(4.1) Let P : R → R be a polynomial which does not vanish identically. Use Fubini’s Theorem to show, by induction on the dimension n, that Z(P ) = {x ∈ R|P (x) = 0} has Lebesgue measure equal to zero. proof By induction on n. For n = 1, consider a polynomial P ∈ R[X], and let d = degP . Then P has at most d roots, and Z(P ) = {x ∈ R|P (x) = 0} is finite, thus has measure 0. Now let n ≥ 1, and a...

we have by construction μ(En) = an. An important observation is that a subset of consecutive intervals En, En+1, . . . , En+k are either pairwise disjoint, or they cover the whole interval [0, 1]. Now for all n, let fn = 1En . Then we have • ‖fn‖1 = ∫ X |1En |dμ = μ(En) = an. • for all x ∈ [0, 1], (fn(x))n does not converge. Indeed, if we suppose by contradiction that (fn(x))n converges for som...

(10.1) Let k ∈ N, and let X = C k ([0, 1] d) to be the set of all functions f ∈ C 0 ([0, 1] d) such that for all α ∈ ∆ k , f (α) = ∂f ∂x α 1 1 ...∂x α d d exists and is continuous on (0, 1) d , and extends continuously on [0, 1] d. Here

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 1. The Endocannabinoid System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 1.1 Cannabinoid CB1 Receptor Expression . . . . . . . . . . . . . . . ....

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(4) Let a > 1 be an integer. Prove that for any positive integers n, d that d divides n if and only if a − 1 divides a − 1 (cf. the previous exercise). Conclude in particular that Fpd ⊆ Fpn if and only if p divides n. Proof. The proof from the previous problem shows that d divides n if and only if a − 1 divides a − 1; the only change is that in the last step we argue by size rather than degree ...

in fact we have equality. We have {is, s ∈ S} is a family of linear operators bounded on X∗, since for all `, sups∈S ix(`) = sups∈S `(s) which is finite by assumption. Therefore by the Uniform Boundedness Principle (X ∗ is a Banach space), {is, s ∈ S} are uniformly bounded, i.e. there exists M such that for all s ∈ S, ‖is‖X∗∗ ≤ M , i.e. ‖s‖X ≤M . (11.3) Let X,Y be Banach spaces. Let L ∈ B(X,Y )...

(5.1) The contraction mapping principle Let 0 ≤ r < 1. Let (X, d) be a nonempty metric space and f : X → X be a function that is a strict contraction, that is, for all x, y ∈ X, d(f(x), f(y)) ≤ rd(x, y). Then f has a unique fixed point. proof Existence of a fixed point: Let x0 ∈ X, and define the sequence (xn)n by: for all n ∈ N, xn+1 = f(xn). Then (xn)n is a Cauchy sequence: first, we have by ...

proof E is a Borel set: for each n ∈ N, On is open as the union of open sets. Therefore E is the countable intersection of open sets, thus is a Borel set. E is of the second category: we first prove that O n is nowhere dense. Indeed, we have for all qk ∈ Q, B(qk, 1 n2k ) ⊂ On, thus qk ∈ int(On). Therefore Q ⊆ int(On), therefore R = cl(Q) ⊆ cl(int(On)), i.e. cl(int(On)) = R. Taking complements, ...