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By Choudhary P.

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Proof. Clearly, the proposition holds for p = 1 or p = ∞. Consequently, it is enough to assume 1 < p < ∞. Clearly u+v p ≤ ( u + v )p + 2p( u p + v p) which yields u + v ∈ L p (Ω ). Now u+v p p = = ≤ Ω Ω Ω u(x) + v(x) p dx u(x) + v(x) p−1 · u(x) − v(x) dx u(x) + v(x) p−1 · u(x) dx + Ω u(x) + v(x) p−1 · v(x) dx. 81) it follows that u+v which yields u + v p ≤ u p p ≤ u+v p+ p−1 p ( u p+ v p ), v p. 83 shows that L p (Ω ) for 1 ≤ p ≤ ∞ is a vector space and hence · p ) is a normed vector space. 84 (Fischer–Riesz).

105. Let Ω ⊂ Rn be an open subset and let H = L2 (Ω ), which we equip with its natural inner product given by u, v 2 := Ω u(x1 , x2 , . . , xn )v(x1 , x2 , . . , xn )dx1 dx2 . . 3 Hilbert Spaces 35 1/2 u 2= Ω |u(x1 , x2 , . . , xn )|2 dx1 dx2 . . dxn The space (L2 (Ω ), ·, · 2 , · 2) . is complete and hence is a Hilbert space. 106. Let Ω ⊂ Rn be an open subset and let H = H k (Ω ), which we equip with its natural inner product defined by u, v H k (Ω ) := ∑ Ω |α |≤k Dα u(x)Dα v(x)dx for all u, v ∈ H k (Ω ), whose associated norm is 1/2 u The space (H k (Ω ), ·, · Hk ∑ = Hk , Ω |α |≤k · Hk ) |Dα u(x)|2 dx .

Equivalently, FT = {x}. ¯ Proof (Existence of a fixed-point). Choose any arbitrary point y ∈ X and consider the sequence (xn )n∈N ⊂ X defined by 18 1 Metric, Banach, and Hilbert Spaces x0 = y, and xn+1 = T (xn ), n = 1, 2, 3, . . Using the fact that T is Lipschitz it follows that d(x2 , x1 ) ≤ Ld(x1 , y), d(x3 , x2 ) ≤ Ld(x2 , x1 ) ≤ L2 d(x1 , y), ... d(xn+1 , xn ) ≤ Ln d(x1 , y), and therefore d(xn+m , xn ) ≤ d(xn+m , xn+m−1 ) + d(xn+m−1 , xn+m−2 ) + · · · + d(xn+1 , xn ) ≤ (Lm−1 + Lm−2 + · · · + L + 1)Ln d(x1 , y) ≤ Ln d(x1 , y).

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