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Since \(d_{n}=\sup\left\{\left|\dfrac{x}{1+nx}\right|:x\in[0,1],n\in\mathbb{N}\right\}\), choose \(d_{n}=\dfrac{1}{1+n}\) which goes to \(0\) as \(n\to\infty\). Therefore, we have uniform convergence.
**Example 83**.: When a sequence of integrable functions \(\{f_{n}(x)\}\) converges pointwise to \(f(x)\) on an interval \([a,b]\), the limit function \(f\) need not to be integrable (Figure 2.14). For example, take \(f_{n}(x)=ng_{n}(x)\) where \(g_{n}(x)\) is the “triangle function” defined as:
\[g_{n}(x)=\begin{cases}nx&\text{ if }x\in[0,1/n]\\ 2-nx&\text{ if }x\in[1/n,2/n]\\ 0&\text{ if }x\in[2/n,1]\end{cases}.\]
The sequence \(\{f_{n}\}\) converges pointwise (but not uniformly) to \(f\equiv 0\), yet
\[\int_{0}^{1}f_{n}(x)\,dx=1,\ \text{ for every n, whereas }\int_{0}^{1}f(x)\,dx=0.\]
The next theorem shows that integrals behave nicely under uniform convergence.
**Theorem 94** (Interchange of limit and integral).: _If the functions \(f_{1},f_{2},\dots\) are integrable over an interval \([a,b]\) and if \(f_{n}\rightrightarrows f\) (uniformly) on \([a,b]\), then \(f\) is integrable and_
\[\int_{a}^{b}f(x)\,dx=\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx.\]
Proof
: We again appeal to uniform convergence. Thus for all \(\epsilon>0\), there exists \(N\in\mathbb{N}\) such that for all \(n>N\), \(|f_{n}(x)-f(x)|<\epsilon\). So we have
\[\left|\int_{a}^{b}f_{n}-\int_{a}^{b}f\right|=\left|\int_{a}^{b}f_{n}-f\right| \leq\int_{a}^{b}|f_{n}-f|<\epsilon(b-a).\]
It remains to show that \(f\) is integrable. We need to show \(U(f,P)-L(f,P)<\epsilon\) where \(U(f,P)\) and \(L(f,P)\) are upper and lower sums as used in the definition of the Riemann integral. However, for any partition \(P\) of \([a,b]\),
\[U(f,P)-L(f,P) \leq |U(f,P)-U(f_{n},P)|+|U(f_{n},P)-L(f_{n},P)|\] \[+|L(f_{n},P)-L(f,P)|\]
holds true. Let \(\epsilon>0\), then by uniform convergence there is an \(N\) such that the inequalities
\[|U(f,P)-U(f_{n},P)|<\frac{\epsilon}{3}\ \ \text{and}\ \ |L(f_{n},P)-L(f,P)|< \frac{\epsilon}{3}\]
hold true for all \(n\geq N\) and for all partitions \(P\) of \([a,b]\). On the other hand, integrability of \(f_{n}\) allows for a partition P for which \(|U(f_{n},P)-L(f_{n},P)|<\frac{\epsilon}{3}\).
To see how uniform convergence behaves with respect to differentiation, consider the sequence of functions \(f_{n}(x)=|x|^{1+\frac{1}{n}}\). Note that \(f_{n}(x)\to f(x)=|x|\) and that \(f_{n}\rightrightarrows f\) (uniformly) on \([-1,1]\). Even though each \(f_{n}\) is differentiable, \(f\) is not differentiable at zero. Thus while continuity is preserved, differentiability is not preserved under uniform convergence. Another common example is the sequence of functions \(f_{n}(x)=\frac{1}{n}\sin(n^{2}x)\). These functions are differentiable and converge uniformly to \(f(x)\equiv 0\), but their derivatives \(f_{n}^{{}^{\prime}}(x)=n\cos(n^{2}x)\) do not even stay bounded if we take their limit as \(n\to\infty\).
**Theorem 95** (Limit of a derivative is the derivative of the limit).: _Suppose \(\{f_{n}\}\) is a sequence of continuous functions on an interval \([a,b]\) which have continuous derivatives on \((a,b)\). Assume that \(f_{n}\to f\) for each \(x\in[a,b]\) and the sequence of derivatives \(\dot{f_{n}^{{}^{\prime}}}\rightrightarrows g\) on \((a,b)\). Then the limit function \(f\) is differentiable on \((a,b)\) and \(g=f^{\prime}\)._
Proof
: Since \(f_{n}^{{}^{\prime}}\rightrightarrows g\) and each \(f_{n}^{{}^{\prime}}(x)\) is continuous we know that \(g(x)\) is a continuous function on \((a,b)\). Furthermore, uniform continuity allows us to interchange the limit and integration, thus
\[\int_{a}^{x}g(t)dt=\lim_{n\to\infty}\int_{a}^{x}f_{n}^{{}^{\prime}}(t)dt= \lim_{n\to\infty}[f_{n}(x)-f_{n}(a)]=f(x)-f(a)\]
holds true for each \(x\in(a,b)\). Thus, by the Fundamental Theorem of Calculus, \(f\) is differentiable and \(f^{\prime}(x)=g(x)\) on \((a,b)\)
### 2.3 Modes of Convergence
**Corollary 16**.: _If the sequence of functions \(g_{1},g_{2},\dots\) are integrable over an interval \([a,b]\) and the infinite series \(\sum_{n=1}^{\infty}g_{n}\rightrightarrows g\) on \([a,b]\), then \(g\) is integrable and_
\[\int_{a}^{b}g(x)\,dx=\sum_{n=1}^{\infty}\int_{a}^{b}g_{n}(x)\,dx.\]
This result is expressed roughly by indicating that a uniformly convergent series of functions may be integrated term by term, or that summing a series and then integrating the sum is equal to integrating each term and then summing up the integrals.
### Weierstrass M-test
The following is a very useful test by which a series of functions may sometimes be shown to be uniformly convergent.
**Theorem 97** (Weierstrass M-test).: _Let \(\{f_{k}\}\) be a sequence of functions with domain \(A\) and let there exist constants \(M_{k}\) with_
\[|f_{k}(x)|\leq M_{k}\]
_for all \(x\in A\) and \(\sum_{k=1}^{\infty}M_{k}\) convergent. Then the series \(\sum_{k=1}^{\infty}f_{k}\) converges uniformly and absolutely._
Proof
: Since we have \(|f_{k}(x)|\leq M_{k}\) for all \(x\in A\), we know that
\[\sum_{k=1}^{\infty}|f_{k}(x)|\leq\sum_{k=1}^{\infty}M_{k}.\]
So by comparison test this series converges absolutely and therefore it converges for each \(x\in A\). Thus, the series \(\sum f_{k}(x)\) converges pointwise to a function which we will call \(s(x)\). Then we look at \(s_{n}=f_{1}+f_{2}+\dots+f_{n}\) and \(t_{n}=M_{1}+M_{2}+\dots+M_{n}\) where each \(s_{n}\) is a function and each \(t_{n}\) is a real number. Say that \(s_{n}\to s\) and \(t_{n}\to t\). Thus, for each \(x\), if \(n>m\) we have
\[|s_{n}(x)-s_{m}(x)|=\left|\sum_{k=m+1}^{n}f_{k}(x)\right|\leq\sum_{k=m+1}^{n} |f_{k}(x)|<\sum_{k=m+1}^{n}M_{k}=t_{n}-t_{m}.\]
Thus we have that the sequence
\[\{|s_{n}(x)-s_{m}(x)|\}_{n=1}^{\infty}\]
and \(\{t_{n}-t_{m}\}\) are both convergent. If we take the limit as \(n\to\infty\), then we know that \(|s(x)-s_{m}(x)|\leq t-t_{m}\) for all \(m\in\mathbb{N}\) and \(x\in A\). When \(m\to\infty\), \(t-t_{m}\to 0\), and this gives the uniform convergence.
**Example 84**.:
1. The series \(\sum_{n=1}^{\infty}\frac{(\sin(nx))^{2}}{n^{2}}\) converges uniformly. This can be deduced by considering the inequality: \[\left|\frac{(\sin(nx))^{2}}{n^{2}}\right|\leq\frac{1}{n^{2}}.\] Let \(M_{n}=\frac{1}{n^{2}}\); since we know that \(\sum\frac{1}{n^{2}}\) is a convergent \(p\)-series (with \(p=2\)), by the Weierstrass M-test, the given series converges uniformly.
2. The series \(\sum_{n=1}^{\infty}e^{-nx}\) converges uniformly on any closed subinterval of \((0,\infty)\). Let \(I=[a,\infty)\). If \(x\in I\), then \(e^{-nx}\leq e^{-na}\). We know that \(\sum_{n=0}^{\infty}e^{-na}\) is a convergent geometric series. Therefore, by the Weierstrass M-test, the series converges uniformly.
**Remark 48**.: A simple and useful test for uniform convergence called _Dirichlet’s test_ is as follows: Suppose that the \(N\)-th partial sum of the series \(\sum f_{n}(x)\) is uniformly bounded with respect to both \(x\) and \(N\) on the interval \(I\) and that \(g_{n}(x)\) is a monotone decreasing sequence converging uniformly to zero. Then the series \(\sum_{n=1}^{\infty}f_{n}(x)g_{n}(x)\) converges uniformly on \(I\). (see, e.g., [26], p. 287, also Exercise 11 below).
The Cauchy criterion for convergence of a numerical sequence can be adapted to a criterion for uniform convergence.
**Definition 77** (Uniform Cauchy Criterion).: Suppose \(\{f_{n}\}\) is a sequence of functions defined on \(D\subset\mathbb{R}\). Then we say the sequence \(f_{n}\) is a _uniform Cauchy sequence_ on \(D\) if for every \(\epsilon>0\), there exists \(N\in\mathbb{N}\) such that \(n,m>N\) implies that
\[|f_{n}(x)-f_{m}(x)|<\epsilon\]
for all \(x\in D\).
It is not hard to observe that every uniform Cauchy sequence is uniformly convergent. To see this first apply the uniform Cauchy property to state
\[|f_{n}(x)-f_{m}(x)|<\epsilon\ \ \text{for all}\ \ x\in D\]
when \(n,m\geq N.\) Now, fix \(n\) and take limit as \(m\to\infty\). Since \(f_{m}\to f\) as \(m\to\infty\), we obtain that
\[|f_{n}(x)-f(x)|<\epsilon\]
for all \(x\in D\) which implies that \(f_{n}\rightrightarrows f\) uniformly. Note that the Weierstrass M-test can be obtained as an application of the uniform Cauchy criterion.
**Remark 49**.: In some of the above proofs that involve integrals we use the following inequalities many times. Perhaps it is useful to list them:
1. If \(f\) and \(g\) are integrable functions on \([a,b]\), and \(f(x)\leq g(x)\) for all \(x\in[a,b]\), then \[\int_{a}^{b}f(x)dx\leq\int_{a}^{b}g(x)dx.\]
2. If \(f\) is integrable on \([a,b]\), so is the function \(|f|\), and \[\left|\int_{a}^{b}f(x)dx\right|\leq\int_{a}^{b}|f(x)|dx.\] The above inequality (b) is the integral version of the well known triangle inequality \(|a+b|\leq|a|+|b|\) for any numbers \(a\) and \(b\). If one assumes the integrability of \(|f|\), then (b) follows from (a) and from the fact that \[-|f(x)|\leq f(x)\leq|f(x)|.\]
**Remark 50**.: There are many excellent texts in analysis and many contain the topics of uniform and pointwise convergence and their consequences. For example [1, 18, 26, 42, 52] are such texts.
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