Schwarz's Inequality

$$\vert\left\langle{\psi_1\vert\psi_2}\right\rangle{}\vert^2 \... ...\right\rangle{}\left\langle{\psi_2\vert\psi_2}\right\rangle{}.$$ (1)

Written out explicitly

$$\left[{\int_a^b \psi_1(x)\psi_2(x)\,dx}\right]^2\leq \int_a^b [\psi_1(x)]^2\,dx \int_a^b [\psi_2(x)]^2\,dx,$$ (2)

with equality Iff $g(x)=\alpha f(x)$ with $\alpha$ a constant. To derive, let $\psi(x)$ be a Complex function and $\lambda$ a Complex constant such that $\psi(x)\equiv f(x)+\lambda g(x)$ for some $f$ and $g$. Since $\int\psi^*\psi\,dx\geq 0$,

$$\int \psi^*\psi\,dx = \int f^*f\,dx+\lambda\int f^*g\,dx+\lambda^*\int g^*f\,dx+\lambda\lambda^* \int g^*g\,dx \geq 0,$$ (3)

with equality when $\psi(x)=0$. Set

$$\lambda=-\int g^*f\,dx$$ (4)

so that

$$\lambda^*=-\int f^*g\,dx.$$ (5)

Plugging (5) and (4) into (3) then gives

$\int f^*f\,dx\int g^*g\,dx-\int g^*f\,dx\int f^*g\,dx$
$-\int f^*g\,dx\int g^*f\,dx + \int g^*f\,dx\int f^*g\,dx\geq 0\quad$	(6)

$$\int g^* f\,dx\int f^*g\,dx\leq \int f^*f\,dx\int g^*g\,dx$$ (7)

$$\left\vert{\int g^*f\,dx}\right\vert = \left\vert{\int f^*g\,dx}\right\vert \leq \int f^*f\,dx\int g^*g\,dx$$ (8)

so

$$\vert\left\langle{f\vert g}\right\rangle{}\vert^2\leq \left\... ...f\vert f}\right\rangle{}\left\langle{g\vert g}\right\rangle{}.$$ (9)

Bessel's Inequality can be derived from this.

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 527-529, 1985.