View source for Cauchy-Schwarz inequality for inner product spaces

{{Stub page|grade=A|msg=Flesh out statement}}
{{M|\newcommand{\ip}[2][]{\langle {#2} \rangle {#1} } }}

==Statement==
Let {{M|\langle\cdot,\cdot\rangle:X\times X\rightarrow\mathbb{K} }} be an [[inner product]] so {{M|(X,\langle\cdot,\cdot\rangle)}} is an [[inner product space]], then{{rW2014LNFARS}}:
* {{M|\forall x,y\in X\left[\vert\ip{x,y}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\right]}}
==Proof==
* Let {{M|x,y\in X}} be given
** We have two cases now, {{M|y\eq 0}} and {{M|y\neq 0}}
**# {{M|y\eq 0}} case
**#* We now have two more cases, {{M|x\eq 0}} and {{M|x\neq 0}}
**#*# {{M|x\eq 0}}
**#*#* Now {{M|\ip{x,y}\eq 0}}, {{M|\ip{x,x}\eq 0}} and {{M|\ip{y,y}\eq 0}} so:
**#*#** {{M|\vert\ip{x,y}\vert\eq 0\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\times 0\eq 0}} is the case
**#*#*** {{M|0\le 0}} obviously holds. We're done in this case
**#*# {{M|x\neq 0}}
**#*#* Now {{M|\ip{x,y}:\eq\ip{x,0}\eq\overline{\ip{0,x} }\eq \overline{0\ip{z,x} } }} for any {{M|z\in X}}
**#*#*: {{M|\eq 0\ip{x,z}\eq 0}}
**#*#** Now {{M|\vert\ip{x,y}\vert\eq 0}} and {{M|\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0\sqrt{\ip{x,x} }\eq 0 }} - as {{M|y\eq 0}} {{M|\ip{y,y}\eq 0}}
**#*#*** {{M|\vert\ip{x,y}\vert\eq 0\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} }\eq 0}} gives us {{M|0\le 0}} which is obviously true (we have equality so we have {{M|\le}})
**# {{M|y\neq 0}} case
**#* Consider {{M|\lambda\in\mathbb{K} }} then:
**#** {{M|0\le\ip{x-\lambda y,x-\lambda y} }} (as for an [[inner product]] {{M|\forall z\in X[\ip{z,z}\in\mathbb{R}\wedge\ip{z,z}\ge 0]}} 
**#*** Then {{M|\ip{x-\lambda y,x-\lambda y}\eq \ip{x,x-\lambda y}-\lambda\ip{y,x-\lambda y} }}
**#***: {{M|\eq\overline{\ip{x-\lambda y,x} }-\lambda\big(\overline{\ip{x-\lambda y,y} }\big)}}
**#***: {{M|\eq\overline{\ip{x,x}-\lambda\ip{y,x} }-\lambda\big(\overline{\ip{x,y} - \lambda\ip{y,y} }\big) }}
**#***: {{M|\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \lambda\overline{\lambda}\ip{y,y} }}
**#***: {{M|\eq\ip{x,x}-\overline{\lambda\ip{y,x} }-\lambda\ip{y,x} + \vert\lambda\vert^2\ip{y,y} }}<ref group="Note">Let {{M|a+bj\in\mathbb{C} }}, then:
* {{M|(a+bj)\overline{(a+bj)}\eq(a+bj)(a-bj) }}
*: {{M|\eq a^2-j^2b^j +j(ab-ab)}}
*: {{M|\eq a^2+b^2\ (\ +0j\ )}}
*: {{M|\eq\vert a+bj\vert^2}} - as required
* If {{M|a+bj}} is the complex representation of {{M|\lambda}} then we see {{M|\lambda\overline{\lambda}\eq\vert\lambda\vert^2}}</ref>
**#***: {{M|\eq\ip{x,x}-\big(\overline{\lambda\ip{y,x} }+\lambda\ip{y,x}\big)+\vert\lambda\vert^2\ip{y,y} }}<ref group="Note">Let {{M|a+bj\in\mathbb{C} }}, then:
* {{M|(a+bj)+\overline{(a+bj)}\eq (a+bj)+(a-bj)\eq 2a}}
* Thus if {{M|a+bj}} is the complex representation of {{M|\lambda\ip{y,x} }} then
** {{M|\overline{\lambda\ip{y,x} }+\lambda\ip{y,x}\eq 2\text{Re}(\lambda\ip{y,x})}}</ref>
**#***: {{M|\eq\ip{x,x}-2\text{Re}(\lambda\ip{y,x})+\vert\lambda\vert^2\ip{y,y} }}
**#***: {{M|\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y } }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } }} - by [[completing the square]] {{XXX|I think}} (without considering the {{M|-2\text{Re}...}})
**#**** So {{M|\ip{x-\lambda y,x-\lambda y}\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }  }}
**#***** Note that {{M|\forall z\in X[\ip{z,z}\ge 0}} holds, so:
**#****** {{M|0\le\ip{x-\lambda y,x-\lambda y}\eq\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }  }}
**#******* Or just {{M|0\le\left(\sqrt{\ip{x,x} }-\vert\lambda\vert\sqrt{\ip{y,y} }\right)^2-2\text{Re}(\lambda\ip{y,x})+2\vert\lambda\vert\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }  }}
**#******* Define {{M|\theta\in[0,2\pi) }} such that {{M|\ip{y,x}\eq\vert\ip{y,x}\vert e^{j\theta} }} (a form of complex number)
**#******** Define {{MM|\lambda:\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta} }} and note that {{MM|\vert\lambda\vert\eq\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } }}
**#********* this is fine to do as {{M|y\neq 0}} so {{M|\ip{y,y}>0}} so {{M|\sqrt{\ip{y,y} }>0}} and the division in the fraction is defined
**#********* we substitute this into our expression to obtain:
**#********** {{MM|0\le\left(\sqrt{\ip{x,x} }-\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{y,y} }\right)^2-2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta}\ip{y,x}\right)+2\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{x,x} }\sqrt{\ip{y,y} }  }}
**#**********: {{MM|\implies 0\le\underbrace{\left(\sqrt{\ip{x,x} }-\sqrt{\ip{x,x} }\right)^2}_{\eq 0}-2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } e^{-j\theta}\vert\ip{y,x}\vert e^{j\theta}\right)+2\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } }}
**#**********: {{MM|\implies 0\le -2\text{Re}\left(\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\vert\ip{y,x}\vert e^0\right)+2\ip{x,x} }} - but {{M|e^0\eq 1}}<ref group="Note">From {{M|e^{-j\theta}e^{j\theta}\eq e^{j\theta-j\theta}\eq e^0}}</ref> so there is no imaginary component of the thing in the {{M|\text{Re} }}
**#**********: {{MM|\implies 0\le 2\ip{x,x}-2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } } }}
**#**********: {{MM|\implies 2\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le 2\ip{x,x} }} {{MM|\implies\vert\ip{y,x}\vert\frac{\sqrt{\ip{x,x} } }{\sqrt{\ip{y,y} } }\le \ip{x,x} }}
**#**********: {{M|\implies\vert{\ip{y,x} }\vert\sqrt{\ip{x,x} }\le \ip{x,x}\sqrt{\ip{y,y} } }}
**#*********** {{M|\implies\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } }} - ''almost'' as required
**#************ Note that {{M|\vert\ip{x,y}\vert\eq\vert\overline{\ip{x,y} }\vert\eq\vert\ip{y,x}\vert }}<ref group="Note">As for {{M|a+bj\in\mathbb{C} }} we see:
* {{M|\vert a+bj\vert:\eq \sqrt{a^2+b^2} }} and
* {{M|\vert \overline{a+bj}\vert\eq\vert a-bj\vert\eq\vert a+(-b)j\vert:\eq \sqrt{a^2+(-b)^2}\eq\sqrt{a^2+b^2}\eq\vert a+bj\vert}} as required</ref>
**#************* So {{M|\vert\ip{x,y}\vert\eq\vert\ip{y,x}\vert\le\sqrt{\ip{x,x} }\sqrt{\ip{y,y} } }}
**#************** Thus {{M|\vert\ip{x,y}\vert\le \sqrt{\ip{x,x} }\sqrt{\ip{y,y} } }} - as required
* Since {{M|x,y\in X}} were arbitrary we have shown the claim holds for all.
This completes the proof.
==References==
<references/>
==Notes==
<references group="Note"/>
{{Theorem Of|Functional Analysis|Analysis|Real Analysis|Topology|Metric Space}}