In Example 8.1 i we have seen that the pointwise limit of a sequence of continuous functions is not necessarily continuous. This is different, if the sequence converges uniformly.
Theorem 8.1 • Uniform limit theorem
Let \(K \subseteq {\mathbb{C}}\) and let \(\left( f_n:K \to {\mathbb{C}}\right)_{n \in {\mathbb{N}}}\) be a function sequence that is uniformly convergent towards \(f:K \to {\mathbb{C}}.\) If \(f_n\) is continuous for all \(n \in {\mathbb{N}},\) then also \(f\) is continuous.
Proof. Let \(x \in K\) be arbitrary. We show with the help of the \(\varepsilon\)-\(\delta\) criterion for continuity (cf. Theorem 4.11) that \(f\) is continuous in \(x.\) Let \(\varepsilon > 0\) be given. Then we aim to find a \(\delta > 0\) such that for all \(\widetilde{x} \in K\) it holds that \[\left| x -\widetilde{x} \right| < \delta \quad \Longrightarrow \quad \left| f(x) - f(\widetilde{x})\right| < \varepsilon.\] For each \(\widetilde{x} \in K\) and \(N \in {\mathbb{N}},\) be adding productive zeros and using the triangle inequality we have \[\tag{8.3} \left| f(x) - f(\widetilde{x})\right| \le |f(x) - f_N(x)| + |f_N(x) - f_N(\widetilde{x})| + |f_N(\widetilde{x}) - f(\widetilde{x})|.\] Next we determine \(\delta > 0\) and \(N \in {\mathbb{N}}\) such that each of the summands on the right-hand side of (8.3) becomes smaller than \(\frac{\varepsilon}{3}.\) Since \({(f_n)}_{n \in {\mathbb{N}}}\) converges uniformly towards \(f,\) there exists an \(N \in {\mathbb{N}}\) such that \[|f_N(y) - f(y)| < \frac{\varepsilon}{3} \quad \text{for all } y \in K.\] Moreover, by assumption, \(f_N\) is continuous in \(x\) and thus, there exists a \(\delta > 0\) such that for all \(\widetilde{x} \in K\) it holds that \[\left| x-\widetilde{x} \right| < \delta \quad \Longrightarrow \quad |f_N(x) - f_N(\widetilde{x})| < \frac{\varepsilon}{3}.\] With this \(\delta\) and the previously chosen \(N,\) from (8.3) we obtain \[\left| x-\widetilde{x} \right| < \delta \quad \Longrightarrow \quad \left| f(x) - f(\widetilde{x})\right| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.\] Hence, \(f\) is continuous in \(x\) and since \(x \in K\) has been chosen arbitrarily, \(f\) is continuous.
Video 8.2. Uniform limit theorem.
In Theorem 8.1 it would have been sufficient to assume that there exists an \(N \in {\mathbb{N}}\) such that all \(f_n\) with \(n \ge N\) are continuous.
Another important property of uniformly convergent sequences is the desired interchangability of integration and limit calculation.
Theorem 8.2
Let \(\left( f_n:[a,b] \to {\mathbb{R}}\right)_{n \in {\mathbb{N}}}\) be a sequence of continuous functions that converges uniformly towards a function \(f:[a,b] \to {\mathbb{R}}.\) Then \[\lim_{n \to \infty} \int_a^b f_n(x)\, \mathrm{d}x = \int_a^b \lim_{n \to \infty} f_n(x)\,\mathrm{d}x = \int_a^b f(x)\,\mathrm{d}x.\]
Proof. By Theorem 8.1, \(f\) is continuous and in particular, integrable (cf. Theorem 7.5). Further, with the linearity and monotonicity of the integral (see Theorem 7.4) we obtain \[\left| \int_a^b f(x)\,\mathrm{d}x - \int_a^b f_n(x)\,\mathrm{d}x \right| \le \int_a^b |f(x)-f_n(x)|\,\mathrm{d}x \le \int_a^b \| f-f_n \| \, \mathrm{d}x,\] where is the last step we have made use of the fact that \(|f(x) - f_n(x)| \le \| f-f_n \|\) for all \(x \in [a,b].\) The integral on the right-hand side is also well-defined since with \(f\) and \(f_n,\) also \(f-f_n\) is continuous. Then with the extreme value theorem (cf.Theorem 4.9), \(f-f_n\) attains its maximum and minimum on \([a,b]\) and so \(\|f-f_n\| < \infty\) for all \(n \in {\mathbb{N}}.\) Moreover, \[\int_a^b \|f-f_n\| \,\mathrm{d}x = (b-a) \cdot \|f-f_n\|.\] Then from the uniform convergence of \({(f_n)}_{n \in {\mathbb{N}}}\) towards \(f\) (i. e., \(\|f-f_n\| \to 0\) for \(n \to \infty\) by Remark 8.2 ii) we obtain \[\lim_{n \to \infty} \left| \int_a^b f(x)\,\mathrm{d}x - \int_a^b f_n(x)\,\mathrm{d}x \right| = 0,\] hence \[\lim_{n \to \infty}\int_a^b f_n(x)\,\mathrm{d}x = \int_a^b f(x)\,\mathrm{d}x.\]
One may ask the question whether, as in the fashion of integration, uniformly convergent sequences of differentiable functions allow the interchangability of differentiation and limit calculation. This is not the case as the following example shows.
Example 8.2
We consider the function sequence \({(f_n)}_{n \ge 1}\) with \(f_n: {\mathbb{R}}\to {\mathbb{R}},\) \(x \mapsto \frac{1}{n}\sin(nx).\) Obviously, we have \[\|f_n\| = \sup\left\{ |f_n(x)| \; | \; x \in {\mathbb{R}}\right\} = \frac{1}{n} \quad \text{for all } n \in {\mathbb{N}}\setminus\{0\},\] i. e., \({(f_n)}_{n \ge 1}\) is uniformly convergent towards the zero function \(f:{\mathbb{R}}\to {\mathbb{R}},\) \(x \mapsto 0.\) As derivatives we obtain \[f_n':{\mathbb{R}}\to {\mathbb{R}}, \quad x \mapsto \cos(nx), \quad \text{for all } n \in {\mathbb{N}}\setminus \{0\}.\] However, the sequence \({(f_n')}_{n \ge 1}\) does not even converge pointwise towards \(f' = 0.\) For example, for the point \(x = \pi\) we obtain \[f_n(\pi) = \begin{cases} 1, & \text{for even $n$}, \\ -1, & \text{for odd $n$}. \end{cases}\]
Indeed, one cannot conclude anything about the convergence of derivatives of a function sequence. Instead one can show the converse under certain assumptions, i. e., that uniform convergence of \({(f_n')}_{n \in {\mathbb{N}}}\) implies uniform convergence of \({(f_n)}_{n \in {\mathbb{N}}}.\)
Theorem 8.3
Let \(f_n:[a,b] \to {\mathbb{R}}\) for \(n \in {\mathbb{N}}\) be continuously differentiable functions and \(x_0 \in [a,b]\) be such that the sequence \(\left(f_n(x_0)\right)_{n \in {\mathbb{N}}}\) is convergent. If the function sequence \({(f_n')}_{n \in {\mathbb{N}}}\) is uniformly convergent towards \(f^*:[a,b] \to {\mathbb{R}},\) then also the function sequence \({(f_n)}_{n \in {\mathbb{N}}}\) is uniformly convergent towards \(f:[a,b] \to {\mathbb{R}}.\) Moreover, \(f\) is continuously differentiable with \(f' = f^*.\) In particular, for all \(x \in [a,b]\) it holds that \[\lim_{n \to \infty} f_n'(x) = \left( \lim_{n \to \infty} f_n \right)'(x) = f'(x).\]
Proof. Since by assumption all \(f_n'\) are continuous, then by the uniform convergence of \({(f_n')}_{n \in {\mathbb{N}}}\) and Theorem 8.1, this is also the case for the uniform limit \(f^*\) and hence, \(f^*\) is integrable on \([a,b].\) We define the function \(f:[a,b] \to {\mathbb{R}}\) by \(f(x_0) := \lim_{n \to \infty} f_n(x_0)\) and \[f(x) := \int_{x_0}^x f^*(t)\,\mathrm{d}t + f(x_0) \quad \text{for } x\in [a,b]\setminus\{x_0\}.\] Then by the fundamental theorem of calculus (Theorem 7.10), \(f\) is differentiable with \(f' = f^*.\) It remains to show that \({(f_n)}_{n \in {\mathbb{N}}}\) is uniformly convergent towards \(f.\) First we observe that for all \(x \in [a,b]\) and \(n \in {\mathbb{N}}\) it holds that \[\tag{8.4} \begin{aligned} |f_n(x) - f(x)| &= |f_n(x) - f_n(x_0) + f_n(x_0) - f(x_0) + f(x_0) - f(x)| \\ &= \left| \int_{x_0}^x f_n'(t)\,\mathrm{d}t + f_n(x_0) - f(x_0) - \int_{x_0}^x f'(t)\,\mathrm{d}t \right| \\ &\le \int_{x_0}^x |f_n'(t) - f'(t)|\,\mathrm{d}t + |f_n(x_0) - f(x_0)|. \end{aligned}\] Let now \(\varepsilon > 0\) be arbitrary. Because of the convergence of \(\left(f_n(x_0)\right)_{n \in {\mathbb{N}}}\) and because of the uniform convergence of \({(f_n')}_{n \in {\mathbb{N}}},\) there exists an \(N_\varepsilon \in {\mathbb{N}}\) such that \[|f_n(x_0) - f(x_0)| < \frac{\varepsilon}{2} \quad \text{and} \quad \left\| f_n' - f' \right\| < \frac{\varepsilon}{2(b-a)} \quad \text{for all } n \ge N_\varepsilon.\] Then (8.4) implies \[|f_n(x) - f(x)| < \frac{\varepsilon}{2(b-a)}\cdot|x-x_0| + \frac{\varepsilon}{2} \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \quad \text{for all } n\ge N_\varepsilon \text{ and } x \in [a,b].\] Since \(\varepsilon > 0\) has been arbitrary, this implies uniform convergence of \({(f_n)}_{n \in {\mathbb{N}}}\) towards \(f.\)
Remark 8.3
The assumption that the function sequence \({(f_n')}_{n \in {\mathbb{N}}}\) is convergent in at least one point has the following background: By the fundamental theorem of calculus (Theorem 7.10), the antiderivative of a continuous function on an interval is only uniquely determined up to a constant. Hence, the function sequence \({(f_n)}_{n \in {\mathbb{N}}}\) could be replaced by a second sequence \({(f_n+c_n)}_{n \in {\mathbb{N}}}\) with arbitrary constants \(c_n \in {\mathbb{R}}\) for \(n \in {\mathbb{N}}\) without changing the sequence of derivatives \({(f_n')}_{n \in {\mathbb{N}}}.\) By our additional assumption of convergence in one point, we restrict the choice of such constants such that “no damage is done”.
Analogously to series of real numbers, one can introduce series of functions.
Definition 8.3 • Function series
Let \(K\) be a set and \(\left( f_n: K \to {\mathbb{C}}\right)_{n \in {\mathbb{N}}}\) be a function sequence. Then the sequence \[\sum_{k=0}^\infty f_k := \left( \sum_{k=0}^n f_k\right)_{n \in {\mathbb{N}}}\] is called a function series.
The terms of pointwise and uniform convergence immediately translate to the case of function series. If \(f\) is the limit function, then in both cases we write \[f = \sum_{k=0}^\infty f_k.\] Then one has to mention which kind of convergence is meant. A criterion for uniform convergence of a function sequence is discussed in the next theorem.
Theorem 8.4 • Weierstraß’ criterion for uniform convergence
Let \(K\) be a set and \(\left(f_n:K \to {\mathbb{C}}\right)_{n \in {\mathbb{N}}}\) be a function sequence with \(\|f_n\| < \infty\) for all \(n \in {\mathbb{N}}.\) If the series \[\sum_{k=0}^\infty \|f_k\|\] is convergent, then the function series \(\sum_{k=0}^\infty f_k\) is uniformly convergent.
Proof. Because \(|f_k(x)| \le \| f_k \|\) for all \(k \in {\mathbb{N}},\) the majorant criterion (Theorem 3.9) implies that the series \(\sum_{k=0}^\infty f_k(x)\) is absolutely convergent for each \(x \in K.\) We define \[f(x) := \sum_{k=0}^\infty f_k(x) \quad \text{for all } x \in K.\] We show that our function series is uniformly convergent towards \(f.\) Let \(\varepsilon > 0.\) Then by Cauchy’s convergence criterion (Theorem 3.3) there exists an \(N_\varepsilon \in {\mathbb{N}}\) such that \[\sum_{k=n+1}^m \|f_k\| < \frac{\varepsilon}{2} \quad \text{for all } m > n \ge N_\varepsilon.\] This implies \[\sum_{k=n+1}^\infty \|f_k\| \le \frac{\varepsilon}{2} < \varepsilon \quad \text{for all } n \ge N_\varepsilon.\] Consequently, \[\left| \sum_{k=0}^n f_k(x) - f(x) \right| = \left| \sum_{k=n+1}^\infty f_k(x) \right| \le \sum_{k=n+1}^\infty |f_k(x)| \le \sum_{k=n+1}^\infty \|f_k\| < \varepsilon.\] Here, the first inequality is a consequence of the majorant criterion. Since \(\varepsilon > 0\) has been chosen arbitrarily, the uniform convergence of the series towards \(f\) follows.
Remark 8.4
By Theorem 8.2, integration and limit calculation can also be interchanged in the case of function series. For continuous functions \(f_n: [a,b] \to {\mathbb{R}}\) for \(n \in {\mathbb{N}}\) such that the series \(\sum_{k=0}^\infty f_k\) is uniformly convergent we obtain \[\begin{aligned} \int_a^b \sum_{k=0}^\infty f_k(x)\,\mathrm{d}x &= \int_a^b \lim_{n \to \infty} \sum_{k=0}^n f_k(x)\,\mathrm{d}x \\ &= \lim_{n \to \infty} \int_a^b \sum_{k=0}^n f_k(x)\,\mathrm{d}x \\ &= \lim_{n \to \infty} \sum_{k=0}^n\int_a^b f_k(x)\,\mathrm{d}x \\ &= \sum_{k=0}^\infty \int_a^b f_k(x)\, \mathrm{d}x. \end{aligned}\] Here, we have used Theorem 8.2 in the second equality and the linearity of the integral in the third equality.