Exercise 13, § 1

Functions of a Real Variable, II, 81

Let $f$ be a real regulated function on a compact interval $[a,\ b]$ . Put

r(n) = \frac{b-a}{n} \sum_{k=1}^n f\left(a + k \frac{b-a}{n}\right) - \int_a^b f(t)\,dt.

a) Show that if $f$ is increasing on $[a,\ b]$ one has

0 \leq r(n) \leq \frac{b-a}{n} \left(f(b) - f(a)\right).

b) If $f$ is continuous and admits a regulated bounded right derivative on $[a,\ b[$ , show that one has $\lim_{n \to \infty} n r(n) = \frac{b-a}{2} (f(b) - f(a))$ (putting $x_k = a + k \frac{b-a}{n}$ , remark that one has

r(n) = \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} \left(f(x_{k+1}) - f(t)\right) dt,

and apply prop. 5 of II, p. 57).

c) Give an example of a function $f$ , increasing and continuous on $[a,\ b]$ , such that $n r(n)$ does not tend to $\frac{b-a}{2} (f(b) - f(a))$ as $n$ grows indefinitely [take for $f$ the limit of a decreasing sequence $(f_n)$ of increasing functions whose graphs are broken lines such that

f_n \left( a + k \frac{b-a}{2^n} \right) = f \left( a + k \frac{b-a}{2^n} \right) \quad \text{for} \quad 0 \leq k \leq 2^n

and

(b-a) \sum_{k=1}^{2^n} f_n \left( a + k \frac{b-a}{2^n} \right) - 2^n \int_a^b f_n(t) \, dt \geq \frac{3}{4}(b-a)(f_n(b) - f_n(a)) \Big].

Solution

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