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 $\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert$

A nonabelian $$p$$-group

Consider a prime number $$p$$ and a finite p-group $$G$$, i.e. a group of order $$p^n$$ with $$n \ge 1$$.

If $$n=1$$ the group $$G$$ is cyclic hence abelian.

For $$n=2$$, $$G$$ is also abelian. This is a consequence of the fact that the center $$Z(G)$$ of a $$p$$-group is non-trivial. Indeed if $$\vert Z(G) \vert =p^2$$ then $$G=Z(G)$$ is abelian. We can’t have $$\vert Z(G) \vert =p$$. If that would be the case, the order of $$H=G / Z(G)$$ would be equal to $$p$$ and $$H$$ would be cyclic, generated by an element $$h$$. For any two elements $$g_1,g_2 \in G$$, we would be able to write $$g_1=h^{n_1} z_1$$ and $$g_2=h^{n_1} z_1$$ with $$z_1,z_2 \in Z(G)$$. Hence $g_1 g_2 = h^{n_1} z_1 h^{n_2} z_2=h^{n_1 + n_2} z_1 z_2= h^{n_2} z_2 h^{n_1} z_1=g_2 g_1,$ proving that $$g_1,g_2$$ commutes in contradiction with $$\vert Z(G) \vert < \vert G \vert$$. However, all $$p$$-groups are not abelian. For example the unitriangular matrix group $U(3,\mathbb Z_p) = \left\{ \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix} \ | \ a,b ,c \in \mathbb Z_p \right\}$ is a $$p$$-group of order $$p^3$$. Its center $$Z(U(3,\mathbb Z_p))$$ is $Z(U(3,\mathbb Z_p)) = \left\{ \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} \ | \ b \in \mathbb Z_p \right\},$ which is of order $$p$$. Therefore $$U(3,\mathbb Z_p)$$ is not abelian.

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.

Subset of elements of finite order of a group

Consider a group $$G$$ and have a look at the question: is the subset $$S$$ of elements of finite order a subgroup of $$G$$?

The answer is positive when any two elements of $$S$$ commute. For the proof, consider $$x,y \in S$$ of order $$m,n$$ respectively. Then $\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e$ where $$e$$ is the identity element. Hence $$xy$$ is of finite order (less or equal to $$mn$$) and belong to $$S$$.

Example of a non abelian group

In that cas, $$S$$ might not be subgroup of $$G$$. Let’s take for $$G$$ the general linear group over $$\mathbb Q$$ (the set of rational numbers) of $$2 \times 2$$ invertible matrices named $$\text{GL}_2(\mathbb Q)$$. The matrices $A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}$ are of order $$2$$. They don’t commute as $AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.$ Finally, $$AB$$ is of infinite order and therefore doesn’t belong to $$S$$ proving that $$S$$ is not a subgroup of $$G$$.

Counterexamples around Cauchy condensation test

According to Cauchy condensation test: for a non-negative, non-increasing sequence $$(u_n)_{n \in \mathbb N}$$ of real numbers, the series $$\sum_{n \in \mathbb N} u_n$$ converges if and only if the condensed series $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ converges.

The test doesn’t hold for any non-negative sequence. Let’s have a look at counterexamples.

A sequence such that $$\sum_{n \in \mathbb N} u_n$$ converges and $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges

Consider the sequence $u_n=\begin{cases} \frac{1}{n} & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ 0 & \text{ else} \end{cases}$ For $$n \in \mathbb N$$ we have $0 \le \sum_{k = 1}^n u_k \le \sum_{k = 1}^{2^n} u_k = \sum_{k = 1}^{n} \frac{1}{2^k} < 1,$ therefore $$\sum_{n \in \mathbb N} u_n$$ converges as its partial sums are positive and bounded above. However $\sum_{k=1}^n 2^k u_{2^k} = \sum_{k=1}^n 1 = n,$ so $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges.

A sequence such that $$\sum_{n \in \mathbb N} v_n$$ diverges and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges

Consider the sequence $v_n=\begin{cases} 0 & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ \frac{1}{n} & \text{ else} \end{cases}$ We have $\sum_{k = 1}^{2^n} v_k = \sum_{k = 1}^{2^n} \frac{1}{k} – \sum_{k = 1}^{n} \frac{1}{2^k} > \sum_{k = 1}^{2^n} \frac{1}{k} -1$ which proves that the series $$\sum_{n \in \mathbb N} v_n$$ diverges as the harmonic series is divergent. However for $$n \in \mathbb N$$, $$2^n v_{2^n} = 0$$ and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges.

Counterexamples around the Cauchy product of real series

Let $$\sum_{n = 0}^\infty a_n, \sum_{n = 0}^\infty b_n$$ be two series of real numbers. The Cauchy product $$\sum_{n = 0}^\infty c_n$$ is the series defined by $c_n = \sum_{k=0}^n a_k b_{n-k}$ According to the theorem of Mertens, if $$\sum_{n = 0}^\infty a_n$$ converges to $$A$$, $$\sum_{n = 0}^\infty b_n$$ converges to $$B$$ and at least one of the two series is absolutely convergent, their Cauchy product converges to $$AB$$. This can be summarized by the equality $\left( \sum_{n = 0}^\infty a_n \right) \left( \sum_{n = 0}^\infty b_n \right) = \sum_{n = 0}^\infty c_n$

The assumption stating that at least one of the two series converges absolutely cannot be dropped as shown by the example $\sum_{n = 0}^\infty a_n = \sum_{n = 0}^\infty b_n = \sum_{n = 0}^\infty \frac{(-1)^n}{\sqrt{n+1}}$ Those series converge according to Leibniz test, as the sequence $$(1/\sqrt{n+1})$$ decreases monotonically to zero. However, the Cauchy product is defined by $c_n=\sum_{k=0}^n \frac{(-1)^k}{\sqrt{k+1}} \cdot \frac{(-1)^{n-k}}{\sqrt{n-k+1}} = (-1)^n \sum_{k=0}^n \frac{1}{\sqrt{(k+1)(n-k+1)}}$ As we have $$1 \le k+ 1 \le n+1$$ and $$1 \le n-k+ 1 \le n+1$$ for $$k = 0 \dots n$$, we get $$\frac{1}{\sqrt{(k+1)(n-k+1)}} \ge \frac{1}{n+1}$$ and therefore $$\vert c_n \vert \ge 1$$ proving that the Cauchy product of $$\sum_{n = 0}^\infty a_n$$ and $$\sum_{n = 0}^\infty b_n$$ diverges.

The Cauchy product may also converge while the initial series both diverge. Let’s consider $\begin{cases} (a_n) = (2, 2, 2^2, \dots, 2^n, \dots)\\ (b_n) = (-1, 1, 1, 1, \dots) \end{cases}$ The series $$\sum_{n = 0}^\infty a_n, \sum_{n = 0}^\infty b_n$$ diverge. Their Cauchy product is the series defined by $c_n=\begin{cases} -2 & \text{ for } n=0\\ 0 & \text{ for } n>0 \end{cases}$ which is convergent.

A linear differential equation with no solution to an initial value problem

Consider a first order linear differential equation $y^\prime(x) = A(x)y(x) + B(x)$ where $$A, B$$ are real continuous functions defined on a non-empty real interval $$I$$. According to Picard-Lindelöf theorem, the initial value problem $\begin{cases} y^\prime(x) = A(x)y(x) + B(x)\\ y(x_0) = y_0, \ x_0 \in I \end{cases}$ has a unique solution defined on $$I$$.

However, a linear differential equation $c(x)y^\prime(x) = A(x)y(x) + B(x)$ where $$A, B, c$$ are real continuous functions might not have a solution to an initial value problem. Let’s have a look at the equation $x y^\prime(x) = y(x) \tag{E}\label{eq:IVP}$ for $$x \in \mathbb R$$. The equation is linear.

For $$x \in (-\infty,0)$$ a solution to \eqref{eq:IVP} is a solution of the explicit differential linear equation $y^\prime(x) = \frac{y(x)}x$ hence can be written $$y(x) = \lambda_-x$$ with $$\lambda_- \in \mathbb R$$. Similarly, a solution to \eqref{eq:IVP} on the interval $$(0,\infty)$$ is of the form $$y(x) = \lambda_+ x$$ with $$\lambda_+ \in \mathbb R$$.

A global solution to \eqref{eq:IVP}, i.e. defined on the whole real line is differentiable at $$0$$ hence the equation $\lambda_- = y_-^\prime(0)=y_+^\prime(0) = \lambda_+$ which means that $$y(x) = \lambda x$$ where $$\lambda=\lambda_-=\lambda_+$$.

In particular all solutions defined on $$\mathbb R$$ are such that $$y(0)=0$$. Therefore the initial value problem $\begin{cases} x y^\prime(x) = y(x)\\ y(0)=1 \end{cases}$ has no solution.

Field not algebraic over an intersection but algebraic over each initial field

Let’s describe an example of a field $$K$$ which is of degree $$2$$ over two distinct subfields $$M$$ and $$N$$ respectively, but not algebraic over $$M \cap N$$.

Let $$K=F(x)$$ be the rational function field over a field $$F$$ of characteristic $$0$$, $$M=F(x^2)$$ and $$N=F(x^2+x)$$. I claim that those fields provide the example we’re looking for.

$$K$$ is of degree $$2$$ over $$M$$ and $$N$$

The polynomial $$\mu_M(t)=t^2-x^2$$ belongs to $$M[t]$$ and $$x \in K$$ is a root of $$\mu_M$$. Also, $$\mu_M$$ is irreducible over $$M=F(x^2)$$. If that wasn’t the case, $$\mu_M$$ would have a root in $$F(x^2)$$ and there would exist two polynomials $$p,q \in F[t]$$ such that $p^2(x^2) = x^2 q^2(x^2)$ which cannot be, as can be seen considering the degrees of the polynomials of left and right hand sides. This proves that $$[K:M]=2$$. Considering the polynomial $$\mu_N(t)=t^2-t-(x^2+x)$$, one can prove that we also have $$[K:N]=2$$.

We have $$M \cap N=F$$

The mapping $$\sigma_M : x \mapsto -x$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$M$$ are fixed under $$\sigma_M$$. Similarly, the mapping $$\sigma_N : x \mapsto -x-1$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$N$$ are fixed under $$\sigma_N$$. Also $(\sigma_N\circ\sigma_M)(x)=\sigma_N(\sigma_M(x))=\sigma_N(-x)=-(-x-1)=x+1.$ An element $$z=p(x)/q(x) \in M \cap N$$ where $$p(x),q(x)$$ are coprime polynomials of $$K=F(x)$$ is fixed under $$\sigma_M \circ \sigma_N$$. Therefore following equality holds $\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)},$ which is equivalent to $p(x)q(x+1)=p(x+1)q(x).$ By induction, we get for $$n \in \mathbb Z$$ $p(x)q(x+n)=p(x+n)q(x).$ Assume $$p(x)$$ is not a constant polynomial. Then it has a root $$\alpha$$ in some finite extension $$E$$ of $$F$$. As $$p(x),q(x)$$ are coprime polynomials, $$q(\alpha) \neq 0$$. Consequently $$p(\alpha+n)=0$$ for all $$n \in \mathbb Z$$ and the elements $$\alpha +n$$ are all distinct as the characteristic of $$F$$ is supposed to be non zero. This implies that $$p(x)$$ is the zero polynomial, in contradiction with our assumption. Therefore $$p(x)$$ is a constant polynomial and $$q(x)$$ also according to a similar proof. Hence $$z$$ is constant as was supposed to be proven.

Finally, $$K=F(x)$$ is not algebraic over $$F=M \cap N$$ as $$(1,x, x^2, \dots, x^n, \dots)$$ is independent over the field $$F$$ which concludes our claims on $$K, M$$ and $$N$$.

Pointwise convergence not uniform on any interval

We provide in this article an example of a pointwise convergent sequence of real functions that doesn’t converge uniformly on any interval.

Let’s consider a sequence $$(a_p)_{p \in \mathbb N}$$ enumerating the set $$\mathbb Q$$ of rational numbers. Such a sequence exists as $$\mathbb Q$$ is countable.

Now let $$(g_n)_{n \in \mathbb N}$$ be the sequence of real functions defined on $$\mathbb R$$ by $g_n(x) = \sum_{p=1}^{\infty} \frac{1}{2^p} f_n(x-a_p)$ where $$f_n : x \mapsto \frac{n^2 x^2}{1+n^4 x^4}$$ for $$n \in \mathbb N$$.

$$f_n$$ main properties

$$f_n$$ is a rational function whose denominator doesn’t vanish. Hence $$f_n$$ is indefinitely differentiable. As $$f_n$$ is an even function, we can study it only on $$[0,\infty)$$.

We have $f_n^\prime(x)= 2n^2x \frac{1-n^4x^4}{(1+n^4 x^4)^2}.$ $$f_n^\prime$$ vanishes at zero (like $$f_n$$) is positive on $$(0,\frac{1}{n})$$, vanishes at $$\frac{1}{n}$$ and is negative on $$(\frac{1}{n},\infty)$$. Hence $$f_n$$ has a maximum at $$\frac{1}{n}$$ with $$f_n(\frac{1}{n}) = \frac{1}{2}$$ and $$0 \le f_n(x) \le \frac{1}{2}$$ for all $$x \in \mathbb R$$.

Also for $$x \neq 0$$ $0 \le f_n(x) =\frac{n^2 x^2}{1+n^4 x^4} \le \frac{n^2 x^2}{n^4 x^4} = \frac{1}{n^2 x^2}$ consequently $0 \le f_n(x) \le \frac{1}{n} \text{ for } x \ge \frac{1}{\sqrt{n}}.$

$$(g_n)$$ converges pointwise to zero

First, one can notice that $$g_n$$ is well defined. For $$x \in \mathbb R$$ and $$p \in \mathbb N$$ we have $$0 \le \frac{1}{2^p} f_n(x-a_p) \le \frac{1}{2^p} \cdot\ \frac{1}{2}=\frac{1}{2^{p+1}}$$ according to previous paragraph. Therefore the series of functions $$\sum \frac{1}{2^p} f_n(x-a_p)$$ is normally convergent. $$g_n$$ is also continuous as for all $$p \in \mathbb N$$ $$x \mapsto \frac{1}{2^p} f_n(x-a_p)$$ is continuous. Continue reading Pointwise convergence not uniform on any interval