8. Sequences of Functions

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 \(A\).

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 \(x\in A\).

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 \(n\in\N\).

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 \(c\in A\).

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 \(\int_a^b f\).

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

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 \(f'(x) = g(x)\).

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.

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 \[ \int_a^b \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int_a^b f_n \]

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.

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*}

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.