# Pointwise convergence and properties of the limit (part 1)

We look here at the continuity of a sequence of functions that converges pointwise and give some counterexamples of what happens versus uniform convergence.

### Recalling the definition of pointwise convergence

We consider here real functions defined on a closed interval $$[a,b]$$. A sequence of functions $$(f_n)$$ defined on $$[a,b]$$ converges pointwise to the function $$f$$ if and only if for all $$x \in [a,b]$$ $$\displaystyle \lim\limits_{n \to +\infty} f_n(x) = f(x)$$. Pointwise convergence is weaker than uniform convergence.

### Pointwise convergence does not, in general, preserve continuity

Suppose that $$f_n \ : \ [0,1] \to \mathbb{R}$$ is defined by $$f_n(x)=x^n$$. For $$0 \le x <1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 0$$, while if $$x = 1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 1$$. Hence the sequence $$f_n$$ converges to the function equal to $$0$$ for $$0 \le x < 1$$ and to $$1$$ for $$x=1$$. Although each $$f_n$$ is a continuous function of $$[0,1]$$, their pointwise limit is not. $$f$$ is discontinuous at $$1$$. We notice that $$(f_n)$$ doesn't converge uniformly to $$f$$ as for all $$n \in \mathbb{N}$$, $$\displaystyle \sup\limits_{x \in [0,1]} \vert f_n(x) - f(x) \vert = 1$$. That's reassuring as uniform convergence of a sequence of continuous functions implies that the limit is continuous! Continue reading Pointwise convergence and properties of the limit (part 1)

# The set of all commutators in a group need not be a subgroup

I here provide a simple example of a group whose set of commutators is not a subgroup. The counterexample is due to P.J. Cassidy (1979).

### Description of the group $$G$$

Let $$k[x,y]$$ denote the ring of all polynomials in two variables over a field $$k$$, and let $$k[x]$$ and $$k[y]$$ denote the subrings of all polynomials in $$x$$ and in $$y$$ respectively. $$G$$ is the set of all upper unitriangular matrices of the form
$A=\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)$ where $$f(x) \in k[x]$$, $$g(y) \in k[y]$$, and $$h(x,y) \in k[x,y]$$. The matrix $$A$$ will also be denoted $$(f,g,h)$$.
Let’s verify that $$G$$ is a group. The products of two elements $$(f,g,h)$$ and $$(f^\prime,g^\prime,h^\prime)$$ is
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & f^\prime(x) & h^\prime(x,y) \\ 0 & 1 & g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$
$=\left(\begin{array}{ccc} 1 & f(x)+f^\prime(x) & h(x,y)+h^\prime(x,y)+f(x)g^\prime(y) \\ 0 & 1 & g(y)+g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$ which is an element of $$G$$. We also have:
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)^{-1} = \left(\begin{array}{ccc} 1 & -f(x) & f(x)g(y) – h(x,y) \\ 0 & 1 & -g(y) \\ 0 & 0 & 1 \end{array}\right)$ proving that the inverse of an element of $$G$$ is also an element of $$G$$. Continue reading The set of all commutators in a group need not be a subgroup

# A topological vector space with no non trivial continuous linear form

We consider here the $$L^p$$- spaces of real functions defined on $$[0,1]$$ for which the $$p$$-th power of the absolute value is Lebesgue integrable. We focus on the case $$0 < p < 1$$. We'll prove that those $$L^p$$-spaces are topological vector spaces on which there exists no continuous non-trivial linear forms (i.e. not vanishing identically). Continue reading A topological vector space with no non trivial continuous linear form

# A nowhere locally bounded function

In that article, I described some properties of Thomae’s function$$f$$. Namely:

• The function is discontinuous on $$\mathbb{Q}$$.
• Continuous on $$\mathbb{R} \setminus \mathbb{Q}$$.
• Its right-sided and left-sided limits vanish at all points.

Let’s modify $$f$$ to get function $$g$$ defined as follow:
$g: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & q \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.$ $$f$$ and $$g$$ both vanish on the set of irrational numbers, while on the set of rational numbers, $$g$$ is equal to the reciprocal of $$f$$. We now consider an open subset $$O \subset \mathbb{R}$$ and $$x \in O$$. As $$f$$ right-sided and left-sided limits vanish at all points, we have $$\lim\limits_{n \to +\infty} f(x_n) = 0$$ for all sequence $$(x_n)$$ of rational numbers converging to $$x$$ (and such a sequence exists as the rational numbers are everywhere dense in the reals). Hence $$\lim\limits_{n \to +\infty} g(x_n) = + \infty$$ as $$f$$ is positive.

We can conclude that $$g$$ is nowhere locally bounded. The picture of the article is a plot of function $$g$$ on the rational numbers $$r = \frac{p}{q}$$ in lowest terms for $$0 < r < 1$$ and $$q \le 50$$.

# A function continuous at all irrationals and discontinuous at all rationals

Let’s discover the beauties of Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.

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

### $$f$$ is periodic with period $$1$$

This is easy to prove as for $$x \in \mathbb{R} \setminus \mathbb{Q}$$ we also have $$x+1 \in \mathbb{R} \setminus \mathbb{Q}$$ and therefore $$f(x+1)=f(x)=0$$. While for $$y=\frac{p}{q} \in \mathbb{Q}$$ in lowest terms, $$y+1=\frac{p+q}{q}$$ is also in lowest terms, hence $$f(y+1)=f(y)=\frac{1}{q}$$. Continue reading A function continuous at all irrationals and discontinuous at all rationals

# Generating the symmetric group with a transposition and a maximal length cycle

Can the symmetric group $$\mathcal{S}_n$$ be generated by any transposition and any $$n$$-cycle for $$n \ge 2$$ integer? is the question we deal with.

We first recall some terminology:

Symmetric group
The symmetric group $$\mathcal{S}_n$$ on a finite set of $$n$$ symbols is the group whose elements are all the permutations of the $$n$$ symbols. We’ll denote by $$\{1,\dots,n\}$$ those $$n$$ symbols.
Cycle
A cycle of length $$k$$ (with $$k \ge 2$$) is a cyclic permutation $$\sigma$$ for which there exists an element $$i \in \{1,\dots,n\}$$ such that $$i, \sigma(i), \sigma^2(i), \dots, \sigma^k(i)=i$$ are the only elements moved by $$\sigma$$. We’ll denote the cycle $$\sigma$$ by $$(s_0 \ s_1 \dots \ s_{k-1})$$ where $$s_0=i, s_1=\sigma(i),\dots,s_{k-1}=\sigma^{k-1}(i)$$.
Transposition
A transposition is a cycle of length $$2$$. We denote below the transposition of elements $$a \neq b$$ by $$(a \ b)$$ or $$\tau_{a,b}$$.