PROBLEM SET 4, 18.155DUE MIDNIGHT FRIDAY 7 OCTOBER, 2011Consider the space B(Rn) of complex Borel measures, defined as thedual space of C00(Rn), the space or bounded continous functions on Rnwhich ‘vanish at infinity’:Given > 0, φ ∈ C00(Rn) ∃ R > 0 s.t. |φ(x)| < in |x| > R.Recall that this is a Banach space with respect to the supremum normand hence B(Rn) is a Banach space with respect to the dual norm. Inthe second last question you only need to show weak convergence.(1) Show that L1(Rn) is mapped injectively into B(Rn) by integra-tion,µf(φ) =Zf(x)φ(x), f ∈ L1(Rn), φ ∈ C00(Rn).(2) [A] Show that convolution with an element ψ ∈ C0c(Rn), de-fined byµ ∗ ψ(y) = µ(ψ(y − ·)), y ∈ Rn, µ ∈ B(Rn)gives an element of C0∞(Rn), the space of bounded continuousfunctions.[B] Show that µ ∗ ψ can also be interpreted as an elementof B(Rn) by proving that for φ continuous of compact supportZ(µ ∗ ψ(y))φ(y)dy = µ(ˇψ ∗ φ),ˇψ(x) = ψ(−x).Hint: You can use without proof the fact that the Riemannintegral of a continuous function of compact support is givenby the limit of the sum over the points at 2−kZn⊂ Rn– theintegral lattice.(3) If χ ∈ C∞c(Rn) has χ(0) = 1 then for a given µ ∈ B(Rn),µk(φ) = µ(χkφ), φ ∈ C00(Rn), χk(x) = χ(xk)is a sequence in B(Rn) which (norm) converges to µ.Hint: Weak convergence should be fairly clear; to get normconvergence one can use a weak formulation of the norm formeasures. First split a measure into real and imaginary parts, so12 PROBLEM SET 4, 18.155 DUE MIDNIGHT FRIDAY 7 OCTOBER, 2011we can assume that µ is real on real function. Then, somewhatless trivially, split into positive and negative parts, whereµ+(ψ) = sup0≤φ≤ψµ(φ), 0 ≤ ψ, φ ∈ C00(Rn).A little work (it is in the old notes somewhere) is required toshow that this extends to be linear where µ+(ψ) = µ+(ψ+) −µ+(ψ−) in terms of splitting continuous functions into positiveand negative parts. Finally then for such a positive measure(non-negative on non-negative functions) one can check thatthe dual normkµ+k = limk→∞µ+(χk)(a bounded increasing sequence). Here, χ is chosen to have0χ ≤ 1 and = 1 near 0 χ(x/j)χ(x/k) = χ(x/k) when j >> kso (χ(x/k)µ)(χ(x/j)) = µ(χ(x/k)) and it follows from the limitformula thatkχ(x/k)µk = µ(χ(x/k))(using positivity). The right side increases to kmuk so|µ − χ(χ(x/k)µk = kµk − µ(χ(x/k)) → 0.The case of general χ with χ(x) = 1 near zero then follows.(4) Show that if γ ∈ C∞cis non-negative andRRnγ = 1 thenµ ∗ γk(ψ) → µ(ψ), where γk(x) = knγ(xk), µ ∈ B(Rn)for each ψ ∈ C00(Rn).(5) Show that restriction to S(Rn) ⊂ C0∞(Rn) gives an injection ofB(Rn) into
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