Let $z_1, . . . , z_n$ and $w_1, . . . , w_n$ be complex numbers. Show that $$|z_1w_1 + ··· + z_n w_n|^2 ≤ \sum ^n _{j=1} |z_j|^2 \sum ^n _{j=1}|w_j|^2$$

I basically tried to use the proof given for real numbers but I feel that something must be wrong. Could somebody look over the proof?

Let $z=(z_1,...,z_n)$ and $w=(w_1,...,w_n)$. Then:

$$0 \leq \sum_j^n\bigg(||z|||w_j|-||w|||z_j|\bigg)^2=||z||^2||w||^2-2||z||||w||\sum_j^n|z_jw_j|+||z||^2||w||^2=2||z||||w||\bigg(||z||||w||-\sum_j^n|z_jw_j|\bigg)$$

But then

$$|\sum_j^nz_jw_j|\leq \sum_j^n|z_jw_j|\leq ||z||||w||$$

And just squaring is left.


  • 1
    $\begingroup$ Why not simply using triangle inequality and then CS for moduli as you did? $\endgroup$
    – S.B.
    Jul 17, 2013 at 22:41
  • $\begingroup$ You have a (small) hole in your proof, you can only deduce $\sum \lvert z_jw_j\rvert \leqslant \lVert z\rVert\cdot\lVert w\rVert$ directly from $0 \leqslant 2\lVert z\rVert\cdot\lVert w\rVert \left(\lVert z\rVert\cdot\lVert w\rVert - \sum \lvert z_jw_j\rvert\right)$ when $\lVert z\rVert \cdot \lVert w\rVert > 0$. But when the product is $0$, then $z = 0$ or $w = 0$, and then the inequality is trivial. But still, that case ought to be mentioned. $\endgroup$ Jul 17, 2013 at 22:46
  • $\begingroup$ You mean, all is missing is just consideration of the trivial case, right? Hence, I only need to assume that $z$ and $w$ are non-zero vectors? $\endgroup$
    – Sarunas
    Jul 17, 2013 at 23:01
  • $\begingroup$ Yes. Start with "If $z = 0$ or $w = 0$, both sides are $0$, otherwise ..." $\endgroup$ Jul 17, 2013 at 23:10
  • 1
    $\begingroup$ A simple proof is to start with the inequality $0 \leq \|z - \text{proj}_w(z) \|^2$. Expand the right hand side and Cauchy-Schwartz pops out. $\endgroup$
    – littleO
    Jul 17, 2013 at 23:50

2 Answers 2


You have a canonical inner product in $\Bbb C^n$ given by $x\cdot y=\displaystyle \sum_{i=1}^n x_i\overline{y_i}$. Note $x\cdot x=\displaystyle \sum_{i=1}^n x_i\overline{x_i}=\displaystyle \sum_{i=1}^n |x_i|^2$

Claim Let $V$ be a $\Bbb C$-vector space, and $\langle \cdot,\cdot\rangle$ an inner product. If we define a norm $\lVert x\rVert :=\langle x,x\rangle^{1/2}$, then we always have $$|\langle x,y\rangle|^2\leqslant \lVert x\rVert ^2\lVert y\rVert^2$$

Proof If $x$ or $y$ are zero, the inequality is true, since $\langle x,0\rangle=0$ and $\lVert 0\rVert =0$. We may assume then that both norms are nonzero. Since $\langle z,z\rangle\geqslant 0$ for any choice of $z$, we may take $$z=x-\frac{\langle x,y\rangle y}{\lVert y\rVert ^2} $$

Using the inner product is (sesqui)linear we get

$$\begin{align} \langle z,z\rangle &= \langle {x,x} \rangle - \frac{1}{\lVert y\rVert^2}\langle {x,\langle x,y \rangle y} \rangle - \frac{1}{\lVert y\rVert^2}\langle {\langle x,y \rangle y,x} \rangle + \frac{{\langle x,y \rangle \overline {\langle x,y \rangle} }}{{\lVert y\rVert^2\lVert y\rVert^2}}\langle {y,y} \rangle \\ &= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} + \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} \\ &= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\| y \|^2} \geqslant 0\end{align} $$ as we wanted. We made repeated use of $$\langle a+b,c\rangle=\langle a,c\rangle+\langle b,c\rangle\\ \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle\\\langle \alpha a,b\rangle=\alpha\langle a,b\rangle\\\langle a,\beta b\rangle=\overline \beta\langle a,b\rangle \\z\overline z=|z|^2$$


Let $\omega = z_1w_1 + z_2w_2 + \cdots + z_nw_n$. In general, $|\omega|^2 = \omega \overline{\omega}$.

$$\omega \overline{\omega} = (z_1w_1+z_2w_2 + \cdots + z_nw_n)(\overline{z_1}\overline{w_1} + \cdots + \overline{z_n}\overline{w_n}) \\ = z_1\overline{z_1}w_1\overline{w_1} + z_1\overline{z_2}w_1\overline{w_2} \cdots + z_n\overline{z_n}w_n\overline{w_n} \\ = |z_1|^2|w_1|^2 + \cdots + |z_n|^2 |w_1|^2 + \textrm{mixed terms}$$

now It should be clear from here.

  • 3
    $\begingroup$ what happens if mixedterms be negative? $\endgroup$ Nov 7, 2017 at 11:30

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