Tag Archives: sequences-and-series

Analysis

Counterexamples around Dini’s theorem

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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 →

Analysis

Counterexamples around differentiation of sequences of functions

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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 →

Analysis

Pointwise convergence and properties of the limit (part 1)

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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) →

Analysis

Counterexamples on real sequences (part 2)

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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) →

Analysis

A continuous function not differentiable at the rationals but differentiable elsewhere

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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 →