Cauchy-Schwarz Inequality: Equality Case

In this note, we will show the necessary conditions for two sequences of reals, aka_k and bkb_k such that equality is reached by the Cauchy-Schwarz Inequality. I believe this fact is already presented in the Cauchy-Schwarz Brilliant Wiki, but I don't believe it was shown in this manner. First, we begin with a statement of the inequality:

(k=1nakbk)2(k=1nak2)(k=1nbk2)\left(\sum_{k=1}^{n} a_kb_k\right)^2 \leq \left(\sum_{k=1}^{n} a_k^{2}\right)\left(\sum_{k=1}^{n} b_k^{2}\right)

Now, subtracting to yield a non-negative result,

(k=1nak2)(k=1nbk2)(k=1nakbk)20\left(\sum_{k=1}^{n} a_k^{2}\right)\left(\sum_{k=1}^{n} b_k^{2}\right) - \left(\sum_{k=1}^{n} a_kb_k\right)^2 \geq 0

This expression can be expanded and then neatly, intuitively compacted into summation notation. Also, we'll be leaving behind the inequality sign in exchange for an equal sign, since we're focusing on the equality case.

(a12+a22+...+an2)(b12+b22+...+bn2)(a1b1+a2b2+...+anbn)2=0(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2) - (a_1b_1 + a_2b_2 + ... + a_nb_n)^2 = 0

(i=1nj=1nai2bj2)(i=1nj=1naibiajbj)=0\left(\sum_{i=1}^{n}\sum_{j=1}^{n} a_i^{2}b_j^{2}\right) - \left(\sum_{i=1}^{n}\sum_{j=1}^{n} a_ib_ia_jb_j \right) = 0

12(i=1nj=1nai2bj2+aj2bi2)(i=1nj=1naibiajbj)=0\frac{1}{2}\left(\sum_{i=1}^{n}\sum_{j=1}^{n} a_i^{2}b_j^{2} + a_j^2b_i^2\right) - \left(\sum_{i=1}^{n}\sum_{j=1}^{n} a_ib_ia_jb_j \right) = 0

Now, combining our double sums:

12(i=1nj=1nai2bj22aibjajbi+aj2bi2)=12(i=1nj=1n(aibjajbi)2)=0\frac{1}{2}\left(\sum_{i=1}^{n}\sum_{j=1}^{n} a_i^{2}b_j^{2} -2a_ib_ja_jb_i + a_j^2b_i^2\right) = \frac{1}{2}\left(\sum_{i=1}^{n}\sum_{j=1}^{n} (a_ib_j - a_jb_i)^2 \right) = 0

If (b1,b2,...,bn)0(b_1, b_2, ... , b_n) \neq 0, then bk0b_k \neq 0 for some kk. For equality to occur, the above expression must equal 00, and so each term of the double sum must equal 00. Considering only the positive terms bkb_k, it follows that equality holds with the Cauchy-Schwarz Inequality if and only if akbi=aibka_kb_i = a_ib_k for all ii such that 1in1 \leq i \leq n and for some value of kk.

Dividing both sides of this relation by bib_i and bkb_k and setting γ=akbk\gamma = \frac{a_k}{b_k}, we reach the fact that aibi=γ\frac{a_i}{b_i} = \gamma. In other words, our sequences aka_k and bkb_k must be proportional\textit{proportional} for the equality case of the Cauchy-Schwarz Inequality.

Note by Ryan Tamburrino
4 years, 9 months ago

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interesting! I love the proof, especially the usage of summations. But I think the last portion of the argument is not valid.

Dividing both sides of this relation by bi and bk.

We only assumed bk to be non-zero. bi could be zero, making division by b_i wrong. Is that not true?

Arun Kumar - 8 months, 3 weeks ago

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Considering only the positive terms b_k

Did you mean non-zero? b_k could be negative, no?

Arun Kumar - 8 months, 3 weeks ago

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