SOLUTIONS PART A

A. Math 25a: Homework 4A

A.1. A regularization operator. Let {x_{n}}_{n≥1} be a sequence of real numbers. Set

y_{n} = \frac{1}{n} (x_{1} + x_{2} + ⋯ + x_{n})

If y_{n} → y for some y ∈ ℝ, we say that {x_{n}}_{n≥1} Cesàro-converges to y.

(1) Show that if x_{n} converges to x, then it also Cesàro-converges to x.

(2) Find the Cesàro-limit of the sequence {x_{n} = (−1)^{n}}.

1. If {x_{n}}_{n≥1} converges, then it is also bounded and we fix M such that |x_{n} − x| ≤ M for all n ≥ 1. Choose ε > 0 and let N be such that |x_{n} − x| < ε if n ≥ N. For m ≥ 1, we have

|y_{N+m} − x| = |\frac{1}{N + m}(x_{1} + ⋯ + x_{N+m}) − x|

= |\frac{1}{N + m}((x_{1} − x) + ⋯ + (x_{N+m} − x))|

= |\frac{1}{N + m}((x_{1} − x) + ⋯ + (x_{N} − x)) + \frac{1}{N + m}((x_{N+1} − x) + ⋯ + (x_{N+m} − x))|

≤ \frac{1}{N + m}(MN) + \frac{1}{N + m}(mε)

This shows that for any m such that MN/(N +m) < ε, we have |y_{N+m} −x| < ε which shows the convergence of y_{n} to x. □

2. Observe that for n ≥ 1, x_{2n+1} = −1 and x_{2n} = 1, which means that

y_{m} = \begin{cases}−\frac{1}{m} & for m = 2n + 1 \\ 0 & for m = 2n. \end{cases}

Hence, the Cesàro-limit of the sequence {x_{n} = (−1)^{n}} is 0. □

A.2. Extending continuous functions. Let f : [0, 1) → ℝ be a continuous function. Show that one can extend f to a continuous function on [0, 1] if and only if f is uniformly continuous on [0, 1).

2 SOLUTIONS PART A

Proof. If f extends to a continuous function f : [0, 1] → ℝ, then f is uniformly continuous on [0, 1] by Heine’s theorem, since [0, 1] is compact. Its restriction to [0, 1) will also be uniformly continuous then.

Suppose now that f is uniformly continuous on [0, 1). We need to define f(1) so that the extended function will be continuous. Note that we proved this in class in more generality! We’ll do it here anyway. We claim:

(1) if (s_{n}) is a sequence in [0, 1) converging to 1, then {f(s_{n})} converges;

(2) if {s_{n}} and {t_{n}} are sequences in [0, 1) converging to 1, then lim f(s_{n}) = lim f(t_{n}). We will prove this claim below, but observe first that if it is true, then one can define f(1) as the limit of any sequence {f(s_{n})} where (s_{n}) is a sequence in [0, 1) converging to 1. The resulting f : [0, 1] → ℝ will be continuous by the sequential definition for continuity.

To prove (1), note that {s_{n}} is a Cauchy sequence, and since f is uniformly continuous, {f(s_{n})} is also a Cauchy sequence (see the more general proof given in class). Hence {f(s_{n})} converges.

To prove (2), we define a third sequence {u_{n}} such that {s_{n}} and {t_{n}} are both subsequences of {u_{n}}. This is easily accomplished by interlacing {s_{n}} and {t_{n}}:

{u_{n}}_{n=1}^{∞} = {s_{1}, t_{1}, s_{2}, t_{2}, s_{3}, t_{3}, …}.

We see that lim u_{n} = 1, which implies that lim f(u_{n}) exists. The two subsequences {f(s_{n})} and {f(t_{n})} of {f(u_{n})} both must converge to lim f(u_{n}), so lim f(s_{n}) = lim f(t_{n}). This proves the two claims. □

A.3. Discontinuities of non-decreasing functions. Let f : ℝ → ℝ be a non-decreasing function.

(1) Show that f is continuous at x ∈ ℝ if and only if sup_{y<x} f(y) = f(x) and inf_{y>x} f(y) = f(x) (draw a picture!).

(2) Prove that the set of x ∈ ℝ at which f is not continuous is countable.

1. (⇒) This implication does not use the fact that f is non-decreasing. Suppose that f is continuous at x ∈ ℝ. Then for any sequence x_{n} → x, we have f(x_{n}) → f(x), so if x_{n} is a sequence of real numbers such that f(x_{n}) → inf_{y>x} f(y), then f(x_{n}) → f(x) and therefore f(x) = inf_{y>x} f(y). The exact same argument will show that sup_{y<x} f(y) = f(x).

(⇐) Let f be a non-decreasing function and x ∈ ℝ such that sup_{y<x} f(y) = f(x) and inf_{y>x} f(y) = f(x). We will prove that f is continuous at x. Choose ε > 0. Because sup_{y<x} f(y) = f(x), there exists z < x with f(x) > f(z) > f(x) − ε. Similarly, because

SOLUTIONS PART A 3

inf_{y>x} f(y) = f(x), there exists w > x with f(x) < f(w) < f(x) + ε. Since f is nondecreasing, we see that for all y in the interval (w, z) we must have |f(y) − f(x)| < ε. This shows that f is continuous. □

2. If f is discontinuous at x, then choose a rational number q_{x} in the interval

(sup_{z<x} f(z), inf_{y>x} f(y)).

This defines an injective map from the set of discontinuities of f to the rational numbers, hence the set of discontinuities of f is countable. □