... (a 2 , b 2 ). But nowwe have a cover of [c, d] by n − 1 intervals:[c, d] ⊂ (a1, b 2 ) ∪ni=3(ai, b i).So by inductiond − c ≤ (b 2 − a1) +ni=3 (b i− ai).But b 2 − a1≤ (b 2 − ... numbers a, b and ζ the inequality (ζa −ζ−1 b) 2 ≥ 0 implies that2ab ≤ ζ 2 a 2 + ζ 2 b 2 . Taking ζ 2 = m 2 +1we can replace the second term on theright in the preceding estimate for |I 2 | ... B] := AB −BA.Notice that [A, B] = − [B, A] and [I, B] = 0 for any B. So if A and B areself adjoint, so is i[A, B] and replacing A by A −AI and B by B − B I doesnot change ∆A, B or i[A, B] .The...