The first one is when $\sum_{k=1}^mz_k\geq0$. Then we have $$\sum_{k=1}^mz_k=\sum_{k=1}^m|z_k|. $$ Then $$0=\sum_{k=1}^m|z_k|-z_k. $$ As every term in the sum is nonnegative, we get that $|z_k|=z_k$ for all $k$, so $z_k>0$ for all $k$. Now take $\alpha_k=z_k/z_1$.

The second case, when $\sum_{k=1}^mz_k<0$, is treated similarly and we get $|z_k|+z_k=0$ for all $k$. Now $z_k=-|z_k|<0$ for all $k$. So we can define $\alpha_k=z_k/z_1>0$ (quotient of two negative numbers).

This problem is definitely more interesting when $z_1,\ldots,z_m$ are vectors. Then the $\|\cdot\|_2$ notation is then justified, and the proof indeed requires the use of the case of equality in the CBS inequality.

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This is just let everybody knows that this answer actually answered of my first format of the question, that's when $z \in \mathbb{R}^m.$ Thanks @Martin –  Zizo Apr 1 '12 at 6:50

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