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Learnt Cauchy–Bunyakovsky–Schwarz Inequality for Definite Integrals

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When I was a secondary school student, I was quite satisfied with this proof of Cauchy–Schwarz Inequality.

\[ \begin{aligned} & \left(\sum_{i = 1}^n a_i b_i\right)^2 \le \left(\sum_{i = 1}^n a_i^2\right) \left(\sum_{j = 1}^n b_j^2\right) \\ \iff& \left(\sum_{i = 1}^n a_i b_i\right) \left(\sum_{j = 1}^n a_j b_j\right) \le \left(\sum_{i = 1}^n a_i^2\right) \left(\sum_{j = 1}^n b_j^2\right) \\ \iff& \sum_{i = 1}^n \sum_{j = 1}^n a_i a_j b_i b_j \le \sum_{i = 1}^n \sum_{j = 1}^n a_i^2 b_j^2 \\ \iff& \sum_{i = 1}^n \sum_{j = 1}^n a_i a_j b_i b_j \le \sum_{i = 1}^n \sum_{j = 1}^n \frac{1}{2} \left(a_i^2 b_j^2 + a_j^2 b_i^2 \right)\\ \iff& \sum_{i = 1}^n \sum_{j = 1}^n \frac{1}{2} (a_i^2 b_j^2 - 2 a_i b_j \cdot a_j b_i + a_j^2 b_i^2) \ge 0 \\ \iff& \sum_{i = 1}^n \sum_{j = 1}^n \frac{1}{2} (a_i b_j - a_j b_i)^2 \ge 0 \end{aligned} \]

Equality holds if and only if $\forall i,j = 1,\dots,n, a_i b_j - a_j b_i = 0$. I was so happy that I didn’t think of further generalisations. I didn’t realise that the inequality can be more concisely written as $\langle \vect{a}, \vect{b} \rangle^2 \le \norm{\vect{a}}^2 \norm{\vect{b}}^2$, where $\vect{a} = (a_1,\cdots,a_n), \vect{b} = (b_1,\cdots,b_n) \in \R^n$

This Friday evening, I did a question about the integral version of the inequality. After spending hours to come up with an idea, I realised why I needed to learn inner product spaces. The very first version of the inequality that I learnt has too many summation signs, and it can’t be easily generalised to other spaces. The second proof of the same inequality that I learnt makes use of the determinant of a quadratic polynomial $p(t) = \norm{\vect{u} - t \vect{v}}^2$. That proof is much more elegant, and it helped me a lot while I was doing that question. When equality holds, it’s very hard to imagine what happens by looking at the original equality. However, if we convert it into the determinant of $p(t)$, then one quickly knows that this is equivalent to $p(t) = 0$, and can easily conclude that equality holds if and only if the two functions $f$ and $g$ satisfy $f \equiv k g$ for some $k \in \R$ for almost all points.1


  1. To be more precise, if one accepts that the space $C([a,b])$ of continuous functions defined on a closed and bounded interval $[a,b]$ in an inner product space, and $f,g \in C([a,b])$, then the condition “for almost all points” can be dropped. If we don’t to want be to so strict on the functions $f$ and $g$ defined on $[a,b]$, and we just say that $f$ and $g$ are integrable functions defined on $[a,b]$, then we have to accept the fact that $\int_a^b f^2 = 0$ doesn’t imply that $f \equiv 0$.

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