2c. Sequences and series

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Our goal now is to extend the material from chapter 1 regarding the numeric sequences and series, to the case of the sequences and series of functions. To start with, we can talk about the convergence of sequences of functions, [math]f_n\to f[/math], as follows:

Definition

We say that [math]f_n[/math] converges pointwise to [math]f[/math], and write [math]f_n\to f[/math], if

[[math]] f_n(x)\to f(x) [[/math]]

for any [math]x[/math]. Equivalently, [math]\forall x,\forall\varepsilon \gt 0,\exists N\in\mathbb N,\forall n\geq N,|f_n(x)-f(x)| \lt \varepsilon[/math].

The question is now, assuming that [math]f_n[/math] are continuous, does it follow that [math]f[/math] is continuous? I am pretty much sure that you think that the answer is “yes”, based on:

[[math]] \begin{eqnarray*} \lim_{y\to x}f(y) &=&\lim_{y\to x}\lim_{n\to\infty}f_n(y)\\ &=&\lim_{n\to\infty}\lim_{y\to x}f_n(y)\\ &=&\lim_{n\to\infty}f_n(x)\\ &=&f(x) \end{eqnarray*} [[/math]]

However, this proof is wrong, because we know well from chapter 1 that we cannot intervert limits, with this being a common beginner mistake. In fact, the result itself is wrong in general, because if we consider the functions [math]f_n:[0,1]\to\mathbb R[/math] given by [math]f_n(x)=x^n[/math], which are obviously continuous, their limit is discontinuous, given by:

[[math]] \lim_{n\to\infty}x^n =\begin{cases} 0&,\quad x\in[0,1)\\ 1&,\quad x=1 \end{cases} [[/math]]

Of course, you might say here that allowing [math]x=1[/math] in all this might be a bit unnatural, for whatever reasons, but there is an answer to this too. We can do worse, as follows:

Proposition

The basic step function, namely the sign function

[[math]] sgn(x)=\begin{cases} -1&,\quad x \lt 0\\ 0&,\quad x=0\\ 1&,\quad x \gt 0 \end{cases} [[/math]]

can be approximated by suitable modifications of [math]\arctan(x)[/math]. Even worse, there are examples of [math]f_n\to f[/math] with each [math]f_n[/math] continuous, and with [math]f[/math] totally discontinuous.

Show Proof

To start with, [math]\arctan(x)[/math] looks a bit like [math]sgn(x)[/math], so to say, but one problem comes from the fact that its image is [math][-\pi/2,\pi/2][/math], instead of the desired [math][-1,1][/math]. Thus, we must first rescale [math]\arctan(x)[/math] by [math]\pi/2[/math]. Now with this done, we can further stretch the variable [math]x[/math], as to get our function closer and closer to [math]sgn(x)[/math], as desired. This proves the first assertion, and the second assertion, which is a bit more technical, and that we will not really need in what follows, is left as an exercise for you, reader.

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Sumarizing, we are a bit in trouble, because we would like to have in our bag of theorems something saying that [math]f_n\to f[/math] with [math]f_n[/math] continuous implies [math]f[/math] continuous. Fortunately, this can be done, with a suitable refinement of the notion of convergence, as follows:

Definition

We say that [math]f_n[/math] converges uniformly to [math]f[/math], and write [math]f_n\to_uf[/math], if:

[[math]] \forall\varepsilon \gt 0,\exists N\in\mathbb N,\forall n\geq N,|f_n(x)-f(x)| \lt \varepsilon,\forall x [[/math]]

That is, the same condition as for [math]f_n\to f[/math] must be satisfied, but with the [math]\forall x[/math] at the end.

And it is this “[math]\forall x[/math] at the end” which makes the difference, and will make our theory work. In order to understand this, which is something quite subtle, let us compare Definition 2.22 and Definition 2.24. As a first observation, we have:

Proposition

Uniform convergence implies pointwise convergence,

[[math]] f_n\to_uf\implies f_n\to f [[/math]]

but the converse is not true, in general.

Show Proof

Here the first assertion is clear from definitions, just by thinking at what is going on, with no computations needed. As for the second assertion, the simplest counterexamples here are the functions [math]f_n:[0,1]\to\mathbb R[/math] given by [math]f_n(x)=x^n[/math], that we met before in Proposition 2.23. Indeed, uniform convergence on [math][0,1)[/math] would mean:

[[math]] \forall\varepsilon \gt 0,\exists N\in\mathbb N,\forall n\geq N,x^n \lt \varepsilon,\forall x\in[0,1) [[/math]]

But this is wrong, because no matter how big [math]N[/math] is, we have [math]\lim_{x\to1}x^N=1[/math], and so we can find [math] x\in[0,1)[/math] such that [math]x^N \gt \varepsilon[/math]. Thus, we have our counterexample.

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Moving ahead now, let us state our main theorem on uniform convergence, as follows:

Theorem

Assuming that [math]f_n[/math] are continuous, and that

[[math]] f_n\to_uf [[/math]]

then [math]f[/math] is continuous. That is, uniform limit of continuous functions is continuous.

Show Proof

As previously advertised, it is the “[math]\forall x[/math] at the end” in Definition 2.24 that will make this work. Indeed, let us try to prove that the limit [math]f[/math] is continuous at some point [math]x[/math]. For this, we pick a number [math]\varepsilon \gt 0[/math]. Since [math]f_n\to_uf[/math], we can find [math]N\in\mathbb N[/math] such that:

[[math]] |f_N(z)-f(z)| \lt \frac{\varepsilon}{3}\quad,\quad\forall z [[/math]]

On the other hand, since [math]f_N[/math] is continuous at [math]x[/math], we can find [math]\delta \gt 0[/math] such that:

[[math]] |x-y| \lt \delta\implies|f_N(x)-f_N(y)| \lt \frac{\varepsilon}{3} [[/math]]

But with this, we are done. Indeed, for [math]|x-y| \lt \delta[/math] we have:

[[math]] \begin{eqnarray*} |f(x)-f(y)| &\leq&|f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|\\ &\leq&\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=&\varepsilon \end{eqnarray*} [[/math]]

Thus, the limit function [math]f[/math] is continuous at [math]x[/math], and we are done.

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Obviously, the notion of uniform convergence in Definition 2.24 is something quite interesting, worth some more study. As a first result, we have:

Proposition

The following happen, regarding uniform limits:

[math]f_n\to_uf[/math], [math]g_n\to_ug[/math] imply [math]f_n+g_n\to_uf+g[/math].
[math]f_n\to_uf[/math], [math]g_n\to_ug[/math] imply [math]f_ng_n\to_ufg[/math].
[math]f_n\to_uf[/math], [math]f\neq0[/math] imply [math]1/f_n\to_u1/f[/math].
[math]f_n\to_uf[/math], [math]g[/math] continuous imply [math]f_n\circ g\to_uf\circ g[/math].
[math]f_n\to_uf[/math], [math]g[/math] continuous imply [math]g\circ f_n\to_ug\circ f[/math].

Show Proof

All this is routine, exactly as for the results for numeric sequences from chapter 1, that we know well, with no difficulties or tricks involved.

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Finally, there is some abstract mathematics to be done as well. Indeed, observe that the notion of uniform convergence, as formulated in Definition 2.24, means that:

[[math]] \sup_x\big|f_n(x)-f(x)\big|\ \longrightarrow_{n\to\infty}\ 0 [[/math]]

This suggests measuring the distance between functions via a supremum as above, and in relation with this, we have the following result:

Theorem

The uniform convergence, [math]f_n\to_uf[/math], means that we have [math]f_n\to f[/math] with respect to the following distance,

[[math]] d(f,g)=\sup_x\big|f(x)-g(x)\big| [[/math]]

which is indeed a distance function.

Show Proof

Here the fact that [math]d[/math] is indeed a distance, in the sense that it satisfies all the intuitive properties of a distance, including the triangle inequality, follows from definitions, and the fact that the uniform convergence can be interpreted as above is clear as well.

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Finally, regarding the series, some general theory can be developed here as well, in connection with the notion of uniform convergence, and in connection with the notion of convergence radius. We will see applications of all this, in a moment.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].