Tag Archives: analysis

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)

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)

Counterexamples on real sequences (part 3)

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 \((u_n)\).

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.

Painter’s paradox

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.