# A nonzero continuous map orthogonal to all polynomials

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 $$\epsilon >0$$ if 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.

# Counterexamples around series (part 1)

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)

# Counterexamples on real sequences (part 3)

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 $$(u_n)$$.

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

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

# A monotonic function whose points of discontinuity form a dense set

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

# A function whose Maclaurin series converges only at zero

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.

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 Garbiel’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 Garbiel’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!

# A non complete normed vector space

Consider a real normed vector space $$V$$. $$V$$ is called complete if every Cauchy sequence in $$V$$ converges in $$V$$. A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like $$(\ell_p, \Vert \cdot \Vert_p)$$ the space of real sequences endowed with the norm $$\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}$$ for $$p \ge 1$$, the space $$(C(X), \Vert \cdot \Vert)$$ of real continuous functions on a compact Hausdorff space $$X$$ endowed with the norm $$\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert$$ or the Lebesgue space $$(L^1(\mathbb R), \Vert \cdot \Vert_1)$$ of Lebesgue real integrable functions endowed with the norm $$\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx$$.

Let’s give an example of a non complete normed vector space. Let $$(P, \Vert \cdot \Vert_\infty)$$ be the normed vector space of real polynomials endowed with the norm $$\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert$$. Consider the sequence of polynomials $$(p_n)$$ defined by
$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.$ For $$m < n$$ and $$x \in [0,1]$$, we have $\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}$ which proves that $$(p_n)$$ is a Cauchy sequence. Also for $$x \in [0,1]$$ $\lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.$ As uniform converge implies pointwise convergence, if $$(p_n)$$ was convergent in $$P$$, it would be towards $$p$$. But $$p$$ is not a polynomial function as none of its $$n$$th-derivative always vanishes. Hence $$(p_n)$$ is a Cauchy sequence that doesn't converge in $$(P, \Vert \cdot \Vert_\infty)$$, proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

# Uniform continuous function but not Lipschitz continuous

Consider the function $\begin{array}{l|rcl} f : & [0,1] & \longrightarrow & [0,1] \\ & x & \longmapsto & \sqrt{x} \end{array}$

$$f$$ is continuous on the compact interval $$[0,1]$$. Hence $$f$$ is uniform continuous on that interval according to Heine-Cantor theorem. For a direct proof, one can verify that for $$\epsilon > 0$$, one have $$\vert \sqrt{x} – \sqrt{y} \vert \le \epsilon$$ for $$\vert x – y \vert \le \epsilon^2$$.

However $$f$$ is not Lipschitz continuous. If $$f$$ was Lipschitz continuous for a Lipschitz constant $$K > 0$$, we would have $$\vert \sqrt{x} – \sqrt{y} \vert \le K \vert x – y \vert$$ for all $$x,y \in [0,1]$$. But we get a contradiction taking $$x=0$$ and $$y=\frac{1}{4 K^2}$$ as $\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert$

# Raabe-Duhamel’s test

The Raabe-Duhamel’s test (also named Raabe’s test) is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number. The Raabe-Duhamel’s test was developed by Swiss mathematician Joseph Ludwig Raabe.

It states that if:

$\displaystyle \lim _{n\to \infty }\left\vert{\frac {a_{n}}{a_{n+1}}}\right\vert=1 \text{ and } \lim _{{n\to \infty }} n \left(\left\vert{\frac {a_{n}}{a_{{n+1}}}}\right\vert-1 \right)=R,$
then the series will be absolutely convergent if $$R > 1$$ and divergent if $$R < 1$$. First one can notice that Raabe-Duhamel's test maybe conclusive in cases where ratio test isn't. For instance, consider a real $$\alpha$$ and the series $$u_n=\frac{1}{n^\alpha}$$. We have $\lim _{n\to \infty } \frac{u_{n+1}}{u_n} = \lim _{n\to \infty } \left(\frac{n}{n+1} \right)^\alpha = 1$ and therefore the ratio test is inconclusive. However $\frac{u_n}{u_{n+1}} = \left(\frac{n+1}{n} \right)^\alpha = 1 + \frac{\alpha}{n} + o \left(\frac{1}{n}\right)$ for $$n$$ around $$\infty$$ and $\lim _{{n\to \infty }} n \left(\frac {u_{n}}{u_{{n+1}}}-1 \right)=\alpha.$ Raabe-Duhamel's test allows to conclude that the series $$\sum u_n$$ diverges for $$\alpha <1$$ and converges for $$\alpha > 1$$ as well known.

When $$R=1$$ in the Raabe’s test, the series can be convergent or divergent. For example, the series above $$u_n=\frac{1}{n^\alpha}$$ with $$\alpha=1$$ is the harmonic series which is divergent.

On the other hand, the series $$v_n=\frac{1}{n \log^2 n}$$ is convergent as can be proved using the integral test. Namely $0 \le \frac{1}{n \log^2 n} \le \int_{n-1}^n \frac{dt}{t \log^2 t} \text{ for } n \ge 3$ and $\int_2^\infty \frac{dt}{t \log^2 t} = \left[-\frac{1}{\log t} \right]_2^\infty = \frac{1}{\log 2}$ is convergent, while $\frac{v_n}{v_{n+1}} = 1 + \frac{1}{n} +\frac{2}{n \log n} + o \left(\frac{1}{n \log n}\right)$ for $$n$$ around $$\infty$$ and therefore $$R=1$$ in the Raabe-Duhamel’s test.