# 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