Proof. Write \(d_n = \sup _{x\in A}|f_n(x) - f(x)|\). Suppose \(f_n\longrightarrow f\) uniformly on \(A\), and let \(\varepsilon >0\). There is an \(N\) such that \(|f_n(x) - f(x)| < \frac {\varepsilon }{2}\) for all \(n> N\) and all \(x\in A\). Taking the supremum over \(x\), \(d_n\leq \frac {\varepsilon }{2}<\varepsilon \) for \(n> N\). Hence \(d_n\longrightarrow 0\).

Conversely, suppose \(d_n\longrightarrow 0\), and let \(\varepsilon >0\). There is an \(N\) with \(d_n<\varepsilon \) for \(n>N\). Then for every \(x\in A\) and \(n>N\), \[|f_n(x) - f(x)| \leq d_n < \varepsilon ,\] and \(N\) does not depend on \(x\). Hence \(f_n\longrightarrow f\) uniformly on \(A\). □

Solution. For \(0\leq x<1\), \(x^n\longrightarrow 0\) and so \(f_n(x)\longrightarrow 0\), while \(f_n(1) = \frac {1}{5}\) for every \(n\). Hence \(f_n \longrightarrow f\) pointwise, where \[ f(x) = \begin {cases} 0, & 0\leq x < 1,\\ \frac {1}{5}, & x = 1. \end {cases} \] To test for uniform convergence, note that \[|f_n(x) - f(x)| = \begin {cases} \frac {x^n}{2 + 3x^n}, & 0\leq x < 1,\\ 0, & x = 1. \end {cases}\] For fixed \(n\), \(\frac {x^n}{2 + 3x^n}\) increases with \(x\) on \([0,1)\) and tends to \(\frac {1}{5}\) as \(x\longrightarrow 1\). Hence \[ \sup _{x\in [0,1]}|f_n(x) - f(x)| = \frac {1}{5} \quad \text {for every } n . \] This does not tend to \(0\), so \(f_n\) does not converge uniformly on \([0,1]\). (This also follows from the next theorem, since each \(f_n\) is continuous but \(f\) is not.)

Proof. Let \(x_0\in E\) and \(\varepsilon >0\). For every \(x\in E\) and every \(n\), \[|f(x) - f(x_0)|\leq |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| .\] By uniform convergence there is an \(N\) such that \(|f_n(x) - f(x)| < \frac {\varepsilon }{3}\) for all \(n> N\) and all \(x\in E\); in particular \(|f_n(x_0) - f(x_0)| < \frac {\varepsilon }{3}\). Fix one \(n>N\). Since \(f_n\) is continuous at \(x_0\), there is a \(\delta > 0\) such that \(|f_n(x) - f_n(x_0)|< \frac {\varepsilon }{3}\) whenever \(x\in E\) and \(|x - x_0| < \delta \). For such \(x\) all three terms are less than \(\frac {\varepsilon }{3}\), so \(|f(x) - f(x_0)| < \varepsilon \). Hence \(f\) is continuous at \(x_0\).

For the limits: \(\lim _{x\rightarrow x_0} f_n(x) = f_n(x_0)\longrightarrow f(x_0)\) as \(n\longrightarrow \infty \), and \(\lim _{x\rightarrow x_0} f(x) = f(x_0)\) by the continuity of \(f\). □

Proof. For each \(x\in E\), \(\{f_n(x)\}\) is a Cauchy sequence of real numbers and hence converges; call its limit \(f(x)\). Let \(\varepsilon >0\) and choose \(N\) such that \(|f_n(x) - f_m(x)| < \frac {\varepsilon }{2}\) for all \(n, m> N\) and all \(x\in E\). Fix \(n>N\) and \(x\in E\), and let \(m\longrightarrow \infty \): this gives \(|f_n(x) - f(x)| \leq \frac {\varepsilon }{2} < \varepsilon \). Since \(N\) does not depend on \(x\), \(f_n \longrightarrow f\) uniformly on \(E\). □

Proof. We may assume that the \(f_n\) decrease (otherwise consider \(-f_n\)). Put \(g_n = f_n - f\). Then \(\{g_n\}\) is a decreasing sequence of non-negative continuous functions converging pointwise to zero.

Let \(\varepsilon > 0\), and let \(V_n = \{x\in E: g_n(x) < \varepsilon \}\). By continuity of \(g_n\), \(V_n\) is open in \(E\), say \(V_n = E\cap O_n\) with \(O_n\) open. If \(x\in E\) then \(g_n(x) < \varepsilon \) for large \(n\), so the sets \(O_n\) cover \(E\). Since \(E\) is compact, \(E\subseteq O_1\cup \dots \cup O_N\) for some \(N\), that is, \(E = V_1\cup \dots \cup V_N\). Since the \(g_n\) decrease, \(V_n \subseteq V_{n + 1}\) for all \(n\), so \(E = V_N\). Thus \(g_N(x) < \varepsilon \) for every \(x\in E\), and for \(n\geq N\), \[0\leq g_n(x) \leq g_N(x) <\varepsilon \qquad \text {for all } x\in E .\] Hence \(g_n \longrightarrow 0\) uniformly on \(E\), that is, \(f_n\longrightarrow f\) uniformly on \(E\). □

Proof. Let \(S_n\) be the \(n^{\text {th}}\) partial sum of the series. For \(m< n\) and every \(x\in E\), \[|S_n(x) - S_m(x)| = \left |\sum ^n_{k = m + 1} f_k(x)\right | \leq \sum ^n_{k = m + 1} |f_k(x)| \leq \sum ^{\infty }_{k=m + 1} M_k ,\] and the right side does not depend on \(x\) and tends to \(0\) as \(m\longrightarrow \infty \), since \(\sum M_n < \infty \). Thus \(\{S_n\}\) has the uniform Cauchy property, and therefore converges uniformly on \(E\). If each \(f_n\) is continuous, so is each \(S_n\), and the uniform limit is continuous by the theorem above. □

Solution. The pointwise limit is \(f = 0\). For \(x>0\) the inequality of arithmetic and geometric means gives \(1 + nx^2\geq 2\sqrt {n}\,x\), so \(|f_n(x)|\leq \frac {1}{2\sqrt {n}}\), with equality at \(x = \frac {1}{\sqrt {n}}\); by symmetry the same bound holds for \(x<0\). Hence \(\sup _{x\in \mathbb {R}}|f_n(x) - 0| = \frac {1}{2\sqrt {n}}\longrightarrow 0\), and the convergence is uniform. For \(\varepsilon = 0.01\) any \(n>2500\) works.

Solution. At \(x = 0\) every term is \(0\); for \(x>0\), \(f_n(x)\leq \frac {nx}{n^2x^2} = \frac {1}{nx}\longrightarrow 0\). But \(f_n\left (\frac {1}{n}\right ) = \frac {1}{2}\) for every \(n\), so \(\sup _{[0,1]}|f_n|\geq \frac {1}{2}\) and the convergence is not uniform on \([0,1]\). The “bump” of height \(\frac {1}{2}\) moves towards \(0\) as \(n\) grows.

On \([a,1]\) we have \(|f_n(x)|\leq \frac {1}{nx}\leq \frac {1}{na}\longrightarrow 0\), uniformly in \(x\).

Solution. On \([-1,1]\), \(\left |\frac {x^n}{n^2}\right |\leq \frac {1}{n^2}\), and on \(\mathbb {R}\), \(\frac {1}{n^2+x^2}\leq \frac {1}{n^2}\). Since \(\sum \frac {1}{n^2}\) converges, both series converge uniformly by the M-test. Their terms are continuous, so both sums are continuous functions. Moreover the tail after \(N\) terms is at most \(\sum _{n>N}\frac {1}{n^2}<\frac {1}{N}\), so \(N = 100\) terms give either sum to within \(0.01\) at every point.

Solution. On the compact set \(\left [0,\frac {1}{2}\right ]\) the functions \(x^n\) are continuous, decrease with \(n\), and converge pointwise to the continuous function \(0\). Dini’s Theorem gives uniform convergence (as also follows from \(\sup x^n = 2^{-n}\)). On \([0,1)\) the limit is still \(0\) and the sequence still decreases, but the convergence is not uniform, since \(\sup _{[0,1)}x^n = 1\). Dini’s Theorem does not apply because \([0,1)\) is not compact.