1.6. Differentiation theorems 115 Exercise 1.6.7. Let f : Rd → C, g : Rd → C be Lebesgue measurable functions such that f is absolutely integrable and g is essentially bounded (i.e. bounded outside of a null set). Show that the convolution f∗g : Rd → C defined by the formula f ∗ g(x) = Rd f(y)g(x − y) dy is well defined (in the sense that the integrand on the right-hand side is absolutely integrable) and that f ∗ g is a bounded, continuous function. The above exercise is illustrative of a more general intuition, which is that convolutions tend to be smoothing in nature the convolution f ∗ g of two functions is usually at least as regular as, and often more regular than, either of the two factors f, g. This smoothing phenomenon gives rise to an important fact, namely the Steinhaus theorem: Exercise 1.6.8 (Steinhaus theorem). Let E ⊂ Rd be a Lebesgue measur- able set of positive measure. Show that the set E − E := {x − y : x, y ∈ E} contains an open neighbourhood of the origin. (Hint: Reduce to the case when E is bounded, and then apply the previous exercise to the convolution 1E ∗ 1−E, where −E := {−y : y ∈ E}.) Exercise 1.6.9. A homomorphism f : Rd → C is a map with the property that f(x + y) = f(x) + f(y) for all x, y ∈ Rd. (i) Show that all measurable homomorphisms are continuous. (Hint: For any disk D centered at the origin in the complex plane, show that f −1(z + D) has positive measure for at least one z ∈ C, and then use the Steinhaus theorem from the previous exercise.) • Show that f is a measurable homomorphism if and only if it takes the form f(x1,...,xd) = x1z1 +...+xdzd for all x1,...,xd ∈ R and some complex coeﬃcients z1,...,zd. (Hint: First establish this for rational x1,...,xd, and then use the previous part of this exercise.) (ii) (For readers familiar with Zorn’s lemma, see §2.4 of An epsilon of room, Vol. I.) Show that there exist homomorphisms f : Rd → C which are not of the form in the previous exercise. (Hint: View Rd (or C) as a vector space over the rationals Q, and use the fact (from Zorn’s lemma) that every vector space—even an infinite- dimensional one—has at least one basis.) This gives an alternate construction of a non-measurable set to that given in previous notes. Remark 1.6.15. One drawback to the density argument is it gives con- vergence results which are qualitative rather than quantitative—there is

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