The limit of a uniformly convergent sequence of functions inherits some useful properties of the functions of the sequence. We will show this for boundedness and continuity.
Let #f_n# #(n=1,2,\ldots)# be a sequence of real-valued functions on #E# that converges uniformly to #f:E \to \mathbb{R}#. Then the following holds.
- If each #f_n# is bounded, then so is #f#.
- If each #f_n# is continuous, then so is #f#.
1. By the definition of uniform convergence, there exists a natural number #n# such that, for all #x\in E#,
\[ |f_n(x)-f(x)|\lt 1 \]
Since #f_n# is bounded, there also exists a natural number #M# such that #|f_n(x)| \leq M# for all #x \in E#. The triangle inequality implies
\[ |f(x)| = |f(x)-f_n(x)+f_n(x)| \lt 1+|f_n(x)| \leq 1+M \]
This proves that #f# is bounded.
2. Assume that each #f_n# is continuous on #E#. Let #a# be an element of #E#. In order to show that #f# is continuous at #a#, it suffices to show \(\lim_{x\to a} \abs{f(x) - f(a)} = 0\). Let #\varepsilon# be an arbitrary positive number. By uniform convergence, there is a natural number #N# such that #\abs{f(x)-f_n(x)}\le \frac{\varepsilon}{3}# for all #x\in E# and #n\ge N#. Since #f_N# is continuous, there is a positive number #\delta# such that #\abs{f_N(x)-f_N(a)}\lt \frac{\varepsilon}{3}# for all #x\in\ivoo{a-\delta}{a+\delta}\cap E#.
Therefore, for each #x# in #\ivoo{a-\delta}{a+\delta}\cap E#, we have
\[\begin{array}{rcl}\abs{f(x)-f(a)} & \le&\displaystyle \abs{f(x)-f_N(x)}+\abs{f_N(x)-f_N(a)}+\abs{f_N(a)-f(a)}\\ &&\phantom{xxx}\color{blue}{\text{triangle inequality applied twice}}\\ & \le&\displaystyle\frac{\varepsilon}{3}+\abs{f_N(x)-f_N(a)}+\frac{\varepsilon}{3}\\ &&\phantom{xxx}\color{blue}{\text{uniform convergence}}\\ & \le&\displaystyle\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &&\phantom{xxx}\color{blue}{\text{continuity }f_N\text{ at } a}\\ & =&{\varepsilon}\end{array}\]
This proves that #f# is continuous at every element #a# of #E#. We conclude that #f# is continuous on #E#.
It is not true that a series #f_n# #(n=1,2,\ldots)# of bounded continuous functions on an interval that converges pointwise to a bounded continuous function #f# also converges uniformly to #f#. To see this, let #a# be a real number and consider the sequence of functions #f_n# #(n=1,2,\ldots)# on the closed interval #\ivcc{0}{1}# given by
\[ f_n(x) = \begin{cases} n^a\cdot x&\text{if } \in \ivco{0}{\frac{1}{2n}}\\
n^a\cdot \left(\frac{1}{n}-x\right)&\text{if } \in \ivco{\frac{1}{2n}}{\frac{1}{n}}\\
0&\text{if } \in \ivcc{\frac{1}{n}}{0}\end{cases}\]
The sequence converges pointwise to the function #f(x) = 0#. The members of the sequence are all continuous and bounded. The absolute value of the difference between #f_n# and the limit satisfies
\[\sup_{x\in\ivcc{0}{1}} \left|f_n(x)-f(x)\right|=\sup_{x\in\ivcc{0}{1}} f_n(x)= n^a\cdot \frac{1}{2n} = \frac12 n^{a-1}\]
This tends to #0# for #n\to\infty# if and only if #a\lt 1#. In other words, the sequence converges to #f# uniformly, if and only if #a\lt 1#. In particular, for #a=1#, we have an example of a sequence of bounded and continuous functions whose limit is also bounded and continuous, although the convergence is not uniform.
We now show that integral and limit may be interchanged when applied to a sequence of functions. We will us the notion of integrability in terms of the upper and lower sums that converge to the same value as we refine the partitions.
Let #a# and #b# be real numbers with #a\lt b#. Suppose that #f_n# #(n=1,2,\ldots)# is a sequence of piecewise continuous functions on #\ivcc{a}{b}# that converges uniformly to #f:\ivcc{a}{b} \rightarrow \mathbb{R}#. Then #f# is integrable and
\[\begin{array}{c}\displaystyle \int_a^b f(x) \, \dd x = \lim_{n\rightarrow \infty} \int_a^b f_n(x) \, \dd x \end{array}\]
The #f_n# are all piecewise continuous on the closed interval #\ivcc{a}{b}#, and so they are bounded. Since the convergence to #f# is uniform, we know by the Uniform limit theorem that #f# will be bounded as well.
We will compute the difference between the upper and lower Riemann sums for #f# and show that it tends to zero as the mesh of the partitions for the Riemann sums tends to zero. This will allow us to conclude that #f# is Riemann integrable. In the following computation, we adopt the notation #M_i(f)# to indicate the supremum of the function #f# on the #i#-th subinterval. Analogously, #m_i(f)# will indicate the infimum of the function #f_n# on the #i#-th subinterval.
Suppose #\varepsilon\gt0#. Since #f_n# converges to #f# uniformly, we can find a natural number #N # such that for all #n \geq N#
\[|f_n(x) - f(x)| \lt \frac{\varepsilon}{4(b-a)} \quad \text{ for all }\quad x \in \ivcc{a}{b} \]
On the other hand, by the Riemann integrability of #f_N#, there is a number #\delta\gt0# such that for any partition #P=x_0,\ldots,x_r# with mesh at most #\delta#,
\[ \sum_{i=1}^{r} \left(M_i(f_{N})- m_i(f_{N})\right)(x_i-x_{i-1})\lt \frac{\varepsilon}{2}\]
taking these bounds into account, for a partition #P=x_0,\ldots,x_{r}# with mesh at most #\delta#, the following estimate holds.
\[\begin{array}{rcl}\displaystyle U_{f,P}-L_{f,P} &=& \displaystyle \sum_{i=1}^{r} M_i(f)(x_i-x_{i-1}) - \sum_{i=1}^{r} m_i(f)(x_i-x_{i-1}) \\ &=& \displaystyle\sum_{i=1}^{r} (M_i(f) -M_i(f_{N}))(x_i-x_{i-1}) -\sum_{i=1}^{r} (m_i(f) -m_i(f_{N}))(x_i-x_{i-1})\\ &&\displaystyle+ \sum_{i=1}^{r} \left(M_i(f_{N})- m_i(f_{N})\right)(x_i-x_{i-1})\\&\le& \displaystyle\sum_{i=1}^{r}\frac{\varepsilon}{4(b-a)} (x_i-x_{i-1}) +\sum_{i=1}^{r}\frac{\varepsilon}{4(b-a)}(x_i-x_{i-1}) +\frac{\varepsilon}{2}\\&\le& \displaystyle\frac{\varepsilon}{4} +\frac{\varepsilon}{4}+\frac{\varepsilon}{2}\\ &=&\varepsilon\end{array}\]
This shows that #f# is integrable. For the second part of the theorem, we compute, for #n\geq N#,
\[\begin{array}{rcl}\displaystyle \left | \int_a^b f(x) \dd x - \int_a^b f_n(x) \dd x \right|& = &\displaystyle \left| \int_a^b f(x)-f_n(x) \right |\\& \le &\displaystyle\int_a^b\left|f(x)-f_n(x) \right |\dd x\\ & \le &\displaystyle \int_a^b\frac{\varepsilon}{4(b-a)}\dd x\\&\le&\displaystyle(b-a)\frac{\varepsilon}{4(b-a)} \\&=& \displaystyle \frac{\varepsilon}{4} \\&\lt& \varepsilon \end{array}\]
This establishes the theorem.
We provide a counterexample to the theorem in case the convergence is pointwise, thus showing the importance of the uniform convergence assumption.
Let #\nu: \mathbb{N} \rightarrow \mathbb{Q} \cap \ivcc{0}{1}# be a bijective function. Such a function can be obtained by enumerating the rational numbers, for example as follows
\[0,1,-1,2,-2,\frac{1}{2},-\frac{1}{2},3,-3,\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3},\frac{3}{2},-\frac{3}{2},4,-4,\frac{1}{4},\ldots\]
in which case #\nu(10)= \frac{1}{3}#. Define
\[f_n(x)= \begin{cases} 1 & \text{ if } x \in \{\nu(1), \ldots, \nu(n)\} \\ 0 & \text{ otherwise} \end{cases} \]
Each #f_n# is piecewise continuous, as the points #\nu(i)# #(i=1,\ldots,\nu(n))# are its only discontinuities and it is constant on each open interval between two subsequent members of #\{\nu(1), \ldots, \nu(n)\}#. As a consequence #f_n# is integrable, and
\[\int_0^1 f_n(x)\,\dd x = 0\]
However, the pointwise limit is
\[f(x)= \begin{cases} 1 & \text{ if } x \in \mathbb{Q} \cap \ivcc{0}{1} \\ 0 & \text{ otherwise}\end{cases}\]
is not integrable. Recall that if a function #f# (not necessarily continuous) defined on #\ivcc{0}{1}# is integrable if for every #\varepsilon \gt 0# there exists a positive number #\delta# such that for every partition #P=\{x_0,x_1,x_2,\ldots,x_n\}# with #\text{mesh}(P) \lt \delta# we have
\[|U_{f,P} - L_{f,P}| \lt \varepsilon\]
where
\[ \begin{array}{rcl}\displaystyle U_{f,P} &=& \displaystyle\sum_{i=1}^n\sup\{f(x)\mid x \in \ivcc{x_{i-1}}{x_i}\}\cdot (x_i - x_{i-1}) \\ &&\text{ and } \\L_{f,P} &=&\displaystyle \sum_{i=1}^n\inf \{ f(x)\mid x \in \ivcc{x_{i-1}}{x_i}\}\cdot (x_i - x_{i-1}) \end{array} \]
Now, for any partition #P# of #\ivcc{0}{1}# we have #U_{f,P}=1# because every subinterval of the partition contains a rational number. Analogously #L_{f,P}=0# because every subinterval of the partition contains an irrational number. This still holds if we take the supremum of #U_{f,P}# and the infimum of #L_{f,P}# over all partitions. Hence
\[U_{f} - L_{f} =1\]
and the integrability criterium fails.
Find an expression of
\[\int_0^\infty{\frac{x^{3}}{\ee^x-1}} \,\dd x\]
that is the product of a rational number and a power of #\pi#.
\(\int_0^\infty{\frac{x^{3}}{\ee^x-1}} \,\dd x =\) \({{\pi^4}\over{15}}\)
\[\begin{array}{rcl}\displaystyle \int_0^\infty{\frac{x^{3}}{\ee^x-1}} \,\dd x &=&\displaystyle\lim_{t\to\infty}\int_0^t{\frac{x^{3}}{\ee^x-1}} \,\dd x\\
&&\phantom{xxx}\color{blue}{\text{definition of definite integral with boundary value }\infty}\\
&=&\displaystyle\lim_{t\to\infty}\int_0^t{\sum_{k=1}^\infty x^{3}\ee^{-k\cdot x}} \,\dd x\\
&&\phantom{xxx}\color{blue}{\text{geometric series: }\sum_{k=1}^{\infty} \ee^{-k\cdot x} =\frac{\ee^{-x}}{1-\ee^{-x}}}\\
&=&\displaystyle\lim_{t\to\infty}\sum_{k=1}^\infty \int_0^t {x^{3}\ee^{-k\cdot x} }\,\dd x\\
&&\phantom{xxx}\color{blue}{\text{Interchange of limit and integral}}\\
&=&\displaystyle\lim_{t\to\infty}\sum_{k=1}^\infty \left(-{{t^3\cdot \euler^ {- k\cdot t }}\over{k}}-{{3\cdot t^2\cdot \euler^ {- k\cdot t }}\over{k^2}}-{{6\cdot t\cdot \euler^ {- k\cdot t }}\over{k^3}}-{{6\cdot \euler^ {- k\cdot t }}\over{k^4}}+{{6}\over{k^4}} \right)\\
&&\phantom{xxx}\color{blue}{\text{definite integral determined}}\\
&=&\displaystyle\lim_{t\to\infty}\left(\sum_{k=1}^\infty {{6}\over{k^4}} -\sum_{k=1}^\infty {{\left(k^3\cdot t^3+3\cdot k^2\cdot t^2+6\cdot k\cdot t+6\right)\cdot \euler^ {- k\cdot t }}\over{k^4}}\right) \\
&&\phantom{xxx}\color{blue}{\text{term without }\e^{-kt}\text{ set aside}}\\
\end{array}\]
For #t\gt 1#, the second term can be bounded as follows.
\[\begin{array}{rcl}\displaystyle 0&\le &\displaystyle\sum_{k=1}^\infty {{\left(k^3\cdot t^3+3\cdot k^2\cdot t^2+6\cdot k\cdot t+6\right)\cdot \euler^ {- k\cdot t }}\over{k^4}}\\
&\le &\displaystyle t^3\cdot\e^{-t} \sum_{k=1}^\infty \frac{k^3+3\cdot k^2+6\cdot k+6}{k^4}\cdot\e^{(1-k)\cdot t}\\
&& \phantom{xxx}\color{blue}{t\ge 1}\\
&\le &\displaystyle t^3\cdot\e^{-t} \sum_{k=1}^\infty \frac{16}{k\cdot\e^{(k-1)\cdot t}}\\
&& \phantom{xxx}\color{blue}{k\ge 1}\\
&\le &\displaystyle t^3\cdot\e^{-t} \sum_{k=1}^\infty \frac{16}{k^2}\\
&& \phantom{xxx}\color{blue}{k\le \e^{(k-1)\cdot t}}\\
&\le &\displaystyle t^3\cdot\e^{-t}\cdot\frac{16\pi^2}{6}\\
&& \phantom{xxx}\color{blue}{\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} }\\
\end{array}\]
This implies that
\[ \displaystyle\lim_{t\to\infty}\sum_{k=1}^\infty {{\left(k^3\cdot t^3+3\cdot k^2\cdot t^2+6\cdot k\cdot t+6\right)\cdot \euler^ {- k\cdot t }}\over{k^4}} = 0\]
We conclude that
\[\begin{array}{rcl}\displaystyle \int_0^{\infty} {\frac{x^{3}}{\ee^x-1}} \,\dd x
&=&\displaystyle\lim_{t\to\infty}\sum_{k=1}^t{{6}\over{k^4}}\\
&& \phantom{xxx}\color{blue}{\text{ second term of expression found above proven to be zero}}\\
&=&\displaystyle {{\pi^4}\over{15}} \\
&& \phantom{xxx}\color{blue}{\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90} }\\
\end{array}\]
Applications of uniform convergence
