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Minkowski’s Inequality. Here the proof is also analogous to that for sequences. Let us estimate the integral Ω | f (x) + g(x)| p dΩ (p > 1) . 22 1 Metric, Banach, and Hilbert Spaces We have Ω | f (x) + g(x)| p dΩ = ≤ ≤ Ω Ω | f (x) + g(x)| | f (x) + g(x)| p−1 dΩ | f (x) + g(x)| p−1 | f (x)| dΩ + Ω | f (x) + g(x)|q(p−1) dΩ + Ω Ω | f (x) + g(x)| p−1 |g(x)| dΩ 1/q | f (x) + g(x)|q(p−1) dΩ Ω 1/q | f (x)| p dΩ Ω 1/p |g(x)| p dΩ 1/p . 6) with f1 = f and f2 = g. Again, only convergence of the integrals of | f (x)| p and |g(x)|q over Ω is assumed.
Proceeding in this way, all the elements of all the sets are put into one-to-one correspondence with the sequence of positive integers. 1. The set of all rational numbers is countable. Proof. A rational number is represented in the form i/ j where i and j are integers. Denoting ai j = i/ j, we obtain the sequence ai j which satisﬁes the condition of the theorem. 1. Show that the set of all polynomials with rational coeﬃcients is countable. 2. The set of real numbers of the segment [0, 1] is not countable.
Xn ) − f 1 (t, x2 , . . , xn ) dt . a Using the identity ∂ fi (t, x2 , . . , xn ) dt ∂t x1 fi (x1 , . . , xn ) − fi (a, . . , xn ) = a we have Δ = [ f (x1 , . . , xn ) − fi (x1 , . . , xn )] − [ f (a, x2 , . . , xn ) − fi (a, x2 , . . , xn )] x1 − f 1 (t, x2 , . . , xn ) − a ∂ fi (t, x2 , . . , xn ) dt . , x1 f (x1 , x2 , . . , xn ) − f (a, x2 , . . , xn ) = f 1 (t, x2 , . . , xn ) dt . a Thus ∂ f (x) = f 1 (x) . ∂x1 6. The space H k,λ (Ω). The H¨older space H k,λ (Ω), 0 < λ ≤ 1, consists of those functions of C (k) (Ω) whose norms in H k,λ (Ω), deﬁned by max |Dα f (x)| + f = 0≤|α|≤k x∈Ω |Dα f (x) − Dα f (y)| , |x − y|λ x,y∈Ω sup |α|=k x y 48 1 Metric, Banach, and Hilbert Spaces are ﬁnite.