Introduction Notes

Consider for t r f t y tw x2 y x tw2 then by

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Unformatted text preview: and x ∈ Cm , then a vector y ∈ V0 satisfies y − x ⊥ V0 if and only if ￿y − x￿ ≤ inf ￿v − x￿ . v ∈V 0 Moreover, there is only one such point y in V0 with these properties. Proof. First we assume v ∈ V0 , y − x ⊥ x. Choose any u ∈ V0 , u ￿= y , then u = y + w with w ￿= 0. Consider for t ∈ R f (t) = ￿y + tw − x￿2 = ￿(y − x) + tw￿2 then by the orthogonality of y − x and w ∈ V0 , f (t) = ￿y − x￿2 + t2 ￿w￿2 which has a global minimum at t = 0. Since w ￿= 0, f (0) = ￿P x − x￿2 < f (1) = ￿u − x + w￿2 = ￿u − x￿2 , so among all u ∈ V0 the unique minimizer for the distance ￿u − x￿ is y . Conversely, assume that there is y ∈ V0 which satisfies that ￿y − x￿ ≤ inf v∈V0 ￿v − x￿. Choosing any w ∈ V0 \ {0} and α ∈ C then gives that f (t) = ￿y + tαw − x￿2 is minimal at t = 0, so 0 = f ￿ (0) = 2Reα￿w, y − x￿. Taking the supremum of the real part over all complex α gives that ￿w, y − x￿ = 0 which me...
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