2.8 Sequence of functions
Let \(A\) and \(B\) denote subsets of \(\mathbb {R}\) and for each natural number \(n\) let \(f_n: A\longrightarrow B\) denote a function with domain \(A\) and Co
domain \(B\). If \(f\) is a function from \(A_0\subseteq A\) into \(B\), the sequence \(\big \{f_n\big \}\) is said to converge in \(A_0\) to \(f\) if the sequence of values \(\big \{f_n(x)\big \}\)
converges to the number \(f(x)\) for every \(x\in A_0\).
More precisely, we say that \(f_n\) converges to \(f\) point wise on \(A_0\) if for each \(x\in A_0\hspace {0.2cm} f_n(x) \longrightarrow f(x)\,\) as \(n\longrightarrow \infty \).
Definition 2.8.1. A sequence of function \(\big \{f_n\big \}\) on a set \(A\) converges to a function \(f\) on \(A_0\subseteq A\) if and only if
for every \(\varepsilon > 0\), and each \(x\in A_0\), there is a number \(N = N(x,\varepsilon )\) such that \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \varepsilon \) whenever \(n>N\).
We write \(\displaystyle {\lim _{n\rightarrow \infty } f_n(x) = f(x)}\, , \hspace {0.2cm} x\in A_0\,\) and say that \(\big \{f_n\big \}\) is convergent on \(A_0\).
Example 2.8.2. Let \(\, f_n : \mathbb {R}\longrightarrow \mathbb {R}\,\) be defined by \(f_n(x) = \dfrac {x}{n}\). Then the limit function \(f\) is such that \(f(x) = 0\) for all \(x\in \mathbb {R}\). Thus this
sequence converges on \(\mathbb {R}\) to \(f\).
Uniform Convergence
In the preceding definition of convergence, the value of \(N\) usually depends on both the choice of \(\varepsilon \) and \(x\).
When \(N\) can be chosen independently of \(x\) then the convergence is said to be uniform.
Definition 2.8.3. A sequence \(f_n: A\longrightarrow B\) of functions converges uniformly to a function \(f\) on \(A_0\subseteq A\) if , for every
\(\varepsilon > 0\) there is an \(N>0\) depending only on \(\varepsilon \) but not on \(x\), such that \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \varepsilon \,\) for all \(n> N\).
Clearly any sequence that is uniformly convergent is also convergent. However, the converse is
not true.
Example 2.8.4. Consider again the sequence of functions \(f_n: \, \mathbb {R} \longrightarrow \mathbb {R}\), defined by \(f_n(x) = \dfrac {x}{n}\). This sequence is
convergent for every \(x\in \mathbb {R}\), but it is not uniformly convergent. To see this pick \(\varepsilon = \dfrac {1}{10}\), then \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \dfrac {1}{10},\hspace {0.2cm}n>10\) at \(x =1\) and \(n>20\) at \(x = 2\)
e.t.c. In short the integer \(N\) depends on the choice of \(x\) and the convergence is not uniform.
Definition 2.8.5. A function \(f: \, A\subseteq \mathbb {R}\longrightarrow \mathbb {R}\) is said to be bounded if and only if there exist a positive number
\(M\), \(\hspace {0.2cm} 0 < M < \infty \,\) such that \(\displaystyle {\sup _{x\in A}\begin {vmatrix} f(x)\\ \end {vmatrix}< M}\).
Theorem 2.8.6. A sequence of bounded function \(\big \{f_n\big \}\) \(\hspace {0.2cm} f:\, A\longrightarrow \mathbb {R},\) converges uniformly on \(A\) to a function \(f\) if
and only if \(\hspace {0.3cm}\displaystyle {\lim _{n\rightarrow \infty } \, \sup _{x\in A}\Big [\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix}\Big ] = 0}\).
Proof. Let \(\big \{f_n\big \}\) converge uniformly to \(f\) on \(A\). Then for every \(\varepsilon > 0\) there is an \(N = N(\varepsilon )\) such that \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \varepsilon \) for \(n> N\), for every \(x\in A\).
Thus \(\big \{\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix},\, x\in A\big \} < \varepsilon \) for \(n> N\) which means that \(\,\displaystyle {\sup _{x\in A}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} }\) converges to zero.
Conversely, suppose that \(\displaystyle {\sup _{x\in A}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} }\,\) converges to zero. Then for any \(\varepsilon > 0\) and \(x\in A\), we have \(\displaystyle {\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} \leq \sup _{x\in A}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \varepsilon }\,\) for \(n> N\) where \(N\) does not
depend on \(x\). Hence \(\big \{f_n\big \}\) converges uniformly on \(A\) to \(f\).
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Example 2.8.7. Let \(\hspace {0.2cm} f_n(x) = n^2\, x\, \big (1 - x\big )^n\hspace {0.2cm}\, \hspace {0.3cm} 0\leq x\leq 1\). Determine whether the sequence of function is converges uniformly on
\(0\leq x\leq 1\).
We first note that \(f_n\) converges point wise to the zero function \(f\). To check for uniform convergence,
we will compute \(\, \displaystyle {\sup _{0\leq x\leq 1}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix}}\,\) and note that the supremum is actually the maximum by continuity of
\(\begin {vmatrix} f_n - f\\ \end {vmatrix}\).
Taking the derivative of \(\, n^2\, x \, \big (1 - x\big )^n\,\) at maximum, we get, \(x = \dfrac {1}{n + 1}\). Hence \begin {align*} \sup _{0\leq x\leq 1}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} & = \dfrac {n^2}{n + 1}\,\Bigg (1 - \dfrac {1}{n + 1}\Bigg )^n\\\\ & = \dfrac {n^2}{n + 1}\Bigg [\Bigg (1 - \dfrac {1}{n + 1}\Bigg )^{n + 1}\Bigg ]^{\dfrac {n}{n + 1}} \end {align*}
Since \(\hspace {0.2cm} \dfrac {n^2}{n + 1}\longrightarrow \infty \, \) and \(\, \Bigg (1 + \dfrac {1}{m}\Bigg )^m\longrightarrow e^r\hspace {0.2cm}\) as \(\, m \longrightarrow \infty \,\) for any \(r\) then
\[ \sup _{0\leq x\leq 1}\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} = \dfrac {n^2}{n + 1}\, \Bigg ( 1 - \dfrac {1}{n}\Bigg )^n\longrightarrow \infty \]
as \(n\longrightarrow \infty \). Thus \(f_n\) does not converge uniformly.
Example 2.8.8. Let \(\, f_n(x) = \dfrac {x^n}{2 + 3\, x^n}\hspace {0.2cm},\hspace {0.2cm} 0\leq x\leq 1\). Show that \(f_n\) converges point wise and determine whether the sequence
converges uniformly on \([0,1]\).
Solution. \(f_n \longrightarrow f\), point wise, where
\(f(x) = 0 \hspace {0.5cm} 0\leq x < 1\)
\(f(1) = \dfrac {1}{5}\)
To check for uniform convergence
\[ \begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} = \begin {cases} \dfrac {x^n}{2 + 3\,x^n} & 0\leq x < 1\\\\ 0 & x = 1\\ \end {cases}\]
Since \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix}\longrightarrow \dfrac {1}{5}\) as \(x\longrightarrow 1\) and \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix}\) increases with \(x< 1\) then
\(\displaystyle {\sup _{x\in [0,1)}\begin {vmatrix} f_(x) - f(x)\\ \end {vmatrix} = \frac {1}{5}}\). Thus \(f_n \longrightarrow f\) uniformly.
Theorem 2.8.9. Let \(E\subset \mathbb {R}\) and let \(f_1, \, f_2, \cdots \cdots \) be a sequence of continuous functions on \(E\). If \(f_n \longrightarrow f\) uniformly on \(E\), then \(f\) is continuous on \(E\). In particular, if \(x_0\in E\), then \[\lim _{n \rightarrow \infty }\, \lim _{x\rightarrow x_0} \, f_n(x) = \lim _{x\rightarrow x_0} \,\lim _{n\rightarrow \infty } \, f_n(x) = f(x_0)\]
Proof. If \(x_0\in E\), then \(\begin {vmatrix} f(x) - f(x_0)\\ \end {vmatrix}\leq \begin {vmatrix} f(x) - f_n(x)\\ \end {vmatrix} + \begin {vmatrix} f_n(x) - f(x_0)\\ \end {vmatrix} + \begin {vmatrix} f_n(x_0) - f(x_0)\\ \end {vmatrix}\). Given \(\varepsilon > 0\), the uniform convergence of \(f_n\) to \(f\) allows us to find a positive integer \(N\)
such that \(\,\begin {vmatrix} f(x) - f_n(x) \\ \end {vmatrix} < \dfrac {\varepsilon }{3}\) for all \(n> N\) and all \(x\in E\). Set \(x = x_0\) to obtain \(\begin {vmatrix} f_n(x_0) - f(x_0)\\ \end {vmatrix} < \dfrac {\varepsilon }{3}\).
Fix \(n>N\), since \(f_n\) is continuous at \(x_0\) there is a \(\delta > 0\) such that \(\begin {vmatrix} f_n(x) - f_n(x_0)\\ \end {vmatrix}< \dfrac {\varepsilon }{3}\) whenever \(\begin {vmatrix} x - x_0\\ \end {vmatrix} < \delta \). Thus \(\begin {vmatrix} x - x_0\\ \end {vmatrix} < \delta \) implies that \(\begin {vmatrix} f(x) - f(x_0)\\ \end {vmatrix} < \varepsilon \).
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Definition 2.8.10. A sequence of functions \(f_1, \, f_2, \cdots \cdots \cdots \) is said to have a uniform Cauchy property on \(E\) if
given \(\varepsilon > 0\) there is a positive integer \(N\) (depending only on \(\varepsilon \)) such that for \(n,\, m > N\), we have \(\, \begin {vmatrix} f_n (x) - f_m(x) \\ \end {vmatrix} < \varepsilon \,\) for all \(x\in E\).
Theorem 2.8.11. Suppose that the real valued functions \(\, f_1, \, f_2, \, \cdots \cdots , \, f_n ,\cdots \cdots \) have a uniform Cauchy property on
\(E\). Then the sequence \(\big \{f_n\big \}\) converges uniformly on \(E\).
Proof. For each \(x\in E,\, \big \{f_n(x)\big \}\) is Cauchy sequence of real numbers hence \(f_n(x)\) converges to a limit \(f(x)\) (since a Cauchy
sequence of real numbers converges).
If \(\begin {vmatrix} f_n(x) - f_m(x)\\ \end {vmatrix} < \varepsilon \) for all \(n,\, m> N\) and all \(x\in E,\,\) fix \(n\) and let \(m\longrightarrow \infty \) to conclude that \(\begin {vmatrix} f_n(x) - f(x)\\ \end {vmatrix} < \varepsilon \) for all \(n>N\) and \(x\in E\). Thus \(f_n \longrightarrow f\) uniformly on \(E\).
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Theorem 2.8.12 (Dini’s Theorem). Let \(f_1,\, f_2,\, \cdots \cdots \cdots \) be continuous functions of real numbers on the
compact set \(E\), and assume that the \(f_n\) form a monotone sequence either \(f_{n+ 1}(x) \leq f_n(x)\) for all \(x\in E\) and all \(n = 1, \, 2,\, \cdots \cdots \) or \(f_{n+1}(x) \geq f_n(x)\) for all
\(x\in E\) and all \(n = 1, \, 2\, \cdots \cdots \). If \(f_n \longrightarrow f\) point wise on \(E\) and \(f\) is continuous on \(E\), then \(f_n\longrightarrow f\) uniformly on \(E\).
Proof. We may assume that the \(f_n\) form a decreasing sequence (in the case of increasing case
consider the functions \(- f_n\)). If \(g_n = f_n - f\), then \(g_n\) form a decreasing sequence of non-negative continuous
functions converging point wise to zero. If \(\varepsilon > 0\), let \(V_n = \big \{x\in E: \, g_n(x) < \varepsilon \big \}\) which is an open set by continuity of \(g_n\). If \(x\in E\) then \(g_n(x) < \varepsilon \)
eventually, hence \(\, \bigcup ^{\infty }_{n=1}\, V_n = E\). Since \(E\) is compact, \(\, \bigcup ^N_{n=1}\, V_n = E\,\) for some \(N\). But since \(g_n\) decreases we have \(V_n \subseteq V_{n + 1}\) for all \(n\). Therefore
\(\, \bigcup ^N_{n=1}\, V_n = V_N\). Thus if \(x\in E\), then \(x\in V_N\); that is \(g_N(x) < \varepsilon \). If \(n>N\), we have \(0\leq g_n(x) \leq g_N(x) <\varepsilon \). Thus \(g_n \longrightarrow 0\) uniformly on \(E\). i.e \(g_n = f_n - f\longrightarrow \,\) uniformly on \(E\) or \(f_n\longrightarrow f\,\) uniformly on
\(E\).
The uniform convergence of a series of functions \(\displaystyle {\sum ^{\infty }_{n = 1} f_n}\,\) means, by definition uniform convergence of
the sequence of \(n^{\text {th}}\) partial sums \(\, \displaystyle {S_n = \sum ^{n}_{k = 1} \, f_k}\)
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Theorem 2.8.13 (Weiestrass M-Test). Let \(\, f_1,\, f_2\, \cdots \cdots \cdots \,\) be a sequence of real valued functions on a set \(E\).
If \(\begin {vmatrix} f_n (x)\\ \end {vmatrix} \leq M_n\,\) for all \(x\in E\) and all \(n = 1, \, 2,\, \cdots \cdots \cdots \) where \(\displaystyle {\sum ^{\infty }_{n = 1} M_n < \infty }\), then the series \(\, \displaystyle {\sum ^{\infty }_{n=1} f_n}\,\) converges uniformly on \(E\). Thus, if each \(f_n\) is continuous
on \(E\), then \(\displaystyle {\sum ^{\infty }_{n = 1} f_n}\,\) is continuous on \(E\).
Proof. If \(S_n\) is the \(n^{\text {th}}\) partial sum of the series, then for \(m< n\) \begin {align*} \begin {vmatrix} S_n(x) - S_m(x)\\ \end {vmatrix} & = \begin {vmatrix} \displaystyle {\sum ^n_{k = m + 1} f_k(x)}\\ \end {vmatrix}\\ & \leq \sum ^n_{k = m + 1} \begin {vmatrix} f_k(x)\\ \end {vmatrix}\\ & \leq \sum ^n_{k = m + 1} M_k\\ & \leq \sum ^{\infty }_{k=m + 1} M_k \end {align*}
which goes to zero as \(m\longrightarrow \infty \) since \(\displaystyle {\sum ^{\infty }_{n = 1}M_n < \infty }\). Thus, the sequence \(\big \{S_n\big \}\) has uniform Cauchy property an therefore converges
uniformly.
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Example 2.8.14. Let \(\, f_n(x) = \dfrac {\sin nx}{x^2}\hspace {0.2cm}, \hspace {0.2cm} x\in \mathbb {R}\).
Then \(\begin {vmatrix} f_n(x)\\ \end {vmatrix}\leq M_n = \dfrac {1}{n^2}\hspace {0.2cm}\) for all \(x\). Then
\[ \sum \begin {vmatrix} f_n(x)\\ \end {vmatrix} = \sum \begin {vmatrix} \dfrac {\sin nx }{n^2}\\ \end {vmatrix} \leq \sum \dfrac {1}{n^2}\]
But we know that the series \(\, \displaystyle {\sum ^{\infty }_{n = 1}\, \dfrac {1}{n^2}}\,\) is convergent. Therefore, \(\, \displaystyle {\sum ^{\infty }_{n=1}\, \dfrac {\sin nx}{n^2}}\,\) converges on \(\mathbb {R}\).
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