Inequalities for inner products
From Maths
Tables
| Equation | Form | Notes |
|---|---|---|
| Cauchy-Schwarz inequality[1] | [math]\vert\langle x,y\rangle\vert\le\Vert x\Vert\Vert y\Vert[/math] for [ilmath]\Vert x\Vert:=\sqrt{\langle x,x\rangle}[/ilmath] (equality if lin dependent) | For any Inner product, note that [ilmath]\Vert\cdot\Vert[/ilmath] is the norm induced by the inner product |
| Parallelogram law[1] | For any i.p.s we have [math]\Vert x+y\Vert^2+\Vert x-y\Vert^2=2\Vert x\Vert^2+2\Vert y\Vert^2[/math] | Page 11 |
| Polarisation identities[1] | For a [ilmath]\mathbb{R} [/ilmath] inner product: [math]\langle x,y\rangle=\frac{1}{4}\Vert x+y\Vert^2-\frac{1}{4}\Vert x-y\Vert^2[/math] | Page 10 |
| For a [ilmath]\mathbb{C} [/ilmath] inner product: [math]\langle x,y\rangle=\frac{1}{4}\Vert x+y\Vert^2-\frac{1}{4}\Vert x-y\Vert^2+j\left[\frac{1}{4}\Vert x+jy\Vert^2-\frac{1}{4}\Vert x-jy\Vert^2\right][/math] | ||
| Pythagorean theorem[1] | If [ilmath]x[/ilmath] is perpendicular to [ilmath]y[/ilmath] in any i.p.s, then: [math]\Vert x+y\Vert^2=\Vert x\Vert^2+\Vert y\Vert^2[/math] | Page 11 |
