I want to prove the following version of the Schwarz Inequality for complex numbers $a_1, a_2, \ldots, a_n \in \mathbb{C}$ and $b_1, b_2, \ldots, b_n \in \mathbb{C}$: $$|\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2$$

Rudin has a nice proof where he shows that $$ \sum (|(\sum |b_j|^2)a_j - (\sum a_j \overline{b_j})b_j|^2) = (\sum |b_j|^2)(\sum |a_j|^2 \sum |b_j|^2 - |\sum a_j \overline{b_j}|^2) $$ Since the left side is $\geq 0$, so is the right side. If $b_1 = b_2 = \ldots = b_n = 0$, the Schwarz inequality is trivial; otherwise, we have $\sum |b_j|^2 > 0$, so that $(\sum |a_j|^2 \sum |b_j|^2 - |\sum a_j \overline{b_j}|^2) \geq 0$.

However, I don't find this proof very intuitive (how the heck did he come up with it?!), and I'd prefer to try it by induction: assume Schwarz holds for $n-1$, then show it's true for $n$. Is it possible to do it this way?

Here's what I have so far: Let $C = |\sum_{j=1}^{n-1} a_j \overline{b_j}|$. Then $$ |\sum_{j=1}^n a_j \overline{b_j}|^2 = |C + a_n \overline{b_n}|^2 = |C|^2 + 2|C||a_n \overline{b_n}| + |a_n \overline{b_n}|^2$$ Using the inductive hypothesis and some basic properties of the complex norm, we have $$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2 + 2|C||a_n||b_n| + |a_n|^2 |b_n|^2$$ We could use the triangle inequality to separate out that $|C|$ in the middle, but I don't think that helps...

  • 3
    $\begingroup$ See rgmia.org/papers/v12e/Cauchy-Schwarzinequality.pdf for a compilation of different proofs of the Cauchy-Schwarz inequality. Proof 6 is by induction. $\endgroup$
    – Jeff
    Feb 2, 2012 at 0:31
  • $\begingroup$ Cool! In Proof 6, they use the fact that $2 a_1 b_1 a_2 b_2 \leq a_1^2 b_2^2 + a_2^2 b_1^2$. Why is this true? $\endgroup$ Feb 2, 2012 at 3:05
  • 1
    $\begingroup$ This is Young's inequality, which is very easy to prove. Noting that $0 \leq (a-b)^2 = a^2 + b^2 - 2ab$, you can rearrange to obtain $2ab \leq a^2 + b^2$. $\endgroup$
    – Jeff
    Feb 2, 2012 at 14:23

1 Answer 1


Well, since no one gave a complete answer yet--and because I wrote one anyway--here's the proof by induction, in a manner which is hopefully easy for students (without much proof experience) to understand. Credit goes to the Wu and Wu paper posted by @Jeff.

Both sides of the Schwarz inequality are real numbers $\geq 0$. If $\sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 = 0$, then it must be that $a_1 = a_2 = \ldots = a_n = 0$ and/or $b_1 = b_2 = \ldots = b_n = 0$, so clearly $|\sum_{j=1}^n a_j \overline{b_j}|^2$ also $= 0$ and we are done. Now we only need to prove the case in which both sides of the inequality are positive.

Base Case. For $n = 1$, we have $$|\sum_{j=1}^1 a_j \overline{b_j}|^2 = |a_j \overline{b_j}|^2 = |a_j|^2 |b_j|^2 = \sum_{j=1}^1 |a_j|^2 \sum_{j=1}^1 |b_j|^2.$$

Inductive Step. The inductive hypothesis is $|\sum_{j=1}^{n-1} a_j \overline{b_j}|^2 \leq \sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2$. Since we only need to worry about the case in which both sides are positive, so we can take the square root to obtain $$|\sum_{j=1}^{n-1} a_j \overline{b_j}| \leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2}.$$

Thus $|\sum_{j=1}^n a_j \overline{b_j}|$

$= |\sum_{j=1}^{n-1} a_j \overline{b_j} + a_n \overline{b_n}|$

$\leq |\sum_{j=1}^{n-1} a_j \overline{b_j}| + |a_n \overline{b_n}|$ (by the triangle inequality)

$\leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2} + |a_n \overline{b_n}|$ (by the inductive hypothesis)

$= \sqrt{\sum_{j=1}^{n-1} |a_j|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2} + |a_n| |b_n|.$

Here we're a little stuck. We want to be able to square $|a_n|$ and $|b_n|$ and bring them into their respective square-rooted sums. So if we label $a = \sqrt{\sum_{j=1}^{n-1} |a_j|^2}$, $b = \sqrt{\sum_{j=1}^{n-1} |b_j|^2}$, $c = |a_n|$, and $d = |b_n|$, we want to be able to say $ab + cd \leq \sqrt{a^2 + c^2} \sqrt{b^2 + d^2}$. In fact, we can say it! This inequality is always true for any $a, b, c, d \in \mathbb{R}$, because

$0 \leq (ad - bc)^2 = a^2 d^2 - 2abcd + b^2 c^2$

$\Rightarrow 2abcd \leq a^2 d^2 + b^2 c^2$

$\Rightarrow a^2 b^2 + 2abcd + c^2 d^2 \leq a^2 b^2 + a^2 d^2 + b^2 c^2 + c^2 d^2$

$\Rightarrow (ab + cd)^2 \leq (a^2 + c^2)(b^2 + d^2),$

and since both sides are positive reals, we can take the square root.

We now use this inequality to obtain

$|\sum_{j=1}^n a_j \overline{b_j}| \leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2} + |a_n| |b_n|$

$\leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 + |a_n|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2 + |b_n|^2}$

$= \sqrt{\sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2},$

and just square both sides to complete the inductive step.


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