Prove (5.4). Hint: Let A be the event that X1, X2 ∈ [0, 4h], 253 5 h ≤ |X1 − X2| ≤ 2h, X3,...,Xn ∈ [6h, 1], and                   Y1 = Y2. Then P{A} > 0 and E  (mn(x) − m(x))2 µ(dx) ≥  E  |mn(x)|dx|A ...

Prove (5.4).

Hint: Let A be the event that X1, X2 ∈ [0, 4h], 253 5 h ≤ |X1 − X2| ≤ 2h, X3,...,Xn ∈ [6h, 1], and

                  Y1 = Y2. Then P{A} > 0 and E  (mn(x) − m(x))2 µ(dx) ≥  E

For the event A and for x ∈ [0, 4h], the quantity |mn(x)| is a ratio of |K((x − X1)/h) − K((x − X2)/h)| and |K((x − X1)/h) + K((x − X2)/h)|, such that there are two x’s for which the denominator is 0 and the numerator is positive, so the integral of |m(x)| is ∞.