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Category: Exercises

Dini’s Mutated Result
Problem 1. Let $\{f_n\}$ be a sequence of functions on $[0, 1]$ that satisfy the following properties:
- each $f_n$ is non-decreasing on $[0, 1]$ ,
- $f_n \to f$ pointwise,
- $f$ is continuous on $[0, 1]$ .
Prove that $f_n \to f$ uniformly. This is a modified convergence criterion from the more famous Dini’s theorem.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . Using the (uniform) continuity of $f$ , for any $k_1 > 0$ , there exists $\delta > 0$ such that

$|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k_1 \cdot \epsilon.$

By the Archimedean property, find $N_1 \in \mathbb N$ so that $1/N_1 < \delta$ . Partition $[0, 1]$ into $\{x_0, x_1,\dots,x_{N_1}\}$ , $x_i = i/N_1$ . For each $i$ , point-wise convergence $f_n \to f$ yields that for any $k_2 > 0$ , there exists $N_2 \in \mathbb N$ such that whenever $n > N_2$ ,

$n > N_2 \quad \Rightarrow \quad |f_n(x_i) - f(x_i)| < k_2 \cdot \epsilon,\quad i=0,1,2,\dots,N_1.$

Finally, fix $x \in [0, 1]$ and suppose $x \in [x_0, x_1]$ without loss of generality. Then for $n > N_2 + 1$ , point-wise monotonicity yields

$\begin{aligned} |f_n(x) - f(x)| &\leq |f_n(x) - f_n(x_1)| + |f_n(x_1) - f(x_1)| + |f(x_1) - f(x)| \\ &< |f_n(x) - f_n(x_1)| + k_2 \cdot \epsilon + k_1 \cdot \epsilon \\&< |f_n(x) - f_n(x_1)| + (k_1 + k_2) \cdot \epsilon .\end{aligned}$

To bound the first term, we use monotonicity to obtain

$\begin{aligned} |f_n(x) - f_n(x_1)| &\leq |f_n(x_0) - f_n(x_1)|\\ &\leq |f_n(x_0) - f(x_0)| + |f(x_0) - f(x_1)| + |f(x_1) - f_n(x_1)| \\ &< k_2 \cdot \epsilon + k_1 \cdot \epsilon + k_2 \cdot \epsilon \\ &= (k_1 + 2 k_2 )\cdot \epsilon. \end{aligned}$

Combining the estimates yields

$\begin{aligned} |f_n(x) - f(x)| &< |f_n(x) - f_n(x_1)| + (k_1 + k_2) \cdot \epsilon \\ &< (2k_1 + 3k_2) \cdot \epsilon.\end{aligned}$

Finally, set $k_1 = 1/4$ , $k_2 = 1/6$ to obtain the desired result.

—Joel Kindiak, 31 Jan 25, 1422H
June 11, 2025
Strengthening Point-wise Convergence
Problem 1. Let $\{f_n\}$ be a sequence of differentiable functions on $[0, 1]$ with the following properties:
- $f_n \to f$ pointwise on $[0, 1]$ ,
- $f$ is continuous on $[0, 1]$ ,
- there exists $M > 0$ such that $\displaystyle \sup \{\|f_n'\|_\infty : n \in \mathbb N\} \leq M$ .
Prove that $f_n \to f$ uniformly.

(Click for Solution)

Solution. The key insight here is that $f$ being continuous on $[0, 1]$ strengthens it to being uniformly continuous, so that we can gain some global control over our estimates.

Fix $\epsilon > 0$ . Since $f$ is (uniformly) continuous on $[a, b]$ , for any $k_1 > 0$ , there exists $\delta > 0$ such that for any $u,v \in [0, 1]$ ,

$|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k_1 \cdot \epsilon.$

By the Archimedean property, find $N_1 \in \mathbb N$ so that $1/N_1 < \delta$ . Partition $[0, 1]$ into $\{x_0, x_1,\dots,x_{N_1}\}$ , $x_i = i/N_1$ . For each $i$ , point-wise convergence yields that for any $k_2 > 0$ , there exists $N_2 \in \mathbb N$ such that whenever $n > N_2$ ,

$n > N_2 \quad \Rightarrow \quad |f_n(x_i) - f(x_i)| < k_2 \cdot \epsilon,\quad i=0,1,2,\dots,N_1.$

Finally, fix $x \in [0, 1]$ and suppose $x\in [x_0, x_1]$ without loss of generality. Using the uniform bound on the derivatives, the mean value theorem yields

$\displaystyle |f_n(x) - f_n(x_0)| \leq M \cdot |x - x_0| \leq M \cdot \frac 1{N_1}.$

Then for any $n > N_2$ , combining the estimates yields

$\displaystyle \begin{aligned} |f_n(x) - f(x)| &\leq |f_n(x)-f_n(x_0)| + |f_n(x_0) - f(x_0)| + |f(x_0) - f(x)|\\ &< M \cdot \frac 1{N_1}+ k_2 \cdot \epsilon + k_1 \cdot \epsilon \\ &=M \cdot \frac 1{N_1} + (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Taking care of the dependencies, set $k_1 = k_2 = 1/3$ and further stipulate $1/N_1 < \epsilon/(3M)$ to yield the desired result.

—Joel Kindiak, 31 Jan 25, 1221H
June 11, 2025
Functional Reverse-Engineering
Define $g : \mathbb R \to \mathbb R$ by $g(x) = x^2 - x + 1 = (x-1/2)^2 + 3/4$ .

Problem 1. Suppose there exists a function $f : \mathbb R \to \mathbb R$ such that $f \circ f = g$ . Evaluate $f(1)$ and $f(0)$ .

(Click for Solution)

Solution. We first observe that $(f \circ f)(1) = g(1) = 1$ . Applying $f$ on both sides,

$\begin{aligned}f((f \circ f)(1)) &= f(1) \\ (f \circ f)(f(1)) &= f(1) \\ f(1)^2 - f(1) + 1 &= f(1) \\ (f(1)-1)^2 &= 0.\end{aligned}$

Hence, $f(1) = 1$ . Similarly, $(f \circ f)(0) = g(0) = 1$ . Applying $f$ on both sides yields

$f(0)^2 - f(0) + 1 = 1 \quad \iff \quad f(0)(f(0) - 1) = 0.$

Hence, $f(0) = 0$ or $f(0) = 1$ . The former yields

$0 = f(0) = f(f(0)) = g(0) = 1,$

a blatant contradiction.

Problem 2. Define the sequence $\{x_n\}$ by $x_0 := 1/2$ , $x_1 := 5/8$ , and $x_{n+2} := g(x_n)$ for integers $n \geq 0$ . Prove that $x_n \to 1$ .
(Click for Solution)

Solution. We first make a few observations:
- $g$ is continuous.
- $g$ is increasing on $[1/2, \infty)$ .
- $g([1/2, 1]) = [3/4, 1]$ .
Thus, $g|_K$ is bijective to its image for any $K \subseteq [1/2, \infty)$ . By construction, $x_0 \leq x_1 \leq 3/4 = g(x_0) = x_2$ . Furthermore, for $n \geq 0$ ,

$x_{n+2} = g(x_n) \leq g(x_{n+1}) = x_{n+3}.$

Thus, by induction, $\{x_n\}$ is an increasing sequence. Since $\{x_n\}$ is bounded above by $1$ , by the monotone convergence theorem, $x_n \to r$ for some $r \in \mathbb R$ . By the continuity of $g$ ,

$\displaystyle r = \lim_{n \to \infty} x_{n+2} = \lim_{n \to \infty}g(x_n) = g \left( \lim_{n \to \infty} x_n \right) = g(r).$

But we have $g(r) = r \iff r = 1$ , so that $x_n \to 1$ , as required.
Problem 3. Construct a continuous function $f: \mathbb R \to \mathbb R$ such that $f \circ f = g$ .
(Click for Solution)

Solution. We follow the construction in this post by user pco. We will construct $f$ on the following subsets of $\mathbb R$ :
- $[1/2, 1)$ ,
- $\{1\}$ ,
- $(1, \infty)$ ,
- $(-\infty, 1/2)$ .
For the interval $[1/2, 1)$ , define the sequence $\{x_n\}$ according to Problem 2. Let $p_{a, b} : [0, 1) \to [a, b)$ denote the canonical continuous bijection defined by

$\displaystyle p_{a,b}(t) = a + t \cdot (b-a).$

Then define the sequence $\{h_n\}$ of bijections $h_n : [x_n, x_{n+1}) \to [x_{n+1}, x_{n+2})$ by

$h_0 = p_{x_1, x_2} \circ p_{x_0,x_1}^{-1},\quad h_{n+1} = g|_{[x_n, x_{n+1})} \circ h_n^{-1}.$

Thus, for $x \in [1/2, 1)$ , we define

$f(x) = h_n(x) \quad \iff \quad x \in [x_n, x_{n+1}).$

By construction,

$\begin{aligned} (f \circ f)(x) &= f(f(x)) \\ &= f(h_n(x)) \\ &= h_{n+1}(h_n(x)) \\ &= (h_{n+1} \circ h_n)(x) = g(x).\end{aligned}$

Thus, we have constructed $f$ on $[1/2, 1)$ . Since we require $f$ to satisfy

$f(f(x)) = x^2 - x + 1,$

Problem 1 stipulates that $f(1) = 1$ . Furthermore, this definition allows $f$ to be continuous by Problem 2. To define $f$ on $(1, \infty)$ , we first define $f$ on $[2, \infty)$ . Define the sequence $\{y_n\}$ similarly to $\{x_n\}$ as in Problem 2, except that we now have the initial conditions $y_0 := 2$ and $y_1 := (2+g(2))/2 = 5/2$ .

Just like before, $y_0 \leq y_1 \leq y_2$ and since $g$ is increasing, $\{ y_n \}$ is an increasing sequence by the argument in Problem 2. For divergence to infinity, we recall that

$y_{n+2} = y_n \cdot (y_n - 1) + 1 \geq y_n + 1.$

If $y_n$ converges to a limit, then $0 \geq 1$ , a blatant contradiction. Thus, $\{y_n\}$ diverges. Since $\{y_n\}$ is strictly increasing, it is unbounded. Thus, $y_n \to \infty$ .

Defining $\{ h_n \}$ similarly as in the $[1/2, 1)$ case, we define

$f(x) = h_n(x)\quad \iff \quad x \in [y_n, y_{n+1})$

and deduce $f \circ f = g$ on that domain too.

We next define $f$ on $(1, 2)$ . Define the sequence $\{z_n\}$ by $z_0 := 5/2$ , $z_1 := 2$ , and $z_{n+2} := g|_{(1,5/2)}^{-1}(z_n)$ . By a similar proof as in Problem 2, since $z_2 \leq z_1 \leq z_0$ and $g_{(1,2)}^{-1}$ is decreasing, $z_n$ decreases to $1$ .

Define the bijections $l_n : [z_{n+2},z_{n+1}) \to [z_{n+1}, z_n)$ in a similar manner to Problem 2, but in the reverse direction:

$l_0 = p_{z_1, z_0} \circ p_{z_2,z_1}^{-1}, \quad l_{n+1} = l_n^{-1} \circ g|_{[z_{n+3}, z_{n+2})}.$

Finally, define

$f(x) = l_n(x)\quad \iff \quad x \in [z_n, z_{n+1}).$

By construction,

$\begin{aligned} (f \circ f)(x) &=(l_n \circ l_{n+1})(x) = g(x).\end{aligned}$

Finally, for the domain $(-\infty, 1/2)$ , define $f(x) := f(1-x)$ , since

$\begin{aligned} (f \circ f)(x) &= f(f(x)) \\ &= f(f(1-x)) \\ &= (f \circ f)(1-x) \\ &= g(1-x) = g(x). \end{aligned}$
—Joel Kindiak, 11 Jun 25, 0008H
June 11, 2025
A Less-Dominated Convergence Theorem
Problem 1. Let $\{f_n\}$ be a sequence of functions defined on $[0,\infty)$ satisfying the following properties:
- for any interval of the form $[0, b]$ , $f_n \to f$ uniformly on $[a, b]$ ,
- the improper integrals $\displaystyle \int_0^\infty f_n$ and $\displaystyle \int_0^\infty f$ converge,
- there exists a nonnegative integrable function $g$ such that $|f_n| \leq g$ ,
- $\displaystyle \int_0^\infty g$ converges.
Prove that $\displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f$ .

Remark 1. This is a weak form of the dominated convergence theorem. The strong form which requires measure-theoretic machinery only requires point-wise convergence.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . Our line of attack is to find sufficiently large $M > 0$ so that integrals of the form $\displaystyle \int_M^\infty$ can be controlled due to the integrals converging, while integrals of the form $\displaystyle \int_0^M$ can be controlled using the uniform convergence of $f_n \to f$ .

To that end, let’s first ascertain that $|f| \leq g$ point-wise. Fix $x \geq 0$ . Applying the uniform convergence $f_n \to f$ on $[0, x]$ , for any $k_1 > 0$ , there exists $N_1 \in \mathbb N$ such that

$n > N_1 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_1 \cdot \epsilon.$

This implies the upper-found

$|f(x)| \leq |f_n(x)| + |f_n(x) - f(x)| < g(x) +k_1 \cdot \epsilon.$

Since $\epsilon > 0$ is arbitrary, $|f| \leq g$ point-wise. Therefore,

$|f_n - f| \leq |f_n|+|f| \leq 2g$

point-wise. The motivation for this step is to do some upper bounds on tail integrals. Since $\displaystyle \int_0^\infty g$ converges, for any $k_2 > 0$ , there exists $M > 0$ such that

$\displaystyle \int_M^\infty g < k_2 \cdot \epsilon.$

Employing our estimate from previous discussions,

$\displaystyle \int_M^\infty |f_n - f| \leq 2 \int_M^\infty g = (2 k_2) \cdot \epsilon.$

Applying the uniform convergence $f_n \to f$ on $[0, M]$ , for any $k_3 > 0$ , there exists $N_2 \in \mathbb N$ such that

$n > N_2 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_3 \cdot \epsilon.$

This means for the infinite integrals, whenever $n > N_2$ ,

$\displaystyle \begin{aligned} \left| \int_0^\infty f_n - \int_0^\infty f \right| &\leq \int_0^\infty |f_n - f|\\ &\leq \int_0^M |f_n - f| + \int_M^\infty |f_n - f| \\ &\leq \int_0^M (k_3 \cdot \epsilon) + (2 k_2) \cdot \epsilon \\ &= (M \cdot k_3 + 2 \cdot k_2) \cdot \epsilon.\end{aligned}$

Taking care of the dependencies, setting $k_2 = 1/4$ , $k_3 = 1/(4M)$ yields the desired final result:

$\displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f.$

—Joel Kindiak, 31 Jan 25, 1213H
June 4, 2025
Relaxing the Fundamental Theorem

Problem 1. Let $f : [a, b] \to\mathbb R$ be differentiable such that $f'$ is Riemann-integrable. Prove that

$\displaystyle \int_a^b f' = f(b) - f(a),\quad \frac{\mathrm d}{\mathrm d x} \int_a^x f' = f'(x).$

(Click for Solution)

Solution. Fix $\epsilon > 0$ . For any $k > 0$ , let $P \subseteq [a, b]$ be any partition satisfying

$\displaystyle \sum_{i=1}^n (M_i(f', P) - m_i(f', P)) \Delta x_i < k \cdot \epsilon.$

For each $i = 1, \dots, n$ , use the mean value theorem to find $c_i \in (x_{i-1}, x_i)$ such that

$f(x_i) - f(x_{i-1}) = f'(c_i)\Delta x_i,$

and furthermore $f'(c_i) \in [m_i(f', P), M_i(f', P)]$ . This implies

$\displaystyle \begin{aligned} f(b) - f(a) - L(f', P)&= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \Delta x_i - L(f', P) \\ &= \sum_{i=1}^n (f'(c_i) - m_i(f',P))\Delta x_i \\ &\leq \sum_{i=1}^n (M_i(f', P) - m_i(f',P))\Delta x_i \in [0, k \cdot \epsilon).\end{aligned}$

Hence,

$L(f', P) \leq f(b) - f(a) < L(f', P) + k \cdot \epsilon \leq \mathcal L_a^b(f') + k \cdot \epsilon.$

Taking the supremum over $P$ ,

$\mathcal L_a^b(f') \leq f(b) - f(a) \leq \mathcal L_a^b(f') + k \cdot \epsilon.$

Setting $k = 1$ then taking $\epsilon \to 0^+$ ,

$\displaystyle f(b) - f(a) = \mathcal L_a^b(f') = \int_a^b f'.$

Adapting the proof to $[a, x] \subseteq [a, b]$ ,

$\displaystyle \int_a^x f' = f(x) - f(a)$

is differentiable with derivative

$\displaystyle \frac{\mathrm d}{\mathrm dx} \int_a^x f' = \frac{\mathrm d}{\mathrm dx}(f(x) - f(a)) = f'(x) - 0 = f'(x).$

—Joel Kindiak, 22 Jan 25, 0114H

May 21, 2025
The Popcorn Function

Problem 1. Prove that the popcorn function by $f : [0, 1] \to \mathbb R$ by

$\displaystyle f(x) = \begin{cases} 1/q, & x=p/q,\ p,q\ \text{coprime,} \\ 0, & \text{otherwise}. \end{cases}$

is Riemann-integrable.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . The key insight is to observe that for any $k > 0$ , there are only finitely many rationals $x$ such that $1/f(x) < 1/(k \cdot \epsilon) \iff f(x) > k \cdot \epsilon$ . Denote these rationals in ascending order by $\{r_1,\dots,r_M\}$ . To “separate” these rationals, define

$\displaystyle \delta := \frac 13 \cdot \min_{2 \leq i \leq M} (r_i - r_{i-1})$

Define the partition

$\displaystyle P := \{0,1\} \cup \bigcup_{i=1} \{r_i - \delta, r_i + \delta\} \subseteq [0, 1].$

Then

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i = 1}^{|P|} (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &\leq (1 - 0) \cdot (k \cdot \epsilon) + M \cdot 1 \cdot 2\delta\\ &= k \cdot \epsilon + 2M \cdot \delta. \end{aligned}$

To obtain the desired result, set $k = 1/4$ and refine the definition of $\delta$ to

$\displaystyle \delta := \min \left\{ \frac{\epsilon}{4M}, \frac 13 \cdot \min_{2 \leq i \leq M} (r_i - r_{i-1}) \right\}.$

—Joel Kindiak, 20 Jan 25, 2143H

May 14, 2025
Flavours of Integrability

In this post, we explore the various settings in which functions can be integrated. Here, we shorten “Riemann-integrable” by “integrable” for brevity.

Problem 1. Let $f : [a, b] \to \mathbb R$ be a bounded function. Suppose for any $c \in (a, b)$ , $f|_{[a,c]} : [a, c] \to \mathbb R$ is integrable. Prove that $f$ is integrable.

Solution. Since $f$ is bounded, there exists $M > 0$ such that for any $x \in [a, b]$ , $|f(x)| \leq M$ .

Fix $\epsilon > 0$ . Fix $\delta \in (0, b-a)$ to be tuned later. Since $f_{[a,b-\delta]}$ is integrable, for any $k > 0$ there exists a partition $P := \{x_0 = a, x_1,\dots,x_n = b-\delta\}$ such that

$U(f_{[a,b-\delta]}, P) - L(f_{[a,b-\delta]}, P) < k \cdot \epsilon.$

Define the partition $Q := \{x_0, x_1,\dots, x_n, x_{n+1} = b\}$ of $[a, b]$ . Expanding the definition,

$\begin{aligned} &U(f, Q) - L(f, Q)\\ &= U(f, P) - L(f, P) + (M_{n+1}(f, Q) - m_{n+1}(f, Q))\Delta_{n+1} \\ &< k \cdot \epsilon + (M+M) \cdot \delta \\ &< k \cdot \epsilon + 2M \cdot \delta. \end{aligned}$

Hence, set $k := 1/2$ and $\delta := \epsilon/(4M)$ to yield the result.

Henceforth, we will say $f : [a, b] \to \mathbb R$ is integrable on $K \subseteq [a, b]$ to mean that $f|_K : K \to \mathbb R$ is integrable.

Problem 2. Let $f : [a, b] \to \mathbb R$ be bounded. Suppose there exists $c \in (a, b)$ such that $f$ is integrable on $[a, c]$ and $[c, b]$ . Then $f$ is integrable on $[a, b]$ . Furthermore,

$\displaystyle \int_a^b f = \int_a^c f + \int_c^b f.$

Solution. Fix $\epsilon > 0$ . For any $k_1 > 0$ , find a partition $P$ of $[a, c]$ such that

$U(f, P) - L(f, P) < k_1 \cdot \epsilon.$

For any $k_2 > 0$ , find a partition $Q$ of $[c, b]$ such that

$U(f, Q) - L(f, Q) < k_2 \cdot \epsilon.$

Then $P \cup Q$ is a partition of $[a, b]$ , and

$\begin{aligned}&U(f, P \cup Q) - L(f, P \cup Q)\\ &= (U(f, P) - L(f, P)) + (U(f, Q) - L(f, Q)) \\ &< k_1 \cdot \epsilon + k_2 \cdot \epsilon \\ &= (k_1 + k_2) \cdot \epsilon.\end{aligned}$

Setting $k_1 = k_2 = 1/2$ yields the desired result.

Finally, we prove the integral identity in two directions. Fix $\epsilon > 0$ . Find a partition $P$ such that

$\displaystyle \int_a^b f < L(f, P) + \epsilon.$

Define the partitions

$\begin{aligned} P_- &:= \{x \in P : x < c\} \cup \{c\} \subseteq [a, c],\\ P_+ &:= \{x \in P : x > c\} \cup \{c\} \subseteq [c, b]. \end{aligned}$

Then

$\displaystyle \int_a^b f < L(f, P) + \epsilon \leq L(f, P_-) + L(f, P_+) + \epsilon \leq \int_a^c f + \int_c^b f+ \epsilon .$

Taking $\epsilon \to 0^+$ yields the inequality

$\displaystyle \int_a^b f \leq \int_a^c f + \int_c^b f.$

For the other direction, fix $\epsilon > 0$ . For $k_1,k_2 > 0$ , find partitions $P_- \subseteq [a, c]$ , $P_+ \subseteq [c, b]$ , such that

$\displaystyle\int_a^c f < L(f, P_-) + k_1 \cdot \epsilon \quad \int_c^b f < L(f, P_+) + k_2 \cdot \epsilon.$

Defining $P := P_- \cup P_+$ ,

$\begin{aligned} \int_a^c f + \int_c^b f & < (L(f, P_-) + L(f, P_+)) + (k_1 + k_2) \cdot \epsilon \\ & = L(f, P) + (k_1 + k_2) \cdot \epsilon \\ &\leq \int_a^b f + (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Setting $k_1 = k_2 = 1/2$ ,

$\displaystyle \int_a^c f + \int_c^b f < \int_a^b f + \epsilon.$

Taking $\epsilon \to 0^+$ ,

$\displaystyle \int_a^c f + \int_c^b f \leq \int_a^b f.$

Combining the results yields the desired result.

Problem 3. Let $f : [a, b] \to \mathbb R$ be continuous except at some $c \in (a, b)$ . Prove that $f$ is integrable.

Solution. For any $u \in (a, c)$ , $f$ is continuous on $[a, u]$ and thus integrable on $[a, u]$ . By Problem 1, $f$ is integrable on $[a, c]$ . Similarly, $f$ is integrable on $[c, b]$ . By Problem 2, $f$ is integrable on $[a, b]$ .

Problem 4. Let $f : [a, b] \to \mathbb R$ be increasing. Prove that $f$ is integrable.

Solution. We prove $[a, b] = [0,1]$ for simplicity. We observe that for any $x \in (a, b)$ , $f(a) < f(x) < f(b)$ so that $f$ is bounded. Fix $\epsilon > 0$ . For any $k > 0$ , use the Archimedean property to find $n \in \mathbb N$ such that $1/n < k \cdot \epsilon$ . Define $P := \{x_0, x_1, \dots, x_n\} \subseteq [0, 1]$ by $x_i = i/n$ . Then

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i=1}^n (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \cdot \frac 1n \\ &= \frac 1n \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \\ &= \frac 1n (f(x_n) - f(x_0)) \\ &= \frac {f(1) - f(0)}n \\ & < (f(1) - f(0)) \cdot k \cdot \epsilon. \end{aligned}$

Since $f(1) - f(0) > 0$ , set $k := 1/(f(1) - f(0)) > 0$ to yield the desired result.

—Joel Kindiak, 20 Jan 25, 1941H

May 7, 2025
Kronecker Approximation

After class one day, one of my students shared with me about this fascinating result. The source material comes from this video, and the writeup is my interpretation of its key ideas.

Let $\mathbb K$ be an ordered field.

Problem 1. Prove that for $\mathbb K$ -sequences $( x_n ), ( y_n )$ , if $x_n \to 0$ and $y_n$ is bounded, then $x_n \cdot y_n \to 0$ .

(Click for Solution)

Solution. Since $(y_n)$ is bounded, find $M > 0$ such that $|y_n| < M$ . Fix $\epsilon > 0$ . Since $x_n \to 0$ , for any $k > 0$ , there exists $N \in \mathbb N$ such that if $n > N$ , then $|x_n| < k \cdot \epsilon$ . By construction,

$|x_n \cdot y_n| = |x_n| \cdot | y_n | < k \cdot \epsilon \cdot M.$

Setting $k := 1/M > 0$ yields the result $x_n \cdot y_n \to 0$ .

For the rest of this exercise, let $\alpha$ be an irrational number.

Definition 1. For any $x \in \mathbb R$ , define the fractional part of $x$ by $\{ x \} := x - \lfloor x \rfloor$ .

Problem 2. Prove that for distinct positive integers $p, q$ , $\{p \alpha\} \neq \{q \alpha\}$ .

(Click for Solution)

Solution. Suppose for a contradiction that there exist distinct positive integers $p, q$ such that $\{p \alpha\} = \{q \alpha \}$ . By the definition of the fractional part,

$\begin{aligned}p \alpha - \lfloor p\alpha \rfloor &= q \alpha - \lfloor q \alpha \rfloor\quad \Rightarrow \quad \alpha = \frac{ \lfloor p\alpha \rfloor - \lfloor q \alpha \rfloor}{p - q} \in \mathbb Q, \end{aligned}$

a contradiction.

Problem 3. Prove that for any $n \in \mathbb N$ , there exist integers $p_n, q_n$ such that

$0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

(Click for Solution)

Solution. Fix $n \in \mathbb N$ . By Problem 2, consider the $n+1$ distinct values

$\{\alpha \},\{2\alpha\},\dots, \{n \alpha\}, \{(n+1)\alpha\}$

and the $n$ equally-spaced sub-intervals of $[0, 1)$ :

$[0, 1/n), [1/n, 2/n), \dots, [(n-1)/n, 1).$

This means there exists integers $i_n, p_n, q_n$ such that

$\{p_n \alpha\}, \{q_n\alpha\} \in [i_n/n, (i_n+1)/n)\quad \Rightarrow \quad 0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

Problem 4. Define

$A := \mathbb Z + \alpha \mathbb Z \equiv \{ m + n \alpha : m, n \in \mathbb Z\}.$

Prove that for any $x \in \mathbb R$ , there exists an $A$ -sequence $( x_n )$ such that $x_n \to x$ .

(Click for Solution)

Solution. Fix $x \in \mathbb R$ . Use Problem 3 to define the sequence $(y_n)$ by $y_n := \{ p_n \alpha \} - \{ q_n \alpha \}$ . It is clear that $y_n \to 0$ and each $y_n \neq 0$ . Define $x_n := y_n \cdot \lfloor x/y_n \rfloor$ . Then

$\displaystyle x_n = y_n \cdot \lfloor x/y_n \rfloor = y_n \cdot \left( \frac x{y_n} - \left\{ \frac x{y_n} \right\}\right)= x - y_n \left\{ \frac x{y_n} \right\}.$

By the definition of the fractional part, the sequence $( \{x / y_n \} )$ is bounded. Since $y_n \to 0$ , by Problem 1, we have $y_n \cdot \{x / y_n \} \to 0$ . Hence, $x_n \to x$ .

It remains to verify that $x_n \in A$ for each $n$ . Indeed,

$\begin{aligned} x_n &= y_n \cdot \lfloor x/y_n \rfloor \\ &= (\{ p_n \alpha \} - \{ q_n \alpha \}) \cdot \lfloor x/y_n \rfloor \\ &= \{ p_n \alpha \} \cdot \lfloor x/y_n \rfloor - \{ q_n \alpha \} \cdot \lfloor x/y_n \rfloor \\ &= (p_n \alpha - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor - (q_n \alpha - \lfloor q_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor \\ &= \underbrace{ (\lfloor q_n \alpha \rfloor - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor }_{\in \mathbb Z} +\, \alpha \cdot \underbrace{ (p_n - q_n) \cdot\lfloor x/y_n \rfloor }_{\in \mathbb Z} \\ &\in \mathbb Z + \alpha \mathbb Z = A, \end{aligned}$

as required.

Problem 5. Define

$B := \sin(\mathbb N) \equiv \{ \sin(n) : n \in \mathbb N\}.$

Prove that for any $y \in [-1, 1]$ , there exists a $B$ -sequence $( y_n )$ such that $y_n \to y$ .

(Click for Solution)

Solution. For any $y \in [-1, 1]$ , find $u \in (0, \infty) \subseteq \mathbb R$ such that $\sin(u) = y$ . Define

$C := \mathbb Z + 2\pi \mathbb Z.$

By Problem 4, there exists a $C$ -sequence $(u_n)$ such that $u_n \to u$ . Denote $u_n = s_n + 2\pi \cdot t_n$ with $s_n \in \mathbb N$ . Defining $y_n = \sin(s_n) \in B$ ,

$y_n = \sin(s_n) = \sin(s_n + 2\pi \cdot t_n) = \sin(u_n) \to \sin(u) = y.$

—Joel Kindiak, 24 Apr 25, 1851H

April 24, 2025
Investigating the Limit Superior

Some analysis courses introduce the idea of the limit superior, denoted $\displaystyle \limsup_{n \to \infty}$ , and its dual cousin the limit inferior, denoted $\displaystyle \liminf_{n \to \infty}$ . Why should we care?

For the rest of the problems, let $\{x_n\}$ be a bounded $\mathbb R$ -sequence satisfying $|x_n| \leq M$ for any $n \in \mathbb N$ .

Problem 1. Let $\mathcal X$ denote the subset of real numbers $x$ where there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k} \to x$ . Then $\sup \mathcal X$ and $\inf \mathcal X$ exist.

(Click for Solution)

Solution. Since $\{x_n\}$ is bounded, by the Bolzano-Weierstrass theorem, $\{x_n\}$ contains a convergent subsequence with limit $L \in \mathcal X$ , so that $\mathcal X$ is nonempty.

Fix $x \in \mathcal X$ , and suppose $x_{n_k} \to x$ . Then

$\displaystyle x = \lim_{k \to \infty} x_{n_k} \in [-M, M].$

Thus, $\mathcal X \subseteq [-M, M]$ and $\mathcal X$ is bounded. By the completeness of $\mathbb R$ , $\sup \mathcal X$ and $\inf \mathcal X$ exist.

Problem 2. Prove that the sequence $\displaystyle \{x_n^{\sup}\}$ defined by $\displaystyle x_n^{\sup} := \sup_{k \geq n} x_k$ converges. Note that $\displaystyle \{x_n^{\sup}\}$ may not be a subsequence of $\{x_{n_k}\}$ .

(Click for Solution)

Solution. By definition, $x_n^{\sup} \geq -M$ is non-increasing. By the monotone convergence theorem, $x_n^{\sup}$ converges.

Problem 3. Prove that $\displaystyle \lim_{n \to \infty} x_n^{\sup} = \sup \mathcal X$ .

(Click for Solution)

Solution. Denote $\displaystyle L_1 := \lim_{n \to \infty} x_n^{\sup}$ and $L_2 := \sup \mathcal X$ .

We aim to prove $L_1 \leq L_2$ and $L_2 \leq L_1$ .

We first prove $L_1 \leq L_2$ . Define $n_0 := 1$ and define $n_k$ for $k \in \mathbb N_0$ inductively as follows. Suppose we know $n_k$ . Then there exists $M_{k+1} \in \mathbb N$ such that for $j > \max\{n_k, M_{k+1}\}$ ,

$\displaystyle |x_j^{\sup} - L_1| < \frac 1{2(k+1)}.$

In particular, the estimate holds for $j = \max\{n_k, M_{k+1}\}+1 > n_k$ . By the definition of $x_j^{\sup}$ , find $n_{k+1} \geq j > n_k$ such that

$\displaystyle x_{n_{k+1}} > x_j^{\sup} - \frac{1}{2(k+1)}.$

Combining the estimates yields

$\displaystyle |x_{n_{k+1}} - L_1| < \frac 1{2(k+1)} \cdot 2 = \frac 1{k+1}.$

Now it is clear that we have constructed a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \to L_1$ , and

$\displaystyle L_1 = \lim_{k \to \infty} x_{n_k} \leq L_2,$

as required.

We next prove $L_2 \leq L_1$ . For any $x \in \mathcal X$ , let $\{x_{n_k}\}$ be a subsequence of $\{x_n\}$ such that $x_{n_k} \to x$ . Then

$\displaystyle x = \lim_{k \to \infty} x_{n_k} \leq \lim_{k \to \infty} \left( \sup_{j \geq n_k} x_j \right) = \lim_{k \to \infty} x_{n_k}^{\sup} = L_1.$

Since $x\in\mathcal X$ was arbitrary, $L_1$ is an upper bound for $\mathcal X$ , so that $L_2 \leq L_1$ , being the least upper bound for $\mathcal X$ .

On the left-hand side, we have the limit of the supremum of the tail sequence. On the right-hand side, we have the supremum of the limits of subsequences. Problem 3 therefore motivates the notation $\displaystyle \limsup_{n \to \infty}$ , since

$\displaystyle \limsup_{n \to \infty}x_n := \lim_{n \to \infty} x_n^{\sup} \equiv \sup \mathcal X.$

Problem 4. Prove that the sequence $\displaystyle \{x_n^{\inf}\}$ defined by $\displaystyle x_n^{\inf} := \inf_{k \geq n} x_k$ converges to $\inf \mathcal X$ .

(Click for Solution)

Solution. We observe that

$\displaystyle x_n^{\inf} := \inf_{k \geq n} x_k = - \sup_{k \geq n} (-x_k) \to -\sup(-\mathcal X) = \inf\mathcal X.$

It is obvious that convergent sequences are bounded, and it is not always true that bounded sequences converge. But, for any bounded sequence $\{x_n\}$ , $\displaystyle \limsup_{n \to \infty} x_n$ , $\displaystyle \liminf_{n \to \infty} x_n$ always exist. We can use their existence to formulate an equivalent criterion (i.e. a characterisation) for bounded sequences to converge.

Problem 5. Prove that a bounded sequence $\{x_n\}$ converges if and only if $\displaystyle \liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n$ .

(Click for Solution)

Solution. The right-hand side is equivalent to $\mathcal X = \{x\}$ for some $x \in \mathbb R$ , which yields the direction $(\Rightarrow)$ trivially.

For the direction $(\Leftarrow)$ , we observe that $x_n^{\inf} \leq x_n \leq x_n^{\sup}$ . Since $x_n^{\inf} \to x$ and $x_n^{\sup} \to x$ , $x_n \to x$ by the squeeze theorem.

—Joel Kindiak, 23 Jan 25, 1212H

April 10, 2025
When Integers Converge

Here’s the key question: when sequences of integers converge, are their limits integers? The answer is yes, but we need to put in a bit of effort to make that happen.

Problem 1. Let $K \subseteq \mathbb Z$ be nonempty and bounded above. Prove that $K$ has a maximum element.

(Click for Solution)

Solution. Since $K \subseteq \mathbb Z \subseteq \mathbb R$ is bounded above, there exists a real number $r$ such that for any $x \in K$ , $x \leq r \leq |r|$ . Use the Archimedean property to find $M \in \mathbb N$ such that $x \leq |r| \leq M \Rightarrow -x \geq -M$ .

Define $L := M + (-K) = \{M + (-x) : x \in K\} \subseteq \mathbb Z$ . Since $K$ is nonempty, so is $L$ . Furthermore, for any $x \in K$ ,

$M + (-x) \geq M + (-M) = 0.$

Thus, $L \subseteq \mathbb N$ . By the well-ordering principle, $L$ contains a minimum element $M + (-m)$ so that for any $x \in K$ ,

$M + (-x) \geq M + (-m) \quad \iff \quad x \leq m.$

By construction, $m \in K$ , so that $m = \max K$ , as required.

Problem 2. Let $x$ be a real number. Prove that there exists a unique integer, denoted $\lfloor x \rfloor$ , such that $\lfloor x \rfloor \leq x <\lfloor x \rfloor + 1$ .

The map $\lfloor \cdot \rfloor : \mathbb R \to \mathbb Z$ is called the floor function.

(Click for Solution)

Solution. Fix $x \in \mathbb R$ . For existence, consider the set

$K := \{n \in \mathbb Z : n \leq x\}.$

Let’s first check that $K$ is nonempty. If $x \geq 0$ , then $0 \in K$ and we are done. If $x < 0$ , then $-x > 0$ . Use the Archimedean property to find $n_0 \in \mathbb Z$ such that $-x \leq n_0 \Rightarrow -n_0 \leq x$ . Thus, $-n_0 \in K$ .

Since $K$ is bounded above and nonempty, by Problem 1, $K$ has a unique maximum element $\lfloor x \rfloor$ . This implies that $\lfloor x \rfloor + 1 \notin K$ . Thus, $\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1$ , as required.

Problem 3. Let $x_n$ be a $\mathbb Z$ -sequence that converges to $x$ . Prove that $x \in \mathbb Z$ .

(Click for Solution)

Solution. Suppose otherwise that $x \notin \mathbb Z$ . Then $x - \lfloor x \rfloor \in (0, 1)$ . Define $\epsilon := \min\{x - \lfloor x \rfloor, 1 - (x - \lfloor x \rfloor ) \} > 0$ . Observe that

$\epsilon \leq -x + \lfloor x \rfloor + 1\quad \Rightarrow \quad x + \epsilon \leq \lfloor x \rfloor + 1$

and

$-\epsilon \geq -(x - \lfloor x \rfloor ) \quad \Rightarrow \quad x -\epsilon \geq \lfloor x \rfloor.$

Since $x_n \to x$ , find $N \in \mathbb N$ such that for $n > N$ ,

$|x_n - x| < \epsilon \quad \Rightarrow \quad \lfloor x \rfloor \leq x - \epsilon < x_n < x + \epsilon \leq \lfloor x \rfloor + 1.$

Subtracting $\lfloor x \rfloor$ on both sides, $0 < x_n - \lfloor x \rfloor < 1$ . Since $x_n$ and $\lfloor x \rfloor$ are integers, $x_n - \lfloor x \rfloor \in \mathbb Z$ , a contradiction.

—Joel Kindiak, 5 Jan 25, 1825H

April 9, 2025

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