Multivariable Advanced Calculus

9.13. EXERCISES 2519. Suppose A is covered by a finite collection of Balls, F. Show that then there existsa disjoint collection of these balls, {B i } p i=1 , such that A ⊆ ∪p i=1 ̂B i where ̂B i hasthe same center as B i but 3 times the radius. Hint: Since the collection of ballsis finite, they can be arranged in order of decreasing radius.10. Let f be a function defined on an interval, (a, b). The Dini derivates are definedasD + f (x) ≡ lim infh→0+D + f (x) ≡ lim suph→0+D − f (x) ≡ lim infh→0+D − f (x) ≡ lim suph→0+f (x + h) − f (x),hf (x + h) − f (x)hf (x) − f (x − h),hf (x) − f (x − h).hSuppose f is continuous on (a, b) and for all x ∈ (a, b), D + f (x) ≥ 0. Show thatthen f is increasing on (a, b). Hint: Consider the function, H (x) ≡ f (x) (d − c)−x (f (d) − f (c)) where a < c < d < b. Thus H (c) = H (d). Also it is easy to seethat H cannot be constant if f (d) < f (c) due to the assumption that D + f (x) ≥ 0.If there exists x 1 ∈ (a, b) where H (x 1 ) > H (c), then let x 0 ∈ (c, d) be the pointwhere the maximum of f occurs. Consider D + f (x 0 ). If, on the other hand,H (x) < H (c) for all x ∈ (c, d), then consider D + H (c).11. ↑ Suppose in the situation of the above problem we only knowD + f (x) ≥ 0 a.e.Does the conclusion still follow? What if we only know D + f (x) ≥ 0 for everyx outside a countable set? Hint: In the case of D + f (x) ≥ 0,consider the badfunction in the exercises for the chapter on the construction of measures whichwas based on the Cantor set. In the case where D + f (x) ≥ 0 for all but countablymany x, by replacing f (x) with ˜f (x) ≡ f (x) + εx, consider the situation whereD + ˜f (x) > 0 for all but ( countably ) many x. If in this situation, ˜f (c) > ˜f (d) forsome c < d, and y ∈ ˜f (d) , ˜f (c) ,letz ≡ sup{x ∈ [c, d] : ˜f}(x) > y 0 .Show that ˜f (z) = y 0 and D + ˜f (z) ≤ 0. Conclude that if ˜f fails to be increasing,then D + ˜f (z) ≤ 0 for uncountably many points, z. Now draw a conclusion aboutf.12. ↑ Let f : [a, b] → R be increasing. Show⎛⎞N pq⎜[ { }} {m ⎝ D + f (x) > q > p > D + f (x) ] ⎟⎠ = 0 (9.23)and conclude that aside from a set of measure zero, D + f (x) = D + f (x). Similarreasoning will show D − f (x) = D − f (x) a.e. and D + f (x) = D − f (x) a.e. and sooff some set of measure zero, we haveD − f (x) = D − f (x) = D + f (x) = D + f (x)