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Question Number 205607 by SANOGO last updated on 25/Mar/24
soit H espace de Hilbert  montrez que:  d(x,{a})^⌊  = ((∣<x,a>∣)/(∣∣a∣∣))
$${soit}\:{H}\:{espace}\:{de}\:{Hilbert} \\ $$$${montrez}\:{que}: \\ $$$${d}\left({x},\left\{{a}\right\}\right)^{\lfloor} \:=\:\frac{\mid<{x},{a}>\mid}{\mid\mid{a}\mid\mid} \\ $$
Answered by Berbere last updated on 25/Mar/24
(a/(∣a∣)),a_2 .....a_n  une base orthonmal  x=Σ_(i=1) ^n <x,(a_i /(∣a∣))>(a_i /(∣a∣))  x=<x,(a/(∣a∣))>(a/(∣a∣))+y;y∈{a}^⊥   d(x,{a})∣<x,(a/(∣a∣))>∣=∣(1/(∣a∣))<x,a>∣=((<x,a>)/(∣∣a∣∣))
$$\frac{{a}}{\mid{a}\mid},{a}_{\mathrm{2}} …..{a}_{{n}} \:{une}\:{base}\:{orthonmal} \\ $$$${x}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}<{x},\frac{{a}_{{i}} }{\mid{a}\mid}>\frac{{a}_{{i}} }{\mid{a}\mid} \\ $$$${x}=<{x},\frac{{a}}{\mid{a}\mid}>\frac{{a}}{\mid{a}\mid}+{y};{y}\in\left\{{a}\right\}^{\bot} \\ $$$${d}\left({x},\left\{{a}\right\}\right)\mid<{x},\frac{{a}}{\mid{a}\mid}>\mid=\mid\frac{\mathrm{1}}{\mid{a}\mid}<{x},{a}>\mid=\frac{<{x},{a}>}{\mid\mid{a}\mid\mid} \\ $$
Commented by SANOGO last updated on 25/Mar/24
thank you
$${thank}\:{you} \\ $$

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