Analysis II (v. 2) by Herbert Amann, Joachim Escher

By Herbert Amann, Joachim Escher

The second one quantity of this advent into research bargains with the mixing thought of services of 1 variable, the multidimensional differential calculus and the idea of curves and line integrals. the fashionable and transparent improvement that begun in quantity I is sustained. during this means a sustainable foundation is created which permits the reader to house fascinating purposes that usually transcend fabric represented in conventional textbooks. this is applicable, for example, to the exploration of Nemytskii operators which permit a clear advent into the calculus of diversifications and the derivation of the Euler-Lagrange equations.

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7 Suppose f ∈ C 1 (I, R) has f (α) = f (β) = 0. Show that f 2 ∞ ≤ 1 2 β f 2 + (f )2 dx . 10) be modified when f ∈ C 1 (I, R) but only f (α) = 0? x (Hint: Suppose x0 ∈ I has f 2 (x0 ) = f 2∞ . ) 8 p β x0 f f dx. Suppose f ∈ C 1 (I, K) has f (α) = 0. Show that β |f f | dx ≤ α β−α 2 β |f |2 dx . α 5 If · 1 and · 2 are norms on a vector space E, we say · 1 is stronger than · 2 if there is a constant K ≥ 1 such that x 2 ≤ K x 1 for all x ∈ E. We say weaker in the opposite case. ) The function f ∈ C 2 (I, R) satisfies f ≤ f and f (α) = f (β) = 0.

Therefore x F (x) := 0 dy 1 − y4 for 0 ≤ x < 1 , is a well-defined antiderivative of f . Because f (x) > 0 for x ∈ (0, 1), F is strictly increasing. 7, F has a well-defined inverse function G. It is known5 that there is a subset M of C that is countable and has no limit points, such that G has a uniquely determined analytic continuation G on C \ M . It is also known that G is doubly periodic, that is, there are two R-linearly independent periods ω1 , ω2 ∈ C such that G(z + ω1 ) = G(z + ω2 ) = G(z) for z ∈ C\M .

54 VI Integral calculus in one variable have ∞ Fx (z) = exz f (z) = xk z k k! k=0 ∞ n n=0 k=0 n ∞ j=0 Bj j z = j! n=0 n k=0 xn−k Bk n z (n − k)! k! n = and, alternately, Fx (z) = ∞ n z Bk xn−k n! k Bn (x)z n /n! The statement (ii) follows immediately from (i). Likewise from (i) we get (iii) because n n+1 Bk X n−k (n + 1 − k) Bn+1 (X) = k k=0 n = (n + 1) k=0 n Bk X n−k = (n + 1)Bn (X) . k Finally, (iv) and (v) follow from Fx+1 (z) − Fx (z) = zexz and F1−x (z) = Fx (−z) by comparing coefficients. 7 Corollary The first four Bernoulli polynomials read B0 (X) = 1 , B1 (X) = X − 1/2 , B2 (X) = X − X + 1/6 , B3 (X) = X 3 − 3X 2 /2 + X/2 .

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