Deduce the Cauchy-Schwarz Inequality from the case m = 1 of Bessel’s Inequality:
the sum of $$\sum_{i=1}^{m}|(v,u_i)|^2 \leq ||v||^2. $$
[Math] Cauchy-Schwarz and Bessel’s Inequalities
inequalityinner-products
inequalityinner-products
Deduce the Cauchy-Schwarz Inequality from the case m = 1 of Bessel’s Inequality:
the sum of $$\sum_{i=1}^{m}|(v,u_i)|^2 \leq ||v||^2. $$
Best Answer
Let $a,b$ be two vectors.
If $b = 0$, the inequality $|(a,b)| \leq \|a\|\|b\|$ is trivially true.
If $b \neq 0$ then $u = \|b\|^{-1}b$ has norm $1$, hence by Bessels's inequality with $m=0$ we have $|(a,u)|^2 \leq \|a\|^2$. Since $(a,b) = (a,\|b\|u) = \|b\| (a,u)$, this yields $|(a,b)| \leq \|a\| \|b\|$.