# Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If $$K$$ is a compact topological space, and $$(f_n)_{n \in \mathbb N}$$ is a monotonically decreasing sequence (meaning $$f_{n+1}(x) \le f_n(x)$$ for all $$n \in \mathbb N$$ and $$x \in K$$) of continuous real-valued functions on $$K$$ which converges pointwise to a continuous function $$f$$, then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

### Cases if $$K$$ is not compact

We take $$K=(0,1)$$, which is not closed equipped with the common distance. The sequence $$f_n(x)=x^n$$ of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all $$n \in \mathbb N$$ $\sup\limits_{x \in (0,1)} x^n = 1$

The set $$K=\mathbb R$$ is closed but unbounded, hence also not compact. The sequence defined by $f_n(x)=\begin{cases} 0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}$ is continuous and monotonically decreasing. It converges to $$0$$. However, the convergence is not uniform as for all $$n \in \mathbb N$$: $$\sup\{f_n(x) : x \in \mathbb R\} =1$$. Continue reading Counterexamples around Dini’s theorem

# Counterexamples around differentiation of sequences of functions

We consider here sequences of real functions defined on a closed interval. Following theorem is the main one regarding the differentiation of the limit.

Theorem: Suppose $$(f_n)$$ is a sequence of functions, differentiable on $$[a,b]$$ and such that $$(f_n(x_0))$$ converges for some point $$x_0 \in [a,b]$$. If $$(f_n^\prime)$$ converges uniformly on $$[a,b]$$, then $$(f_n)$$ converges uniformly on $$[a,b]$$ to a function $$f$$ and for all $$x \in [a,b]$$ $f^\prime(x)=\lim\limits_{n \to \infty} f_n^\prime(x)$ What happens if we drop some hypothesis of the theorem? Continue reading Counterexamples around differentiation of sequences of functions

# 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)

# Counterexamples on real sequences (part 2)

In that article, I provide basic counterexamples on sequences convergence. I follow on here with some additional and more advanced examples.

#### If $$(u_n)$$ converges then $$(\vert u_n \vert )$$ converge?

This is true and the proof is based on the reverse triangle inequality: $$\bigl| \vert x \vert – \vert y \vert \bigr| \le \vert x – y \vert$$. However the converse doesn’t hold. For example, the sequence $$u_n=(-1)^n$$ is such that $$\lim \vert u_n \vert = 1$$ while $$(u_n)$$ diverges.

#### If for all $$p \in \mathbb{N}$$ $$\lim\limits_{n \to +\infty} (u_{n+p} – u_n)=0$$ then $$(u_n)$$ converges?

The assertion is wrong. A simple counterexample is $$u_n= \ln(n+1)$$. It is well known that $$(u_n)$$ diverges. However for any $$p \in \mathbb{N}$$ we have $$\lim\limits_{n \to +\infty} (u_{n+p} – u_n) =\ln(1+\frac{p}{n+1})=0$$.
The converse proposition is true. Assume that $$(u_n)$$ is a converging sequence with limit $$l$$ and $$p \ge 0$$ is any integer. We have $$\vert u_{n+p}-u_n \vert = \vert (u_{n+p}-l)-(u_n-l) \vert \le \vert u_{n+p}-l \vert – \vert u_n-l \vert$$ and both terms of the right hand side of the inequality are converging to zero. Continue reading Counterexamples on real sequences (part 2)

# Counterexamples on real sequences (part 1)

I will come back later on with more complex cases. Unless otherwise stated, $$(u_n)_{n \in \mathbb{N}}$$ and $$(v_n)_{n \in \mathbb{N}}$$ are two real sequences. Continue reading Counterexamples on real sequences (part 1)
We build here a continuous function of one real variable whose derivative exists on $$\mathbb{R} \setminus \mathbb{Q}$$ and doesn’t have a left or right derivative on each point of $$\mathbb{Q}$$.
As $$\mathbb{Q}$$ is (infinitely) countable, we can find a bijection $$n \mapsto r_n$$ from $$\mathbb{N}$$ to $$\mathbb{Q}$$. We now reuse the function $$f$$ defined here. Recall $$f$$ main properties: Continue reading A continuous function not differentiable at the rationals but differentiable elsewhere