Tag Archives: analysis

A semi-continuous function with a dense set of points of discontinuity

August 20, 2017 Jean-Pierre Merx Leave a comment

Let’s come back to Thomae’s function which is defined as:
$f: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.$

We proved here that $f$ right-sided and left-sided limits vanish at all points. Therefore $\limsup\limits_{x \to a} f(x) \le f(a)$ at every point $a$ which proves that $f$ is upper semi-continuous on $\mathbb R$ . However $f$ is continuous at all $a \in \mathbb R \setminus \mathbb Q$ and discontinuous at all $a \in \mathbb Q$ .

Analysis

Converse of fundamental theorem of calculus

July 23, 2017 Jean-Pierre Merx Leave a comment

The fundamental theorem of calculus asserts that for a continuous real-valued function $f$ defined on a closed interval $[a,b]$ , the function $F$ defined for all $x \in [a,b]$ by
$F(x)=\int _{a}^{x}\!f(t)\,dt$ is uniformly continuous on $[a,b]$ , differentiable on the open interval $(a,b)$ and $F^\prime(x) = f(x)$
for all $x \in (a,b)$ .

The converse of fundamental theorem of calculus is not true as we see below.

Consider the function defined on the interval $[0,1]$ by $f(x)= \begin{cases} 2x\sin(1/x) – \cos(1/x) & \text{ for } x \neq 0 \\ 0 & \text{ for } x = 0 \end{cases}$ $f$ is integrable as it is continuous on $(0,1]$ and bounded on $[0,1]$ . Then $F(x)= \begin{cases} x^2 \sin \left( 1/x \right) & \text{ for } x \neq 0 \\ 0 & \text{ for } x = 0 \end{cases}$ $F$ is differentiable on $[0,1]$ . It is clear for $x \in (0,1]$ . $F$ is also differentiable at $0$ as for $x \neq 0$ we have $\left\vert \frac{F(x) – F(0)}{x-0} \right\vert = \left\vert \frac{F(x)}{x} \right\vert \le \left\vert x \right\vert.$ Consequently $F^\prime(0) = 0$ .

However $f$ is not continuous at $0$ as it does not have a right limit at $0$ .

Analysis

Counterexamples around series (part 2)

July 9, 2017 Jean-Pierre Merx Leave a comment

We follow the article counterexamples around series (part 1) providing additional funny series examples.

If $\sum u_n$ converges and $(u_n)$ is non-increasing then $u_n = o(1/n)$ ?

This is true. Let’s prove it.
The hypotheses imply that $(u_n)$ converges to zero. Therefore $u_n \ge 0$ for all $n \in \mathbb N$ . As $\sum u_n$ converges we have $\displaystyle \lim\limits_{n \to \infty} \sum_{k=n/2}^{n} u_k = 0.$ Hence for $\epsilon \gt 0$ , one can find $N \in \mathbb N$ such that $\epsilon \ge \sum_{k=n/2}^{n} u_k \ge \frac{1}{2} (n u_n) \ge 0$ for all $n \ge N$ . Which concludes the proof.

$\sum u_n$ convergent is equivalent to $\sum u_{2n}$ and $\sum u_{2n+1}$ convergent?

Is not true as we can see taking $u_n = \frac{(-1)^n}{n}$ . $\sum u_n$ converges according to the alternating series test. However for $n \in \mathbb N$ $\sum_{k=1}^n u_{2k} = \sum_{k=1}^n \frac{1}{2k} = 1/2 \sum_{k=1}^n \frac{1}{k}.$ Hence $\sum u_{2n}$ diverges as the harmonic series diverges.

$\sum u_n$ absolutely convergent is equivalent to $\sum u_{2n}$ and $\sum u_{2n+1}$ absolutely convergent?

This is true and the proof is left to the reader.

$\sum u_n$ is a positive convergent series then $(\sqrt[n]{u_n})$ is bounded?

Is true. If not, there would be a subsequence $(u_{\phi(n)})$ such that $\sqrt[\phi(n)]{u_{\phi(n)}} \ge 2$ . Which means $u_{\phi(n)} \ge 2^{\phi(n)}$ for all $n \in \mathbb N$ and implies that the sequence $(u_n)$ is unbounded. In contradiction with the convergence of the series $\sum u_n$ .

If $(u_n)$ is strictly positive with $u_n = o(1/n)$ then $\sum (-1)^n u_n$ converges?

It does not hold as we can see with $u_n=\begin{cases} \frac{1}{n \ln n} & n \equiv 0 [2] \\ \frac{1}{2^n} & n \equiv 1 [2] \end{cases}$ Then for $n \in \mathbb N$ $\sum_{k=1}^{2n} (-1)^k u_k \ge \sum_{k=1}^n \frac{1}{2k \ln 2k} – \sum_{k=1}^{2n} \frac{1}{2^k} \ge \sum_{k=1}^n \frac{1}{2k \ln 2k} – 1.$ As $\sum \frac{1}{2k \ln 2k}$ diverges as can be proven using the integral test with the function $x \mapsto \frac{1}{2x \ln 2x}$ , $\sum (-1)^n u_n$ also diverges.

Analysis

A nonzero continuous map orthogonal to all polynomials

June 25, 2017 Jean-Pierre Merx 1 Comment

Let’s consider the vector space $\mathcal{C}^0([a,b],\mathbb R)$ of continuous real functions defined on a compact interval $[a,b]$ . We can define an inner product on pairs of elements $f,g$ of $\mathcal{C}^0([a,b],\mathbb R)$ by $\langle f,g \rangle = \int_a^b f(x) g(x) \ dx.$

It is known that $f \in \mathcal{C}^0([a,b],\mathbb R)$ is the always vanishing function if we have $\langle x^n,f \rangle = \int_a^b x^n f(x) \ dx = 0$ for all integers $n \ge 0$ . Let’s recall the proof. According to Stone-Weierstrass theorem, for all $\epsilon >0$ it exists a polynomial $P$ such that $\Vert f – P \Vert_\infty \le \epsilon$ . Then $\begin{aligned} 0 &\le \int_a^b f^2 = \int_a^b f(f-P) + \int_a^b fP\\ &= \int_a^b f(f-P) \le \Vert f \Vert_\infty \epsilon(b-a) \end{aligned}$ As this is true for all $\epsilon > 0$ , we get $\int_a^b f^2 = 0$ and $f = 0$ .

We now prove that the result becomes false if we change the interval $[a,b]$ into $[0, \infty)$ , i.e. that one can find a continuous function $f \in \mathcal{C}^0([0,\infty),\mathbb R)$ such that $\int_0^\infty x^n f(x) \ dx$ for all integers $n \ge 0$ . In that direction, let’s consider the complex integral $I_n = \int_0^\infty x^n e^{-(1-i)x} \ dx.$ $I_n$ is well defined as for $x \in [0,\infty)$ we have $\vert x^n e^{-(1-i)x} \vert = x^n e^{-x}$ and $\int_0^\infty x^n e^{-x} \ dx$ converges. By integration by parts, one can prove that $I_n = \frac{n!}{(1-i)^{n+1}} = \frac{(1+i)^{n+1}}{2^{n+1}} n! = \frac{e^{i \frac{\pi}{4}(n+1)}}{2^{\frac{n+1}{2}}}n!.$ Consequently, $I_{4p+3} \in \mathbb R$ for all $p \ge 0$ which means $\int_0^\infty x^{4p+3} \sin(x) e^{-x} \ dx =0$ and finally $\int_0^\infty u^p \sin(u^{1/4}) e^{-u^{1/4}} \ dx =0$ for all integers $p \ge 0$ using integration by substitution with $x = u^{1/4}$ . The function $u \mapsto \sin(u^{1/4}) e^{-u^{1/4}}$ is one we were looking for.

Analysis

Counterexamples around series (part 1)

June 11, 2017 Jean-Pierre Merx 1 Comment

The purpose of this article is to provide some basic counterexamples on real series. Counterexamples are provided as answers to questions.

Unless otherwise stated, $(u_n)_{n \in \mathbb{N}}$ and $(v_n)_{n \in \mathbb{N}}$ are two real sequences.

If $(u_n)$ is non-increasing and converges to zero then $\sum u_n$ converges?

Is not true. A famous counterexample is the harmonic series $\sum \frac{1}{n}$ which doesn’t converge as $\displaystyle \sum_{k=p+1}^{2p} \frac{1}{k} \ge \sum_{k=p+1}^{2p} \frac{1}{2p} = 1/2,$ for all $p \in \mathbb N$ .

If $u_n = o(1/n)$ then $\sum u_n$ converges?

Does not hold as can be seen considering $u_n=\frac{1}{n \ln n}$ for $n \ge 2$ . Indeed $\int_2^x \frac{dt}{t \ln t} = \ln(\ln x) – \ln (\ln 2)$ and therefore $\int_2^\infty \frac{dt}{t \ln t}$ diverges. We conclude that $\sum \frac{1}{n \ln n}$ diverges using the integral test. However $n u_n = \frac{1}{\ln n}$ converges to zero. Continue reading Counterexamples around series (part 1) →

Analysis

Counterexamples on real sequences (part 3)

May 21, 2017 Jean-Pierre Merx Leave a comment

This article is a follow-up of Counterexamples on real sequences (part 2).

Let $(u_n)$ be a sequence of real numbers.

If $u_{2n}-u_n \le \frac{1}{n}$ then $(u_n)$ converges?

This is wrong. The sequence
$u_n=\begin{cases} 0 & \text{for } n \notin \{2^k \ ; \ k \in \mathbb N\}\\ 1- 2^{-k} & \text{for } n= 2^k\end{cases}$
is a counterexample. For $n \gt 2$ and $n \notin \{2^k \ ; \ k \in \mathbb N\}$ we also have $2n \notin \{2^k \ ; \ k \in \mathbb N\}$ , hence $u_{2n}-u_n=0$ . For $n = 2^k$ $0 \le u_{2^{k+1}}-u_{2^k}=2^{-k}-2^{-k-1} \le 2^{-k} = \frac{1}{n}$ and $\lim\limits_{k \to \infty} u_{2^k} = 1$ . $(u_n)$ does not converge as $0$ and $1$ are limit points.

If $\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$ then $(u_n)$ has a finite or infinite limit?

This is not true. Let’s consider the sequence
$u_n=2+\sin(\ln n)$ Using the inequality $\vert \sin p – \sin q \vert \le \vert p – q \vert$
which is a consequence of the mean value theorem, we get $\vert u_{n+1} – u_n \vert = \vert \sin(\ln (n+1)) – \sin(\ln n) \vert \le \vert \ln(n+1) – \ln(n) \vert$ Therefore $\lim\limits_n \left(u_{n+1}-u_n \right) =0$ as $\lim\limits_n \left(\ln(n+1) – \ln(n)\right) = 0$ . And $\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$ because $u_n \ge 1$ for all $n \in \mathbb N$ .

I now assert that the interval $[1,3]$ is the set of limit points of $(u_n)$ . For the proof, it is sufficient to prove that $[-1,1]$ is the set of limit points of the sequence $v_n=\sin(\ln n)$ . For $y \in [-1,1]$ , we can pickup $x \in \mathbb R$ such that $\sin x =y$ . Let $\epsilon > 0$ and $M \in \mathbb N$ , we can find an integer $N \ge M$ such that $0 < \ln(n+1) - \ln(n) \lt \epsilon$ for $n \ge N$ . Select $k \in \mathbb N$ with $x +2k\pi \gt \ln N$ and $N_\epsilon$ with $\ln N_\epsilon \in (x +2k\pi, x +2k\pi + \epsilon)$ . This is possible as $(\ln n)_{n \in \mathbb N}$ is an increasing sequence and the length of the interval $(x +2k\pi, x +2k\pi + \epsilon)$ is equal to $\epsilon$ . We finally get $\vert u_{N_\epsilon} - y \vert = \vert \sin \left(\ln N_\epsilon \right) - \sin \left(x + 2k \pi \right) \vert \le \left(\ln N_\epsilon - (x +2k\pi)\right) \le \epsilon$ proving that $y$ is a limit point of $(u_n)$ .

Analysis

A strictly increasing continuous function that is differentiable at no point of a null set

May 8, 2017 Jean-Pierre Merx Leave a comment

We build in this article a strictly increasing continuous function $f$ that is differentiable at no point of a null set $E$ . The null set $E$ can be chosen arbitrarily. In particular it can have the cardinality of the continuum like the Cantor null set.

A set of strictly increasing continuous functions

For $p \lt q$ two real numbers, consider the function $f_{p,q}(x)=(q-p) \left[\frac{\pi}{2} + \arctan{\left(\frac{2x-p-q}{q-p}\right)}\right]$ $f_{p,q}$ is positive and its derivative is $f_{p,q}^\prime(x) = \frac{2}{1+\left(\frac{2x-p-q}{q-p}\right)^2}$ which is always strictly positive. Hence $f_{p,q}$ is strictly increasing. We also have $\lim\limits_{x \to -\infty} f_{p,q}(x) = 0 \text{ and } \lim\limits_{x \to \infty} f_{p,q}(x) = \pi(q-p).$ One can notice that for $x \in (p,q)$ , $f_{p,q}^\prime(x) \gt 1$ . Therefore for $x, y \in (p,q)$ distinct we have according to the mean value theorem $\frac{f_{p,q}(y)-f_{p,q}(x)}{y-x} \ge 1$ .

Covering $E$ with an appropriate set of open intervals

As $E$ is a null set, for each $n \in \mathbb N$ one can find an open set $O_n$ containing $E$ and measuring less than $2^{-n}$ . $O_n$ can be written as a countable union of disjoint open intervals as any open subset of the reals. Then $I=\bigcup_{m \in \mathbb N} O_m$ is also a countable union of open intervals $I_n$ with $n \in \mathbb N$ . The sum of the lengths of the $I_n$ is less than $1$ . Continue reading A strictly increasing continuous function that is differentiable at no point of a null set →

Analysis

A monotonic function whose points of discontinuity form a dense set

April 30, 2017 Jean-Pierre Merx Leave a comment

Consider a compact interval $[a,b] \subset \mathbb R$ with $a \lt b$ . Let’s build an increasing function $f : [a,b] \to \mathbb R$ whose points of discontinuity is an arbitrary dense subset $D = \{d_n \ ; \ n \in \mathbb N\}$ of $[a,b]$ , for example $D = \mathbb Q \cap [a,b]$ .

Let $\sum p_n$ be a convergent series of positive numbers whose sum is equal to $p$ and define $\displaystyle f(x) = \sum_{d_n \le x} p_n$ .

$f$ is strictly increasing

For $a \le x \lt y \le b$ we have $f(y) – f(x) = \sum_{x \lt d_n \le y} p_n \gt 0$ as the $p_n$ are positive and dense so it exists $p_m \in (x, y]$ .

$f$ is right-continuous on $[a,b]$

We pick-up $x \in [a,b]$ . For any $\epsilon \gt 0$ is exists $N \in \mathbb N$ such that $0 \lt \sum_{n \gt N} p_n \lt \epsilon$ . Let $\delta > 0$ be so small that the interval $(x,x+\delta)$ doesn’t contain any point in the finite set $\{p_1, \dots, p_N\}$ . Then $0 \lt f(y) – f(x) \le \sum_{n \gt N} p_n \lt \epsilon,$ for any $y \in (x,x+\delta)$ proving the right-continuity of $f$ at $x$ . Continue reading A monotonic function whose points of discontinuity form a dense set →

Analysis

A function whose Maclaurin series converges only at zero

April 11, 2017 Jean-Pierre Merx Leave a comment

Let’s describe a real function $f$ whose Maclaurin series converges only at zero. For $n \ge 0$ we denote $f_n(x)= e^{-n} \cos n^2x$ and $f(x) = \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty e^{-n} \cos n^2 x.$ For $k \ge 0$ , the $k$ th-derivative of $f_n$ is $f_n^{(k)}(x) = e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right)$ and $\left\vert f_n^{(k)}(x) \right\vert \le e^{-n} n^{2k}$ for all $x \in \mathbb R$ . Therefore $\displaystyle \sum_{n=0}^\infty f_n^{(k)}(x)$ is normally convergent and $f$ is an indefinitely differentiable function with $f^{(k)}(x) = \sum_{n=0}^\infty e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right).$ Its Maclaurin series has only terms of even degree and the absolute value of the term of degree $2k$ is $\left(\sum_{n=0}^\infty e^{-n} n^{4k}\right)\frac{x^{2k}}{(2k)!} > e^{-2k} (2k)^{4k}\frac{x^{2k}}{(2k)!} > \left(\frac{2kx}{e}\right)^{2k}.$ The right hand side of this inequality is greater than $1$ for $k \ge \frac{e}{2x}$ . This means that for any nonzero $x$ the Maclaurin series for $f$ diverges.

Analysis

Painter’s paradox

March 25, 2017 Jean-Pierre Merx Leave a comment

Can you paint a surface with infinite area with a finite quantity of paint? For sure… let’s do it!

Consider the 3D surface given in cylindrical coordinates as $S(\rho,\varphi):\begin{cases} x &= \rho \cos \varphi\\ y &= \rho \sin \varphi\\ z &= \frac{1}{\rho}\end{cases}$ for $(\rho,\varphi) \in [1,\infty) \times [0, 2 \pi)$ . The surface is named Gabriel’s horn.

Volume of Gabriel’s horn

The volume of Gabriel’s horn is $V = \pi \int_1^\infty \left( \frac{1}{\rho^2} \right) \ d\rho = \pi$ which is finite.

Area of Gabriel’s horn

The area of Gabriel’s horn for $(\rho,\varphi) \in [1,a) \times [0, 2 \pi)$ with $a > 1$ is: $A = 2 \pi \int_1^a \frac{1}{\rho} \sqrt{1+\left( -\frac{1}{\rho^2} \right)^2} \ d\rho \ge 2 \pi \int_1^a \frac{d \rho}{\rho} = 2 \pi \log a.$ As the right hand side of inequality above diverges to $\infty$ as $a \to \infty$ , we can conclude that the area of Gabriel’s horn is infinite.

Conclusion

Gabriel’s horn could be filled with a finite quantity of paint… therefore painting a surface with infinite area. Unfortunately the thickness of the paint coat is converging to $0$ as $z$ goes to $\infty$ , leading to a paint which won’t be too visible!

Math Counterexamples

Tag Archives: analysis

A semi-continuous function with a dense set of points of discontinuity

Converse of fundamental theorem of calculus

Counterexamples around series (part 2)

If $\sum u_n$ converges and $(u_n)$ is non-increasing then $u_n = o(1/n)$ ?

$\sum u_n$ convergent is equivalent to $\sum u_{2n}$ and $\sum u_{2n+1}$ convergent?

$\sum u_n$ absolutely convergent is equivalent to $\sum u_{2n}$ and $\sum u_{2n+1}$ absolutely convergent?

$\sum u_n$ is a positive convergent series then $(\sqrt[n]{u_n})$ is bounded?

If $(u_n)$ is strictly positive with $u_n = o(1/n)$ then $\sum (-1)^n u_n$ converges?

A nonzero continuous map orthogonal to all polynomials

Counterexamples around series (part 1)

If $(u_n)$ is non-increasing and converges to zero then $\sum u_n$ converges?

If $u_n = o(1/n)$ then $\sum u_n$ converges?

Counterexamples on real sequences (part 3)

If $u_{2n}-u_n \le \frac{1}{n}$ then $(u_n)$ converges?

If $\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$ then $(u_n)$ has a finite or infinite limit?

A strictly increasing continuous function that is differentiable at no point of a null set

A set of strictly increasing continuous functions

Covering $E$ with an appropriate set of open intervals