Continuous Function With a Sequence of Values That Converge

In the previous sections, we have considered real-number sequences, that is, sequences \((x_n)\) such that \(x_n \in \real\) for each \(n\in\N\). In this section, we consider sequences whose terms are functions. Sequences of functions arise naturally in many applications in physics and engineering. A typical way that sequences of functions arise is in the problem of solving an equation in which the unknown is a function \(f\). In many of these types of problems, one is able to generate a sequence of functions \((f_n)=(f_1,f_2,f_3,\ldots)\) through some algorithmic process with the intention that the sequence of functions \((f_n)\) converges to the solution \(f\). Moreover, it would be desirable that the limiting function \(f\) inherit as many properties possessed by each function \(f_n\) such as, for example, continuity, differentiability, or integrability. We will see that this latter issue is rather delicate. In this section, we develop a notion of the limit of a sequence of functions and then investigate if the fundamental properties of boundedness, continuity, integrability, and differentiability are preserved under the limit operation.

Pointwise Convergence

Let \(A\subset\real\) be a non-empty subset and suppose that for each \(n\in\N\) we have a function \(f_n:A\rightarrow\real\). We then say that \((f_n) = (f_1,f_2,f_3,\ldots,)\) is a sequence of functions on \(A\).

Let \(A=[0,1]\) and let \(f_n(x) = x^n\) for \(n\in \N\) and \(x\in A\). Then \((f_n) = (f_1, f_2, f_3, \ldots)\) is a sequence of functions on \(A\). As another example, for \(n\in \N\) and \(x\in A\) let \(g_n(x) = nx(1-x^2)^n\). Then \((g_n) = (g_1,g_2,g_3,\ldots)\) is a sequence of functions on \(A\). Or how about \[ f_n(x) = a_n \cos(nx) + b_n \sin(nx) \] where \(a_n, b_n \in \real\) and \(x\in [-\pi, \pi]\).

Let \((f_n)\) be a sequence of functions on \(A\). For each fixed \(x\in A\) we obtain a sequence of real numbers \((x_n)\) by simply evaluating each \(f_n\) at \(x\), that is, \(x_n = f_n(x)\). For example, if \(f_n(x) = x^n\) and we fix \(x=\tfrac{3}{4}\) then we obtain the sequence \(x_n = f_n(\tfrac{3}{4}) = \left(\tfrac{3}{4}\right)^n\). If \(x\in A\) is fixed we can then easily talk about the convergence of the sequence of numbers \((f_n(x))\) in the usual way. This leads to our first definition of convergence of function sequences.

Let \((f_n)\) be a sequence of functions on \(A\subseteq\real\). We say that \((f_n)\) converges pointwise on \(A\) to the function \(f:A\rightarrow\real\) if for each \(x\in A\) the sequence \((f_n(x))\) converges to the number \(f(x)\), that is, \[ \lim_{n\rightarrow\infty} f_n(x) = f(x). \] In this case, we call the function \(f\) the pointwise limit of the sequence \((f_n)\).

By uniqueness of limits of sequences of real numbers (Theorem 3.1.12), the pointwise limit of a sequence \((f_n)\) is unique. Also, when the domain \(A\) is understood, we will simply say that \((f_n)\) converges pointwise to \(f\).

Consider the sequence \((f_n)\) defined on \(\real\) by \(f_n(x) = (2xn+(-1)^nx^2)/n\). For fixed \(x\in \real\) we have \[ \limi f_n(x) = \limi \frac{2xn+(-1)^n x^2}{n} = 2x. \] Hence, \((f_n)\) converges pointwise to \(f(x) = 2x\) on \(\real\). In Figure 8.1, we graph \(f_n(x)\) for the values \(n=1,2,3,4\) and the function \(f(x)=2x\). Notice that \(f'_n(x) = (2n+2(-1)^nx)/n\) and therefore \(\limi f'_n(x) = 2\), and for the limit function \(f(x)=2x\) we have \(f'(x) = 2\). Hence, the sequence of derivatives \((f'_n)\) converges pointwise to \(f'\). Also, after some basic computations, \begin{align*} \int_{-1}^1 f_n(x)\,dx &= \int_{-1}^1 \frac{2xn+(-1)^n x^2}{n}\,dx\\ & = \frac{2(-1)^n}{3n} \end{align*} and therefore \begin{align*} \limi \int_{-1}^1 f_n(x)\,dx &= \limi \frac{2(-1)^n}{3n}\\ & = 0. \end{align*} On the other hand it is clear that \(\int_{-1}^1 f(x)\,dx = 0\).

Graph of \(f_n(x)=\frac{2xn+(-1)^n x^2}{n}\) for \(n=1,2,3,4\) and \(f(x) = 2x\)

Before considering more examples, we state the following result which is a direct consequence of the definition of the limit of a sequence of numbers and the definition of pointwise convergence.

Let \((f_n)\) be a sequence of functions on \(A\). Then \((f_n)\) converges pointwise to \(f:A\rightarrow\real\) if and only if for each \(x\in A\) and each \(\eps \gt 0\) there exists \(K\in\N\) such that \(|f_n(x) - f(x)| \lt \eps\) for all \(n\geq K\).

As the following example shows, it is important to note that the \(K\) in Lemma 8.1.4 depends not only on \(\eps \gt 0\) but in general will also depend on \(x\in A\).

Consider the sequence \((f_n)\) defined on \(A=[0,1]\) by \(f_n(x) = x^n\). For all \(n\in \N\) we have \(f_n(1) = 1^n = 1\) and therefore \(\lim_{n\rightarrow\infty} f_n(1) = 1\). On the other hand if \(x\in [0,1)\) then \[ \lim_{n\rightarrow\infty} f_n(x) = \limi x^n = 0. \] Therefore, \((f_n)\) converges pointwise on \(A\) to the function \[ f(x) = \begin{cases} 0, & \text{if \(x\in [0,1)\)} \\[2ex] 1, & \text{if \(x=1\).} \end{cases} \] In Figure 8.2, we graph \(f_n(x)=x^n\) for various values of \(n\). Consider a fixed \(x \in (0,1)\). Since \(\limi x^n = 0\) it follows that for \(\eps \gt 0\) there exists \(K\in\N\) such that \(|x^n - 0| \lt \eps\) for all \(n\geq K\). For \(\eps \lt 1\), in order for \(|x^K| = x^K \lt \eps\) we can choose \(K \gt \ln(\eps)/\ln(x)\). Notice that \(K\) clearly depends on both \(\eps\) and \(x\), and in particular, as \(x\) get closer to \(1\) then a larger \(K\) is needed. We note that each \(f_n\) is continuous while \(f\) is not.

Graph of \(f_n(x)=x^n\) for \(n=1,3,8,12,25\)

Example 8.1.5 also illustrates a weakness of pointwise convergence, namely, that if \((f_n)\) is a sequence of continuous functions on \(A\) and \((f_n)\) converges pointwise to \(f\) on \(A\) then \(f\) is not necessarily continuous on \(A\).

Consider the sequence \((f_n)\) defined on \(A=[-1,1]\) by \(f_n(x) = \sqrt{\frac{nx^2+1}{n}}\). For fixed \(x\in A\) we have \begin{align*} \limi f_n(x) &= \limi \sqrt{\frac{nx^2+1}{n}}\\ & = \limi \sqrt{x^2+\frac{1}{n}} \\ &= \sqrt{x^2}\\ & = |x|. \end{align*} Hence, \((f_n)\) converges pointwise on \(A\) to the function \(f(x)=|x|\). Notice that each function \(f_n\) is continuous on \(A\) and the pointwise limit \(f\) is also continuous. After some basic calculations we find that \[ f'_n(x) = \frac{x}{\sqrt{\frac{nx^2+1}{n}}} \] and \(f'_n(x)\) exists for each \(x\in [-1,1]\), in other words, \(f_n\) is differentiable on \(A\). However, \(f(x) = |x|\) is not differentiable on \(A\) since \(f\) does not have a derivative at \(x=0\). In Figure 8.3, we graph \(f_n\) for various values of \(n\).

Graph of \(f_n(x)=\sqrt{\frac{nx^2+1}{n}}\) for \(n=1,3,8,12,25\) and \(f(x)=|x|\)

Example 8.1.6 illustrates another weakness of pointwise convergence, namely, that if \((f_n)\) is a sequence of differentiable functions on \(A\) and \((f_n)\) converges pointwise to \(f\) on \(A\) then \(f\) is not necessarily differentiable on \(A\).

Consider the sequence \((f_n)\) on \(A=[0,1]\) defined by \(f_n(x) = 2nx e^{-nx^2}\). For fixed \(x\in [0,1]\) we find (using l'H\^{o}pital's rule) that \[ \limi f_n(x) = \limi \frac{2nx}{e^{nx^2}} = 0. \] Hence, \((f_n)\) converges pointwise to \(f(x) = 0\) on \(A\). Consider \begin{align*} \int_0^1 f_n(x)\, dx &= \int_0^1 2nx e^{-nx^2}\,dx\\ & = - e^{-nx^2}\Big|_0^1\\ & = 1 - e^{-n} \end{align*} and therefore \[ \limi \int_0^1 f_n(x) = \limi (1-e^{-n}) = 1. \] On the other hand, \(\int_0^1 f(x)\, dx = 0\). Therefore, \[ \int_0^1 f(x) \, dx \neq \limi \int_0^1 f_n(x)\, dx \] or another way to write this is \[ \limi \int_0^1 f_n(x)\, dx \neq \int_0^1 \limi f_n(x) \, dx. \]

Examples 8.1.5- 8.1.6 illustrate that the pointwise limit \(f\) of a sequence of functions \((f_n)\) does not always inherit the properties of continuity and/or differentiability, and Example 8.1.7 illustrates that unexpected (or surprising) results can be obtained when combining the operations of integration and limits, and in particular, one cannot in general interchange the limit operation with integration.

Exercises

Suppose that \(f_n:[a,b]\rightarrow\real\) is a sequence of functions such that \(f_n\) is increasing for each \(n\in \N\). Suppose that \(f(x)=\lim_{n\rightarrow\infty} f_n(x)\) exists for each \(x\in [a,b]\). Is \(f\) an increasing function?

Let \((a_n)\) be a sequence of positive numbers and define \(f_n:[0,1]\rightarrow\real\) as \[ f_n(x) = \begin{cases} 2na_nx, & 0\leq x\leq 1/(2n),\\[2ex] 2a_n - 2na_nx, & 1/(2n) \leq x \leq 1/n,\\[2ex] 0, & 1/n \leq x \leq 1. \end{cases} \]

1. Find the pointwise limit \(f:[0,1]\rightarrow\real\) of the sequence \((f_n)\).
2. Find \(\int_0^1 f(x)\, dx\).
3. If \(a_n = 4n\), find \[ \lim_{n\rightarrow\infty} \int_0^1 f_n(x)\, dx. \]

Recall that \(\mathbb{Q}\) is countable and thus there exists a bijection \(r:\mathbb{N}\rightarrow\mathbb{Q}\). Define the sequence \((r_n)\) by letting \(r_n = r(n)\). Now define \(f_n:\real\rightarrow\real\) as \[ f_n(x) = \begin{cases} 1, & x \in \{r_1, r_2, \ldots, r_n\}\\[2ex] 0, & \text{ otherwise.} \end{cases} \]

1. Find the pointwise limit \(f:\real\rightarrow\real\) of the sequence \((f_n)\).
2. Is \(f_n\) Riemann integrable? Explain.
3. Is \(f\) Riemann integrable? Explain.

Uniform Convergence

In the previous section we saw that pointwise convergence is a rather weak form of convergence since the limiting function will not in general inherit any of the properties possessed by the terms of the sequence. Examining the concept of pointwise convergence one observes that it is a very localized definition of convergence of a sequence of functions; all that is asked for is that \((f_n(x))\) converge for each \(x\in A\). This allows the possibility that the speed of convergence of \((f_n(x))\) may differ wildly as \(x\) varies in \(A\). For example, for the sequence of functions \(f_n(x)=x^n\) and \(x\in (0,1)\), convergence of \((f_n(x))\) to zero is much faster for values of \(x\) near \(0\) than for values of \(x\) near \(1\). What is worse, as \(x\rightarrow 1\) convergence of \((f_n(x))\) to zero is arbitrarily slow. Specifically, recall in Example 8.1.5 that \(|x^K - 0| \lt \eps\) if and only if \(K \gt \ln(\eps)/\ln(x)\). Thus, for a fixed \(\eps \gt 0\), as \(x\rightarrow 1\) we have \(K\rightarrow\infty\). Hence, there is no single \(K\) that will work for all values of \(x\in (0,1)\), that is, the convergence is not uniform.

Let \((f_n)\) be a sequence of functions on \(A\subseteq\real\). We say that \((f_n)\) converges uniformly on \(A\) to the function \(f:A\rightarrow\real\) if for any \(\eps \gt 0\) there exists \(K\in \N\) such that if \(n\geq K\) then \(|f_n(x) - f(x)| \lt \eps\) for all \(x\in A\).

Notice that in Definition 8.2.1, the \(K\) only depends on the given (but fixed) \(\eps \gt 0\) and the inequality \(|f_n(x)-f(x)| \lt \eps\) holds for all \(x\in A\) provided \(n\geq K\). The inequality \(|f_n(x)-f(x)| \lt \eps\) for all \(x\in A\) is equivalent to \[ f(x) - \eps \lt f_n(x) \lt f(x)+\eps \] for all \(x\in A\) and can therefore be interpreted as saying that the graph of \(f_n\) lies in the tube of radius \(\eps \gt 0\) and centered along the graph of \(f\), see Figure 8.4.

\(\eps\)-tubular neighborhood along the graph of \(f\); if \(|f_n(x)-f(x)| \lt \eps\) for all \(x\in A\) then the graph of \(f_n\) is within the \(\eps\)-tubular neighborhood of \(f\)

The following result is a direct consequence of the definitions but it is worth stating anyhow.

If \((f_n)\) converges uniformly to \(f\) then \((f_n)\) converges pointwise to \(f\).

Let \(A=[-5,5]\) and let \((f_n)\) be the sequence of functions on \(A\) defined by \(f_n(x) = (2xn+(-1)^nx^2)/n\). Prove that \((f_n)\) converges uniformly to \(f(x)=2x\).

We compute that \[ \limi f_n(x) = \limi \frac{2xn+(-1)^nx^2}{n} = 2x \] and thus \((f_n)\) converges pointwise to \(f(x)=2x\) on \(A\). To prove that the convergence is uniform, consider \begin{align*} |f_n(x) - f(x)| &= \left|\frac{2xn+(-1)^nx^2}{n} -2x \right| \\ &= \left| \frac{(-1)^n x^2}{n} \right|\\ & = \frac{|x|^2}{n} \\ &\leq \frac{5^2}{n}. \end{align*} For any given \(\eps \gt 0\) if \(K\in\N\) is such that \(\frac{5^2}{K} \lt \eps\) then if \(n\geq K\) then for any \(x\in A\) we have \begin{align*} |f_n(x) - f(x)| &\leq \frac{5^2}{n}\\ & \leq \frac{5^2}{K} \\ & \lt \eps. \end{align*} This proves that \((f_n)\) converges uniformly to \(f(x)=2x\) on \(A=[-5,5]\). Note that a similar argument will not hold if we take \(A=\real\).

Show that the sequence of functions \(f_n(x) = \sin(nx)/ \sqrt{n}\) converges uniformly to \(f(x)=0\) on \(\real\).

We compute \begin{align*} |f_n(x) - f(x)| &= \left| \frac{\sin(nx)}{\sqrt{n}} \right|\\ & = \frac{|\sin(nx)|}{\sqrt{n}}\\ & \leq \frac{1}{\sqrt{n}} \end{align*} and therefore if \(K\in\N\) is such that \(\frac{1}{\sqrt{K}} \lt \eps\) then if \(n\geq K\) then \(|f_n(x)-0| \lt \eps\) for all \(x\in \real\). Hence, \((f_n)\) converges uniformly to \(f=0\) on \(\real\).

On a close examination of the previous examples on uniform convergence, one observes that in proving that \((f_n)\) converges uniformly to \(f\) on \(A\), we used an inequality of the form: \[ |f_n(x) - f(x)|\leq M_n, \; \forall x \in A \] for some sequence \((M_n)\) of non-negative numbers such that \(\limi M_n = 0\). It follows that \[ \sup_{x\in A} |f_n(x) - f(x)| \leq M_n. \] This observation is worth formalizing.

Let \(f_n:A\rightarrow\real\) be a sequence of functions. Then \((f_n)\) converges uniformly to \(f\) on \(A\) if and only if there exists a sequence \((M_n)\) of non-negative numbers converging to zero such that \(\sup_{x\in A} |f_n(x) - f(x)| \leq M_n\) for \(n\) sufficiently large.

Suppose that \((f_n)\) converges uniformly on \(A\) to \(f\). There exists \(N\in\N\) such that \(|f_n(x)-f(x)| \lt 1\) for all \(n\geq N\) and \(x\in A\). Hence, \(M_n = \sup_{x\in A}|f_n(x)-f(x)|\geq 0\) is well-defined for all \(n\geq N\). Define \(M_n\geq 0\) arbitrarily for \(1\leq n\leq N-1\). Given an arbitrary \(\eps \gt 0\), there exists \(K\in \N\) such that if \(n\geq K\) then \(|f_n(x)-f(x)| \lt \eps\) for all \(x\in A\). We can assume that \(K\geq N\). Therefore, if \(n\geq K\) then \(M_n=\sup_{x\in A}|f_n(x)-f(x)| \leq \eps\). This prove that \(\limi M_n = 0\). Conversely, suppose that there exists \(M_n\geq 0\) such that \(\limi M_n = 0\) and \(\sup_{x\in A} |f_n(x) - f(x)| \leq M_n\) for all \(n\geq N\). Let \(\eps \gt 0\) be arbitrary. Then there exists \(K\in \N\) such that if \(n\geq K\) then \(M_n \lt \eps\). Hence, if \(n\geq K\geq N\) then \(\sup_{x\in A}|f_n(x) - f(x)|\leq M_n \lt \eps\). This implies that if \(n\geq K\) then \(|f_n(x)-f(x)| \lt \eps\) for all \(x\in A\), and thus \((f_n)\) converges uniformly to \(f\) on \(A\).

Let \(f\) be a continuous function on \([a, b]\). Prove that there exists a sequence of step functions \((s_n)\) on \([a,b]\) that converges uniformly to \(f\) on \([a,b]\).

We end this section by stating and proving a Cauchy criterion for uniform convergence.

The sequence \((f_n)\) converges uniformly on \(A\) if and only if for every \(\eps \gt 0\) there exists \(K\in \N\) such that if \(n, m\geq K\) then \(|f_m(x)-f_n(x)| \lt \eps\) for all \(x\in A\).

Suppose that \((f_n)\rightarrow f\) uniformly on \(A\) and let \(\eps \gt 0\). There exists \(K\in\N\) such that if \(n\geq K\) then \(|f_n(x)-f(x)| \lt \eps/2\) for all \(x\in A\). Therefore, if \(n,m \geq K\) then for all \(x\in A\) we have \begin{align*} |f_n(x)-f_m(x)| &= |f_n(x)-f(x)+f(x)-f_m(x)|\\ & \leq |f_n(x)-f(x)|+|f(x)-f_m(x)| \\ & \lt \eps/2 + \eps/2\\ &= \eps. \end{align*} To prove the converse, suppose that for every \(\eps \gt 0\) there exists \(K\in \N\) such that if \(n, m\geq K\) then \(|f_m(x)-f_n(x)| \lt \eps\) for all \(x\in A\). Therefore, for each \(x\in A\) the sequence \((f_n(x))\) is a Cauchy sequence and therefore converges. Let \(f:A\rightarrow \real\) be defined by \(f(x) = \limi f_n(x)\). If \(\eps \gt 0\) let \(K\in \N\) be such that \(|f_m(x)-f_n(x)| \lt \eps\) for all \(x\in A\) and \(n,m\geq K\). Fix \(m\geq K\) and consider the sequence \(z_n = |f_m(x)-f_n(x)|\) and thus \(z_n \lt \eps\). Now since \(\limi f_n(x) = f(x)\) then \(\lim z_n\) exists and \(\lim z_n \leq \eps\), that is, \begin{align*} \limi z_n &= \limi |f_m(x) - f_n(x)|\\ & = |f_m(x) - f(x)| \\ &\leq \eps. \end{align*} Therefore, if \(m\geq K\) then \(|f_m(x)-f(x)|\leq \eps\) for all \(x\in A\).

Exercises

Let \(f_n:A\rightarrow\real\) be a sequence of functions converging uniformly to \(f:A\rightarrow\real\). Let \(g:A\rightarrow\real\) be a function and let \(g_n = g f_n\) for each \(n\in\N\). Under what condition on \(g\) does the sequence \((g_n)\) converge uniformly? Prove it. What is the uniform limit of \((g_n)\)?

Prove that if \((f_n)\) converges uniformly to \(f\) on \(A\) and \((g_n)\) converges uniformly to \(g\) on \(A\) then \((f_n+g_n)\) converges uniformly to \(f+g\) on \(A\).

Let \(f_n:[0,1]\rightarrow\) be the sequence defined in Exercise 8.1.2. Show that if \(\lim_{n\rightarrow\infty} a_n = 0\) then \((f_n)\) converges uniformly.

Let \(f_n(x) = \sin(nx)/\sqrt{n}\) for \(x\in\real\). Prove that \((f_n)\) converges uniformly on \(\real\).

Properties of Uniform Convergence

A sequence \((f_n)\) on \(A\) is said to be uniformly bounded on \(A\) if there exists a constant \(M \gt 0\) such that \(|f_n(x)| \lt M\) for all \(x\in A\) and for all \(n\in\N\).

Suppose that \((f_n)\rightarrow f\) uniformly on \(A\). If each \(f_n\) is bounded on \(A\) then the sequence \((f_n)\) is uniformly bounded on \(A\) and \(f\) is bounded on \(A\).

By definition, there exists \(K\in\N\) such that \begin{align*} |f(x)| &\leq |f_n(x)-f(x)| + |f_n(x)|\\ & \lt 1 + |f_n(x)| \end{align*} for all \(x\in A\) and all \(n\geq K\). Since \(f_K\) is bounded, then \(|f(x)| \leq 1 + \max_{x\in A}|f_K(x)|\) for all \(x\in A\) and thus \(f\) is bounded on \(A\) with upper bound \(M'=1+\max_{x\in A}|f_K(x)|\). Therefore, \(|f_n(x)|\leq |f_n-f(x)| + |f(x)| \lt 1 + M'\) for all \(n\geq K\) and all \(x\in A\). Let \(M_n\) be an upper bounded for \(f_n\) on \(A\) for each \(n\in \N\). Then if \(M=\max\{M_1,\ldots,M_{K-1}, 1 + M'\}\) then \(|f_n(x)| \lt M\) for all \(x\in A\) and all \(n\in\N\).

Give an example of a set \(A\) and a sequence of functions \((f_n)\) on \(A\) such that \(f_n\) is bounded for each \(n\in \N\), \((f_n)\) converges pointwise to \(f\) but \((f_n)\) is not uniformly bounded on \(A\).

Unlike the case with pointwise convergence, a sequence of continuous functions converging uniformly does so to a continuous function.

Let \((f_n)\) be a sequence of functions on \(A\) converging uniformly to \(f\) on \(A\). If each \(f_n\) is continuous on \(A\) then \(f\) is continuous on \(A\).

To prove that \(f\) is continuous on \(A\) we must show that \(f\) is continuous at each \(c\in A\). Let \(\eps \gt 0\) be arbitrary. Recall that to prove that \(f\) is continuous at \(c\) we must show there exists \(\delta \gt 0\) such that if \(|x-c| \lt \delta\) then \(|f(x)-f(c)| \lt \eps\). Consider the following: \begin{align*} |f(x) - f(c)| &= |f(x) - f_n(x) + f_n(x)-f_n(c) + f_n(c) - f(c)|\\[2ex] & \leq |f(x)-f_n(x)| + |f_n(x)-f_n(c)| + |f_n(c) - f(c)|. \end{align*} Since \((f_n)\rightarrow f\) uniformly on \(A\), there exists \(K\in \N\) such that \(|f(x)-f_K(x)| \lt \eps /3\) for all \(x\in A\). Moreover, since \(f_K\) is continuous there exists \(\delta \gt 0\) such that if \(|x-c| \lt \delta\) then \(|f_K(x)-f_K(c)| \lt \eps/3\). Therefore, if \(|x-c| \lt \delta\) then \begin{align*} |f(x) - f(c)| & \leq |f(x)-f_K(x)| + |f_K(x)-f_K(c)| + |f_K(c) - f(c)| \\[2ex] & \lt \eps/3 + \eps/3 + \eps/3 \\ &= \eps. \end{align*} This proves that \(f\) is continuous at \(c\in A\).

A direct consequence of Theorem 8.3.3 is that if \((f_n)\rightarrow f\) pointwise and each \(f_n\) is continuous then if \(f\) is discontinuous then the convergence cannot be uniform.

Let \(\displaystyle g_n(x) = \frac{x}{\sqrt{x^2 + \frac{1}{n}}}\) for \(x\in [-1,1]\) and \(n\in\N\). Each function \(g_n\) is clearly continuous. Now \(g_n(0) = 0\) and thus \(\limi g_n(0) = 0\). If \(x\neq 0\) then \begin{align*} \limi g_n(x) &= \limi \frac{x}{\sqrt{x^2 + \frac{1}{n}}}\\ & = \frac{x}{\sqrt{x^2}}\\ & = \frac{x}{|x|} \\ &= \begin{cases} 1, & x \gt 0 \\ -1, & x \lt 0,\end{cases} \end{align*} Therefore, \((g_n)\) converges pointwise to the function \[ g(x) = \begin{cases} -1, & -1\leq x \lt 0 \\ 0, & x=0\\ 1, & 0 \lt x\leq 1.\end{cases} \] The function \(g\) is discontinuous and therefore, by Theorem 8.3.3, \((g_n)\) does not converge uniformly to \(g\).

The next property that we can deduce from uniform convergence is that the limit and integration operations can be interchanged. Recall from Example 8.1.7 that if \((f_n)\rightarrow f\) pointwise then it is not necessarily true that \[ \limi \int_A f_n= \int_A f \] Since \(\limi f_n(x) = f(x)\), then it in general we can say that \[ \limi \int_A f_n \neq \int_A \limi f_n \] However, when the convergence is uniform we can indeed interchange the limit and integration operations.

Let \((f_n)\) be a sequence of Riemann integrable functions on \([a,b]\). If \((f_n)\) converges uniformly to \(f\) on \([a,b]\) then \(f\in\mathcal{R}[a,b]\) and \[ \limi \int_a^b f_n = \int_a^b f. \]

Let \(\eps \gt 0\) be arbitrary. By uniform convergence, there exists \(K\in\N\) such that if \(n\geq K\) then for all \(x\in [a,b]\) we have \[ |f_n(x) - f(x)| \lt \frac{\eps}{4(b-a)} \] or \[ f_n(x) - \frac{\eps}{4(b-a)} \lt f(x) \lt f_n(x) + \frac{\eps}{4(b-a)}. \] By assumption, \(f_n \pm \frac{\eps}{4(b-a)}\) is Riemann integrable and thus if \(n\geq N\) then \[ \int_a^b \left[(f_n+\eps/4(b-a)) - (f_n - \eps/4(b-a))\right] = \frac{\eps}{2} \lt \eps. \] By the Squeeze Theorem of Riemann integration (Theorem 7.2.3), \(f\) is Riemann integrable. Moreover, if \(n\geq N\) then \[ - \frac{\eps}{4(b-a)} \lt f_n(x) - f(x) \lt \frac{\eps}{4(b-a)} \] implies (by monotonicity of integration) \[ -\frac{\eps}{4} \lt \int_a^b f_n - \int_a^b f \lt \frac{\eps}{4} \] and thus \[ \left| \int_a^b f_n - \int_a^b f\right| \lt \frac{\eps}{4}. \] This proves that the sequence \(\int_a^b f_n \) converges to \(\int_a^b f\).

The following corollary to Theorem 8.3.5 is worth noting.

Let \((f_n)\) be a sequence of continuous functions on the interval \([a,b]\). If \((f_n)\) converges uniformly to \(f\) then \(f\in\mathcal{R}[a,b]\) and \[ \limi \int_a^b f_n = \int_a^b f. \]

If each \(f_n\) is continuous then \(f_n\in\mathcal{R}[a,b]\) and Theorem 8.3.5 applies.

Consider the sequence of functions \((f_n)\) defined on \([0,1]\) given by \[ f_n(x) = \begin{cases} (n+1)^2x, & 0\leq x\leq \frac{1}{n+1} \\ -(n+1)^2\left(x-\frac{2}{n+1}\right), & \frac{1}{n+1}\leq x\leq \frac{2}{n+1}\\ 0, & \frac{2}{n+1} \lt x \leq 1. \end{cases} \]

1. Draw a typical function \(f_n\).
2. Prove that \((f_n)\) converges pointwise.
3. Use Theorem 8.3.5 to show that the convergence is not uniform.

We now consider how the operation of differentiation behaves under uniform convergence. One would hope, based on the results of Theorem 8.3.3, that if \((f_n)\rightarrow f\) uniformly and each \(f_n\) is differentiable then \(f'\) is also differentiable and maybe even that \((f_n')\rightarrow f'\) at least pointwise and maybe even uniformly. Unfortunately, the property of differentiability is not generally inherited under uniform convergence. An example of this occurred in Example 8.1.6 where \(f_n(x) = \sqrt{(nx^2+1)/n}\) and \((f_n)\rightarrow f\) where \(f(x)=|x|\) for \(x\in [-1,1]\). The convergence in this case is uniform on \([-1,1]\) but although each \(f_n\) is differentiable the limit function \(f(x)=|x|\) is not. It turns out that the main assumption needed for all to be well is that the sequence \((f'_n)\) converge uniformly.

Let \((f_n)\) be a sequence of differentiable functions on \([a,b]\). Assume that \(f'_n\) is Riemann integrable on \([a,b]\) for each \(n\in\N\) and suppose that \((f_n')\) converges uniformly to \(g\) on \([a,b]\). Suppose there exists \(x_0\in [a,b]\) such that \((f_n(x_0))\) converges. Then the sequence \((f_n)\) converges uniformly on \([a,b]\) to a differentiable function \(f\) and \(f'=g\).

Let \(x\in [a,b]\) be arbitrary but with \(x\neq x_0\). By the Mean Value theorem applied to the differentiable function \(f_m - f_n\), there exists \(y\) in between \(x\) and \(x_0\) such that \[ \frac{(f_m(x)-f_n(x)) - (f_m(x_0)-f_n(x_0))}{x-x_0} = f'_m(y)-f'_n(y) \] or equivalently \[ f_m(x) - f_n(x) = f_m(x_0) - f_n(x_0) + (x-x_0) (f'_m(y)-f'_n(y)) \] Therefore, \[ |f_m(x) - f_n(x)| \leq |f_m(x_0)-f_n(x_0)| + (b-a) |f'_m(y) - f'_n(y)|. \] Since \((f_n(x_0))\) converges and \((f'_n)\) is uniformly convergent, by the Cauchy criterion, for any \(\eps \gt 0\) there exists \(K\in\N\) such that if \(n,m\geq K\) then \(|f_m(x_0)-f_n(x_0)| \lt \eps/2\) and \(|f'_m(y)-f'_n(y)| \lt (\eps/2)/(b-a)\) for all \(y\in [a,b]\). Therefore, if \(m,n\geq K\) then \begin{align*} |f_m(x) - f_n(x)| &\leq |f_m(x_0)-f_n(x_0)| + (b-a) |f'_m(y) - f'_n(y)|\\ & \lt \eps \end{align*} and this holds for all \(x\in[a,b]\). By the Cauchy criterion for uniform convergence, \((f_n)\) converges uniformly. Let \(f\) be the uniform limit of \((f_n)\). We now prove that \(f\) is differentiable and \(f'=g\). By the Fundamental theorem of Calculus (FTC), we have that \[ f_n(x) = f_n(a) + \int_a^x f'_n(t)\,dt \] for each \(x\in [a,b]\). Since \((f_n)\) converges to \(f\) and \((f'_n)\) converges uniformly to \(g\) we have \begin{align*} f(x) &= \limi f_n(x)\\ &= \limi \left( f_n(a) + \int_a^x f'_n(t)\,dt\right) \\[2ex] &= \limi f_n(a) + \limi \int_a^x f'_n(t)\,dt \\[2ex] &= f(a) + \int_a^x g(t) \, dt. \end{align*} Thus \(f(x) = f(a) + \int_a^x g(t)\, dt\) and by the FTC we obtain \(f'(x) = g(x)\).

Notice that in the statement of Theorem 8.3.8, all that is required is that \((f_n(x_0))\) converge for one \(x_0\in [a,b]\). The assumption that \((f'_n)\) converges uniformly then guarantees that in fact \((f_n)\) converges uniformly.

Consider the sequence \((f_n)\) defined on \([-1,1]\) by \(f_n(x) = (2xn+(-1)^nx^2)/n\). We compute that \(f'_n(x) =(2n+2(-1)^nx)/n\) and clearly \(f'_n\) is continuous on \([-1,1]\) for each \(n\in \N\). Now \(\limi f'_n(x) = 2\) for all \(x\) and thus \((f'_n)\) converges pointwise to \(g(x)=2\). To prove that the convergence is uniform we note that \begin{align*} |f'_n(x) - g(x)| &= |2+(-1)^n \frac{x}{n} - 2|\\ & = \frac{|x|}{n}\\ & \leq \frac{1}{n}. \end{align*} Therefore, \((f'_n)\) converges uniformly to \(g\) on \([-1,1]\). Now \(f_n(0)=0\) and thus \((f_n(0))\) converges to \(0\). By Theorem 8.3.8, \((f_n)\) converges uniformly to say \(f\) with \(f(0)=0\) and \(f'=g\). Now by the FTC, \(f(x) = \int g(x)\, dx + C = 2x + C\) and since \(f(0) = 0\) then \(f(x) = 2x\).

Exercises

Give an example of a set \(A\) and a sequence of functions \((f_n)\) on \(A\) such that \(f_n\) is bounded for each \(n\in \N\), \((f_n)\) converges pointwise to \(f\) but \((f_n)\) is not uniformly bounded on \(A\).

Suppose that \((f_n)\rightarrow f\) uniformly on \(A\) and \((g_n)\rightarrow g\) uniformly on \(A\). Prove that if \((f_n)\) and \((g_n)\) are uniformly bounded on \(A\) then \((f_ng_n)\) converges uniformly to \(fg\) on \(A\). Then give an example to show that if one of \((f_n)\) or \((g_n)\) is not uniformly bounded then the result is false.

Let \[ f_n(x) = \frac{nx^2}{1+nx^2} \] for \(x\in \real\) and \(n\in \N\).

1. Show that \((f_n)\) converges pointwise on \(\real\).
2. Show that \((f_n)\) does not converge uniformly on any closed interval containing \(0\).
3. Show that \((f_n)\) converges uniformly on any closed interval not containing \(0\). For instance, take \([a,b]\) with \(0 \lt a \lt b\).

Suppose that \(f:\real\rightarrow\real\) has that property that \(|f(x) - f(y)|\leq K |x-y|\) for all \(x, y, \in \real\) and some \(K \gt 0\). Prove that if \((g_n)\) converges uniformly on \(\real\) to \(g\) then the sequence \((f\circ g_n)\) converges uniformly to \(f\circ g\) on \(\real\). Note: \(f\circ g_n\) and \(f\circ g\) are compositions of functions and not function multiplication.

Let \(f_n(x) = nx / (nx + 1)\) for \(n\in\mathbb{N}\) and \(x\in [a, 1]\) where \(0 \lt a \lt 1\).

1. Prove directly that the sequence \((f_n)\) is uniformly Cauchy.
2. If \(f\) is the uniform limit of \((f_n)\), find \(\int_a^1 f\) without computing \(f\).

Consider the sequence of functions \((f_n)\) on \(A=[0, \infty)\) defined as follows: \[ f_n(x) = \begin{cases} 1/n, & 0\leq x\leq n^2,\\[2ex] 0, & x \gt n^2. \end{cases} \]

1. Prove that \((f_n)\) converges uniformly to \(f=0\) on \(A\).
2. For each fixed \(n\in\mathbb{N}\), find the improper integral \[ \int_0^\infty f_n \] and show that \[ \lim_{n\rightarrow\infty} \int_0^\infty f_n = \infty. \]
3. The results above seem to contradict Theorem 8.3.5. Explain why there is no contradiction.

Infinite Series of Functions

In this section, we consider series whose terms are functions. You have already encountered such objects when studying power series in Calculus. An example of an infinite series of functions (more specifically a power series) is \[ \sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n)!}. \] In this case, if we set \(f_n(x) = \frac{(-1)^n x^n}{(2n)!}\) then the above infinite series is \(\sum_{n=0}^\infty f_n(x)\). Let us give the general definition.

Let \(A\) be a non-empty subset of \(\real\). An infinite series of functions on \(A\) is a series of the form \(\sum_{n=1}^\infty f_n(x)\) for each \(x\in A\) where \((f_n)\) is a sequence of functions on \(A\). The sequence of partial sums generated by the series \(\sum f_n\) is the sequence of functions \((s_n)\) on \(A\) defined as \(s_n(x) = f_1(x) + \cdots + f_n(x)\) for each \(x\in A\).

Recall that a series of numbers \(\sum x_n\) converges if the sequence of partial sums \((t_n)\), defined as \(t_n=x_1+x_2+\cdots+x_n\), converges. Hence, convergence of an infinite series of functions \(\sum f_n\) is treated by considering the convergence of the sequence of partial sums \((s_n)\) (which are functions). For example, to say that the series \(\sum f_n\) converges uniformly to a function \(f\) we mean that the sequence of partial sums \((s_n)\) converges uniformly to \(f\), etc. It is now clear that our previous work in Sections 8.1-8.3 translate essentially directly to infinite series of functions. As an example:

Let \((f_n)\) be a sequence of functions on \(A\) and suppose that \(\sum f_n\) converges uniformly to \(f\). If each \(f_n\) is continuous on \(A\) then \(f\) is continuous on \(A\).

By assumption, the sequence of functions \(s_n(x) = \sum_{k=1}^n f_k(x)\) for \(x\in A\) converges uniformly to \(f\). Since each function \(f_n\) is continuous, and the sum of continuous functions is continuous, it follows that \(s_n\) is continuous. The result now follows by Theorem 8.3.3.

The following translate of Theorem 8.3.5 is worth explicitly writing out.

Let \((f_n)\) be a sequence of functions on \([a,b]\) and suppose that \(\sum f_n\) converges uniformly to \(f\). If each \(f_n\) is Riemann integrable on \([a,b]\) then \(f\in\mathcal{R}[a,b]\) and \[ \int_a^b \left( \sum_{n=1}^\infty f_n(x)\right)\, dx = \sum_{n=1}^\infty \int_a^b f_n(x)\,dx. \]

By assumption, the sequence \((s_n)\) defined as \(s_n(x) = f_1(x) + \cdots + f_n(x)\) converges uniformly to \(f\). Since each \(f_n\) is Riemann integrable then \(s_n\) is Riemann integrable and therefore \(f=\lim s_n = \sum f_n\) is Riemann integrable by Theorem 8.3.5. Also by Theorem 8.3.5, we have \[ \int_a^b f = \lim_{n\rightarrow\infty} \int_a^b s_n \] or written another way is \[ \int_a^b \sum_{n=1}^\infty f_n = \lim_{n\rightarrow\infty} \int_a^b \sum_{k=1}^n f_k \] or \[ \int_a^b \sum_{n=1}^\infty f_n = \lim_{n\rightarrow\infty} \sum_{k=1}^n \int_a^b f_k \] or \[ \int_a^b \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int_a^b f_n \]

We now state the derivative theorem (similar to Theorem 8.3.8) for infinite series of functions.

Let \((f_n)\) be a sequence of differentiable functions on \([a,b]\) and suppose that \(\sum f_n\) converges at some point \(x_0\in [a,b]\). Assume further that \(\sum f'_n\) converges uniformly on \([a,b]\) and each \(f_n'\) is continuous. Then \(\sum f_n\) converges uniformly to some differentiable function \(f\) on \([a,b]\) and \(f' = \sum f'_n\).

We now state a useful theorem for uniform convergence of infinite series of functions.

Let \((f_n)\) be a sequence of functions on \(A\) and suppose that there exists a sequence of non-negative numbers \((M_n)\) such that \(|f_n(x)|\leq M_n\) for all \(x\in A\), and all \(n\in\N\). If \(\sum M_n\) converges then \(\sum f_n\) converges uniformly on \(A\).

Let \(\eps \gt 0\) be arbitrary. Let \(t_n = \sum_{k=1}^n M_k\) be the sequence of partial sums of the series \(\sum M_n\). By assumption, \((t_n)\) converges and thus \((t_n)\) is a Cauchy sequence. Hence, there exists \(K\in\N\) such that \(|t_m - t_n| \lt \eps\) for all \(m \gt n\geq K\). Let \((s_n)\) be the sequence of partial sums of \(\sum f_n\). Then if \(m \gt n\geq K\) then for all \(x\in A\) we have \begin{align*} |s_m(x) - s_n(x)| &= |f_m(x) + f_{m-1}(x) + \cdots + f_{n+1}(x)| \\[2ex] &\leq |f_m(x)| + |f_{m-1}(x)| + \cdots + |f_{n+1}(x)| \\[2ex] &\leq M_m + M_{m-1} + \cdots + M_{n+1}\\[2ex] &= |t_m - t_n| \\ & \lt \eps. \end{align*} Hence, the sequence \((s_n)\) satisfies the Cauchy Criterion for uniform convergence (Theorem 8.2.7) and the proof is complete.

Prove that \[ \int_0^\pi \left( \sum_{n=1}^\infty \frac{n\sin(nx)}{e^n} \right) = \frac{2e}{e^2-1} \]

For any \(x\in\real\) it holds that \[ \left| \frac{n\sin(nx)}{e^n}\right| \leq \frac{n}{e^n}. \] A straightforward application of the Ratio test shows that \(\sum_{n=1}^\infty \frac{n}{e^n}\) is a convergent series. Hence, by the \(M\)-Test, the given series converges uniformly on \(A=\real\), and in particular on \([0,\pi]\). By Theorem 8.4.3, \begin{align*} \int_0^\pi \sum_{n=1}^\infty \frac{n\sin(nx)}{e^n} \, dx &= \sum_{n=1}^\infty \int_0^\pi \frac{n\sin(nx)}{e^n} \, dx \\[2ex] &= \sum_{n=1}^\infty -\frac{\cos(nx)}{e^n}\Big|_0^\pi \\[2ex] &= \sum_{n=1}^\infty \left[ \left(\tfrac{1}{e}\right)^n - \left(\tfrac{-1}{e} \right)^n\right]\\[2ex] &= \left(\tfrac{1}{1-1/e} - 1\right) - \left(\tfrac{1}{1+1/e}-1\right)\\ &= \frac{2e}{e^2-1} \end{align*}

Consider the function \(r(x)\) whose graph is given in Figure 8.5; one can write down an explicit expression for \(r(x)\) but the details are unimportant. Consider the series \[ \sum_{n=1}^\infty \frac{r(nx)}{n^2}. \] Since \[ \left| \frac{r(nx)}{n^2} \right| \leq \frac{1/2}{n^2} \] and \(\sum_{n=1}^\infty \frac{1}{2n^2}\) converges, then by the \(M\)-test the above series converges uniformly on any interval \([a,b]\). Let \(f\) be the function defined by the series on \([a,b]\). Now, on \([a,b]\), the function \(f_n(x) = \frac{r(nx)}{n^2}\) has only a finite number of discontinuities and thus \(f_n\) is Riemann integrable. Therefore, by Theorem 8.3.5, the function \(f\) is Riemann integrable. The graph of \(f\) is shown in Figure 8.6. One can show that \(f\) has discontinuities at the rational points \(x= \frac{p}{2q}\) where \(\gcd(p,q)=1\).

The function \(r(x)\)

The function \(f(x) = \sum_{n=1}^\infty \frac{r(nx)}{n^2}\)

Recall that a power series is a series of the form \[ \sum_{n=0}^\infty c_n (x-a)^n \] where \(c_n\in\real\) and \(a\in\real\). Hence, in this case if we write the series as \(\sum f_n(x)\) then \(f_n(x) = c_n (x-a)^n\) for each \(n\in\N\) and \(f_0(x)=c_0\). In calculus courses, the main problem you were asked to solve is to find the interval of convergence of the given power series. The main tool is to apply the Ratio test (Theorem 3.7.23): \begin{align*} \lim_{n\rightarrow\infty} \frac{|c_{n+1}| |x-a|^{n+1}}{|c_n| |x-a|^n} = |x-a| \lim_{n\rightarrow\infty} \frac{|c_{n+1}|}{|c_n|}. \end{align*} Suppose that \(\lim_{n\rightarrow\infty} \frac{|c_{n+1}|}{|c_n|}\) exists and is non-zero and \(\lim_{n\rightarrow\infty} \frac{|c_{n+1}|}{|c_n|} = \frac{1}{R}\) (a similar argument can be done when the limit is zero). Then by the Ratio test, the power series converges if \(|x-a| \frac{1}{R} \lt 1\), that is, if \(|x-a| \lt R\). The number \(R \gt 0\) is called the radius of convergence and the interval \((a-R, a+R)\) is the interval of convergence (if \(\lim_{n\rightarrow\infty} \frac{|c_{n+1}|}{|c_n|}\) is zero then \(R \gt 0\) can be chosen arbitrarily and the argument that follows is applicable). Let \(\rho \lt r \lt R\) and consider the closed interval \([a-\rho, a+\rho] \subset (a-R, a+R)\). Then if \(x\in [a-\rho, a+\rho]\) then \begin{align*} |f_n(x)| &= |c_n| |x-a|^n\\ & = |c_n| r^n \frac{|x-a|^n}{r^n} \\ &\leq |c_n| r^n \left(\frac{\rho}{r}\right)^n. \end{align*} Now if \(x=a+r \in (a-R, a+R)\) then by assumption the series \(\sum c_n (x-a)^n = \sum c_n r^n\) converges, and in particular the sequence \(|c_n| r^n\) is bounded, say by \(M\). Therefore, \[ |f_n(x)| \leq M \left(\frac{\rho}{r}\right)^n. \] Since \(\rho/ r \lt 1\), the geometric series \(\sum \left(\frac{\rho}{r}\right)^n\) converges. Therefore, by the \(M\)-test, the series \(\sum c_n (x-a)^n\) converges uniformly on the interval \([a-\rho, a+\rho]\). Let \(f(x) = \sum f_n(x)\) for \(x\in [a-\rho,a+\rho]\). Now consider the series of the derivatives \[ \sum_{n=1}^\infty f'_n(x) = \sum_{n=1}^\infty c_n n (x-a)^{n-1}. \] Applying the Ratio test again we conclude that the series of the derivatives converges for each \(x\in (a-R, a+R)\) and a similar argument as before shows that the series of derivatives converges uniformly on any interval \([a-\rho, a+\rho]\) where \(\rho \lt R\). It follows from the Term-by-Term Differentiation theorem that \(f\) is differentiable and \(f'(x) = \sum c_n n (x-a)^{n-1}\). By the Term-by-Term Integration theorem, we can also integrate the series and \[ \int_I \left(\sum f_n (x)\right) \, dx = \sum \int_I f_n(x)\, dx \] where \(I\subset (a-R,a+R)\) is any closed and bounded interval.

Consider the power series \[ \sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}  \text{ and }   \sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}. \]

1. Prove that the series converges at every \(x\in\real\).
2. Let \(f\) denote the function defined by the series on the left and let \(g\) denote the function defined by the series on the right. Justifying each step, show that \(f'\) exists and that \(f' = g\).
3. Similarly, show that \(g'\) exists and \(g' = - f\).

A Fourier series is a series of the form \[ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(nx) + b_n\sin(nx)) \] where \(a_n, b_n \in \real\).

1. Suppose that for a given \((a_n)\) and \((b_n)\), the associated Fourier series converges pointwise on \([-\pi, \pi]\) and let \(f\) be the pointwise limit. Prove that in fact the Fourier series converges on \(\real\). Hint: For any \(y\in\real\) there exists \(x\in [-\pi,\pi]\) such that \(y=x+2\pi\).
2. Prove that if \(\sum |a_n|\) and \(\sum |b_n|\) are convergent series then the associated Fourier series converges uniformly on \(\real\).
3. Suppose that for a given \((a_n)\) and \((b_n)\), the associated Fourier series converges uniformly on \([-\pi, \pi]\) and let \(f\) be the uniform limit. Prove the following: \begin{align*} a_0 &= \int_{-\pi}^\pi f(x)dx \\[2ex] a_n &= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx) dx\\[2ex] b_n &= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx) dx \end{align*} You will need the following identities: \begin{gather*} \int_{-\pi}^\pi\sin(nx)\cos(mx)dx = 0,  \forall\; n, m \in N \\[2ex] \int_{-\pi}^\pi\sin(nx)\sin(mx)dx = \int_{-\pi}^\pi\cos(nx)\cos(mx)dx = \begin{cases}\pi, & m=n\\ 0, & m\neq n\end{cases} \end{gather*}

1. By an open cover of an interval \([a,b]\), we mean a collection of open intervals \(\{I_\mu\}_{\mu \in X}\) such that \(I_\mu\cap [a,b]\neq \emptyset\) for each \(\mu \in X\) and \[ [a,b] \subset \bigcup_{\mu \in X} I_\mu. \] Here \(X\) is some set, possibly uncountable. Prove that if \(\{I_\mu\}_{\mu \in X}\) is any open cover of \([a,b]\) then there there exists finitely many \(\mu_1,\mu_2,\ldots,\mu_N\in X\) such that \[ [a,b] \subset \bigcup_{k=1}^N I_{\mu_k}. \]
2. Let \(f_n\) be continuous and suppose that \(f_{n+1}(x) \leq f_n(x)\) for all \(x\in [a,b]\) and all \(n\in \N\). Suppose that \((f_n)\) converges pointwise to a continuous function \(f\). Prove that the convergence is actually uniform. Give an example to show that if \(f\) is not continuous then we only have pointwise convergence.

We first prove (a). For convenience, write \(I_\mu = (a_\mu, b_\mu)\) for each \(\mu \in X\) and assume without loss of generality that \(\{b_\mu\;|\; \mu\in X\}\) is bounded above. Let \(b_0 = a\) and let \(I_{\mu_1}\) be such that \(b_0 \in I_{\mu_1}=(a_1, b_1)\) and \[ b_1 = \sup\{ b_\mu \;|\; b_0 \in I_\mu\}. \] If \(b_1 \gt b\) then we are done because then \([a,b]\subset I_{\mu_1}\). By induction, having defined \(b_{k-1}\in [a,b]\), let \(I_{\mu_k}=(a_k, b_k)\) be such that \(b_{k-1} \in I_{\mu_k}\) and \(b_k = \sup\{b_\mu \; |\; b_{k-1} \in I_\mu\}\). We claim that \(b_N \gt b\) for some \(N\in\N\) and thus \([a,b] \subset \cup_{k=1}^N I_{\mu_k}\). To prove the claim, suppose that \(b_k \leq b\) for all \(k\in\N\). Then the increasing sequence \((b_k)\) converges by the Monotone Convergence theorem, say to \(L = \sup\{b_1,b_2,\ldots\}\). Since \(L\in [a,b]\) then \(L\in I_\mu\) for some \(\mu\in X\) and thus by convergence there exists \(k\in \N\) such that \(b_k \in I_\mu=(a_\mu,b_\mu)\). However, by definition of \(b_{k+1}\) we must have that \(L \lt b_\mu \leq b_{k+1}\) which is a contradiction to the definition of \(L\). This completes the proof. Now we prove (b). First of all since \(f_m(x)\leq f_n(x)\) for all \(m\geq n\) it holds that \(f(x)\leq f_n(x)\) for all \(n\in\N\) and all \(x\in [a,b]\). Fix \(\tilde{x}\in[a,b]\) and let \(\eps \gt 0\). By pointwise convergence, there exists \(N\in \N\) such that \(|f_n(\tilde{x}) - f(\tilde{x})| \lt \eps/3\) for all \(n\geq N\). By continuity of \(f_N\) and \(f\), there exists \(\delta_{\tilde{x}} \gt 0\) such that \(|f_N(x)-f_N(\tilde{x})| \lt \eps/3\) and \(|f(x)-f(\tilde{x})| \lt \eps/3\) for all \(x \in I_{\tilde{x}} = (\tilde{x}-\delta_{\tilde{x}}, \tilde{x} + \delta_{\tilde{x}})\). Therefore, if \(n\geq N\) then \begin{align*} |f_n(x)-f(x)| &= f_n(x)-f(x)\\ &\leq f_N(x) - f(x)\\ &\leq |f_N(x)-f_N(\tilde{x})| + |f_N(\tilde{x}-f(\tilde{x})| + |f(\tilde{x})-f(x)|\\ & \lt \eps \end{align*} for all \(x\in I_{\tilde{x}}\). Hence, \((f_n)\) converges uniformly to \(f\) on the interval \(I_{\tilde{x}}\). It is clear that \(\{I_{\tilde{x}}\}_{\tilde{x}\in [a,b]}\) is an open cover of \([a,b]\). Therefore, by part (a), there exists \(\tilde{x}_1,\tilde{x}_2,\ldots,\tilde{x}_k\) such that \([a,b] \subset I_{\tilde{x}_1} \cup \cdots \cup I_{\tilde{x}_k}\). Hence, for arbitrary \(\eps \gt 0\) there exists \(N_j\in \N\) such that if \(n\geq N_j\) then \(|f_n(x)-f(x)| \lt \eps\) for all \(x\in I_{\tilde{x}_j}\). If \(N=\max\{N_1,N_2,\ldots,N_k\}\) then \(|f_n(x)-f(x)| \lt \eps\) for all \(n\geq N\) and all \(x\in [a,b]\). This completes the proof. The sequence \(f_n(x)=x^n\) on \([0,1]\) satisfies \(f_{n+1}(x)\leq f_n(x)\) for all \(n\) and all \(x\in [0,1]\), and \((f_n)\) converges to \(f(x)=0\) if \(x\in [0,1)\) and \(f(1)=1\). Since \(f\) is not continuous on \([0,1]\), the convergence is not uniform.

Let \(g_n\) be continuous and suppose that \(g_n \geq 0\) for all \(n\in\N\). Prove that if \(\sum_{n=1}^\infty g_n\) converges pointwise to a continuous function \(f\) on \([a,b]\) then in fact the convergence is uniform.

Exercises

Show that \(\sum_{n=0}^\infty x^n\) converges uniformly on \([-a,a]\) for every \(a\) such that \(0 \lt a \lt 1\). Then show that the given series does not converge uniformly on \((-1,1)\). Hint: This is an important series and you should know what function the series converges uniformly to.

If \(\sum_{n=1}^\infty |a_n| \lt \infty\) prove that \(\sum_{n=1}^\infty a_n \sin(nx)\) converges uniformly on \(\real\).

Prove, justifying each step, that \[ \int_1^2 \left(\sum_{n=1}^\infty n e^{-nx} \right) dx = \frac{e}{e^2-1} \]

For any number \(q\in \real\) let \(\chi_q:\real\rightarrow\real\) be the function defined as \(\chi_q(x) = 1\) if \(x=q\) and \(\chi_q(x)=0\) if \(x\neq q\). Let \(\{q_1, q_2, q_3, \ldots\}=\mathbb{Q}\) be an enumeration of the rational numbers. Define \(f_n:\real\rightarrow\real\) as \[ f_n = \chi_{q_1} + \chi_{q_2} + \cdots + \chi_{q_n}. \] Find the pointwise limit \(f\) of the sequence \((f_n)\). Is the convergence uniform? Explain.

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Source: https://www.geneseo.edu/~aguilar/public/notes/Real-Analysis-HTML/ch8-sequences-of-functions.html

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