Extra questions

Fix \(x_0\in(0,1)\) and \(\varepsilon>0\). For \(x\in(0,1)\), \[ \begin{aligned} |f(x)-f(x_0)| &=\left|\frac{x}{1-x^2}-\frac{x_0}{1-x_0^2}\right|\\ &=\frac{|x-x_0|\,|1+x x_0|}{(1-x^2)(1-x_0^2)}. \end{aligned} \] Choose \[ \delta :=\min\left\{\frac{1-x_0}{2},\frac{\varepsilon(1-x_0)^2}{4}\right\}. \] If \(|x-x_0|<\delta\), then \[ x<x_0+\frac{1-x_0}{2}=\frac{1+x_0}{2}, \] so \[ 1-x>\frac{1-x_0}{2}. \] Also \(1+x\ge1\), \(1+x_0\ge1\), and \(|1+x x_0|\le2\). Hence \[ (1-x^2)(1-x_0^2) =(1-x)(1+x)(1-x_0)(1+x_0) \ge \frac{(1-x_0)^2}{2}. \] Therefore \[ |f(x)-f(x_0)| \le \frac{4|x-x_0|}{(1-x_0)^2} <\varepsilon. \] Thus \(f\) is continuous at \(x_0\). Since \(x_0\) was arbitrary, \(f\) is continuous on \((0,1)\).

For \(n\ge1\), set \[ x_n=\frac{n}{n+1}, \qquad y_n=\frac{n+1}{n+2}. \] Then \[ |x_n-y_n|=\frac{1}{(n+1)(n+2)}\longrightarrow0. \] On the other hand, \[ f(x_n)=\frac{n(n+1)}{2n+1}, \qquad f(y_n)=\frac{(n+1)(n+2)}{2n+3}, \] so \[ |f(y_n)-f(x_n)| =\frac{2(n+1)^2}{(2n+1)(2n+3)} \longrightarrow\frac12. \] If \(f\) were uniformly continuous, taking \(\varepsilon=1/4\) would force \[ |f(y_n)-f(x_n)|<\frac14 \] for all sufficiently large \(n\), because \(|x_n-y_n|\to0\). This contradicts the displayed limit. Therefore \[ \boxed{f\text{ is not uniformly continuous on }(0,1)}. \]

Fix \(s\in E\). By definition of \(E\), there is \(N\) such that \[ a_k<s\qquad(k\ge N). \] Therefore \[ \sup_{k\ge N}a_k\le s. \] Since the tail suprema decrease to \(L\), \[ L\le \sup_{k\ge N}a_k\le s. \] Thus \(L\) is a lower bound for \(E\), and hence \(L\le\inf E\).

Let \(\varepsilon>0\). Since \[ \sup_{k\ge N}a_k\downarrow L, \] there is \(N\) such that \[ \sup_{k\ge N}a_k<L+\varepsilon. \] Hence \[ a_k<L+\varepsilon\qquad(k\ge N), \] so only the finitely many indices \(k<N\) can satisfy \(a_k\ge L+\varepsilon\). Therefore \[ L+\varepsilon\in E. \] It follows that \[ \inf E\le L+\varepsilon. \] Letting \(\varepsilon\downarrow0\) gives \(\inf E\le L\).

compact metric spaces are sequentially compact.

<1>1. <2>2. \(x_n \to x\).

for \(\eps > 0\), choose \(N\) with \(d(x_n, x_m) < \eps/2\) for \(n, m \ge N\) (Cauchy), and \(k\) with \(n_k \ge N\) and \(d(x_{n_k}, x) < \eps/2\); then \(d(x_n, x) \le d(x_n, x_{n_k}) + d(x_{n_k}, x) < \eps\) for \(n \ge N\).

<1>2 is the definition.

\(\{B(x, 1/m)\}_{x \in X}\) is an open cover of the compact space \(X\), so it has a finite subcover; take the centers.

\(D\) is a countable union of finite sets, hence countable; given \(x \in X\) and \(\eps > 0\), choose \(m > 1/\eps\) and \(j\) with \(x \in B(x^{(m)}_j, 1/m)\); then \(d(x, x^{(m)}_j) < 1/m < \eps\).

iterated triangle inequality.

<1>6 bounds \(d(x_n, x_m)\) by the tail \(\sum_{k \ge n} d(x_k, x_{k+1})\) of a convergent series, which tends to \(0\).

<1>6 and <1>7 show \((x_k)\) is Cauchy; completeness (<1>3) gives convergence.

given.

Mean Value Theorem on \([0, c]\); since \(0 < c < 1\), \(1/c > 1\).

Mean Value Theorem on \([c, 2]\); since \(0 < 2 - c < 1\), \(1/(2-c) > 1\).

Mean Value Theorem on \([0, x]\); \(f(x) > x\) gives \(f(x)/x > 1\).

Mean Value Theorem on \([x, 1]\) (note \(f(1) = f(c) = 1\)); \(f(x) < x\) implies \(1 - f(x) > 1 - x > 0\), so the ratio is \(> 1\).

<2>1 and <2>2 exhaust \(f(x) > x\) and \(f(x) < x\).

if \(f(x) > 2 - x\) for some \(x \in (1,2)\), MVT on \([x,2]\) gives \(|f'(u)| = \left|\frac{f(2) - f(x)}{2-x}\right| = \frac{f(x)}{2-x} > 1\); if \(f(x) < 2 - x\), MVT on \([1, x]\) gives \(|f'(v)| = \left|\frac{f(x) - f(1)}{x - 1}\right| = \frac{|f(x) - 1|}{x-1} > 1\) since \(|f(x) - 1| > |x - 1|\).

the left derivative of \(x\) at \(1\) is \(1\), the right derivative of \(2 - x\) at \(1\) is \(-1\); they differ, so \(f\) is not differentiable at \(1\).

<1>2, <1>3, and <1>4 (cases \(c < 1\), \(c > 1\), \(c = 1\)) are exhaustive.

the factor \(x\) vanishes.

\((1 - x^2)^n\) decays exponentially in \(n\) for \(x > 0\) (\(1 - x^2 \in [0,1)\)), which beats the polynomial growth \(n^\beta\); more precisely \(n^\beta(1-x^2)^n = n^\beta e^{n\ln(1-x^2)} \to 0\) since \(\ln(1 - x^2) < 0\).

<2>1 and <2>2.

derivative computation; \(g' > 0\) for \(x < x_n\) and \(g' < 0\) for \(x > x_n\); \(g(0) = g(1) = 0\).

\(1 - x_n^2 = 1 - \frac{1}{2n+1} = \frac{2n}{2n+1}\).

\(\|f_n\|_\infty = n^\beta g(x_n)\).

\(\frac{1}{\sqrt{2n+1}} \sim \frac{1}{\sqrt 2}n^{-1/2}\); the factor \(\left(\frac{2n}{2n+1}\right)^n\) has a positive limit \(e^{-1/2}\).

\(n^{\beta - 1/2} \to 0\).

\(n^{\beta - 1/2} \not\to 0\) (it diverges for \(\beta > 1/2\) and is bounded away from \(0\) for \(\beta = 1/2\)), and uniform convergence to the pointwise limit \(0\) would force \(\|f_n\|_\infty \to 0\).

<2>2 and <2>3.

The integrator \(\alpha\) is nondecreasing and has a single jump of size \(2\) immediately to the right of \(0\).

Fix \(\varepsilon>0\). By continuity of \(f\) at \(0\), choose \(r\in(0,1)\) such that \[ |f(x)-f(0)|<\frac{\varepsilon}{4} \qquad(0\le x\le r). \] Take any partition containing the points \[ -1<0<r<1. \] All increments of \(\alpha\) are zero except the one on the subinterval \([0,r]\), where \[ \alpha(r)-\alpha(0)=1-(-1)=2. \] Consequently, if \(M\) and \(m\) are the supremum and infimum of \(f\) on \([0,r]\), then \[ U(f,P,\alpha)-L(f,P,\alpha)=2(M-m). \] Since \(|f(x)-f(0)|<\varepsilon/4\) on \([0,r]\), \[ M-m<\frac{\varepsilon}{2}, \] and hence \[ U(f,P,\alpha)-L(f,P,\alpha)<\varepsilon. \] Thus \(f\in\mathcal R(\alpha)[-1,1]\).

Moreover, \[ 2m\le \int_{-1}^1f\,d\alpha\le2M, \] and both \(m\) and \(M\) can be made arbitrarily close to \(f(0)\) by shrinking \(r\). Therefore \[ \boxed{\int_{-1}^1f\,d\alpha=2f(0)}. \]

For every polynomial \(p\), \[ x^2p(x^3) \] is a finite linear combination of the functions \(x^{3k+2}\). Hence the hypothesis gives \[ \int_0^1 g(x)x^2p(x^3)\,dx=0. \]

Define \[ h(t):=g(t^{1/3}),\qquad t\in[0,1]. \] Then \(h\) is continuous. By the Weierstrass approximation theorem there are polynomials \(p_n\) such that \[ p_n\to h \] uniformly on \([0,1]\). Substituting \(t=x^3\) gives \[ p_n(x^3)\to g(x) \] uniformly on \([0,1]\).

Since \(x^2g(x)\) is bounded, we may pass to the limit in the identities from Step 2: \[ 0 =\lim_{n\to\infty}\int_0^1 g(x)x^2p_n(x^3)\,dx =\int_0^1x^2g(x)^2\,dx. \] The integrand is continuous and nonnegative, so it vanishes identically. Thus \(g(x)=0\) for every \(x\in(0,1]\), and continuity gives \(g(0)=0\). Hence \[ \boxed{g\equiv0}. \]

\(d_{M\times N} = d_M(x_1,x_2) + d_N(y_1,y_2) \ge 0\) with equality iff both \(d_M = 0\) and \(d_N = 0\), i.e. \(x_1 = x_2\) and \(y_1 = y_2\) (metrics separate points).

\(d_M, d_N\) are symmetric, so the sum is.

\(d_{M\times N}(p_1,p_3) = d_M(x_1,x_3) + d_N(y_1,y_3) \le (d_M(x_1,x_2) + d_M(x_2,x_3)) + (d_N(y_1,y_2) + d_N(y_2,y_3))\) by the triangle inequalities in \(M\) and \(N\); regroup.

<1>1–<1>3 are the metric axioms.

\(S\) is compact (metric, hence sequentially compact).

\(T\) is compact.

\(d_{M\times N}((s_{k_{j_l}},t_{k_{j_l}}),(s,t)) = d_M(s_{k_{j_l}}, s) + d_N(t_{k_{j_l}}, t) \to 0\).

<1>5 is sequential compactness, equivalent to compactness in metric spaces. (Alternatively: \(S \times T\) is closed and bounded in the product of complete spaces, or use the open-cover/finite-subcover argument.)

in \(\RR\) (a metric space), the singletons \(K_n = \{n\}\) are compact (finite, hence compact). Their union \(\cup_n K_n = \NN\) is not compact: the open cover \(\{(n - 1/2, n + 1/2) : n \in \NN\}\) has no finite subcover, since each open interval contains exactly one point of \(\NN\). (Equivalently, \(\NN\) is closed but unbounded, hence not compact.)

by the triangle inequality, \(|d(x_0,x) - d(x_0,y)| \le d(x,y)\), so the map is \(1\)-Lipschitz.

\(K\) is compact and \(d(x_0, \cdot)\) is continuous (<1>2), so \(d(x_0,\cdot)\) attains its minimum on \(K\): there is \(y_0 \in K\) with \(d(x_0, y_0) = \min_{y \in K} d(x_0, y) = d(x_0, K)\), where the last equality is the definition of \(d(x_0,K)\) as an infimum, attained by compactness.

\(E = \{q \in \QQ : q > 0,\ 2 < q^2 < 3\} = \{q \in \QQ : \sqrt2 < q < \sqrt3\}\), and \(\sqrt3 < 2\), so \(E \subseteq (0,2)\) is bounded.

\(E = \QQ \cap (\sqrt2, \sqrt3) = \QQ \cap [\sqrt2, \sqrt3]\), since \(\sqrt2, \sqrt3 \notin \QQ\).

choose rationals \(q_n\) with \(\sqrt2 < q_n < \sqrt2 + 1/n\) (possible by density of \(\QQ\)); then \(q_n \in E\) for all \(n\), and \(q_n \to \sqrt2\) in \(\RR\).

for \(x, x' \in X\) and any \(f \in F\), \(d(x,f) \le d(x,x') + d(x',f)\), so \(\delta(x) \le d(x,x') + \delta(x')\); symmetrically \(\delta(x') \le d(x,x') + \delta(x)\), hence \(|\delta(x) - \delta(x')| \le d(x,x')\).

\(\delta\) is continuous (<1>1), and a continuous function on a compact set attains its minimum: choose \(k_n \in K\) with \(\delta(k_n) \to \inf_K \delta\); a subsequence \(k_{n_j} \to k \in K\) (sequential compactness in the metric space), and continuity gives \(\delta(k) = \inf_K\delta\).

by <1>2 the infimum is \(\delta(k)\) for some \(k \in K\).

take \(\eps \da \inf_{k\in K} d(k,F) > 0\) from <1>3; then \(d(k,f) \ge d(k,F) \ge \eps\) for all \(k \in K\), \(f \in F\).

the triangle inequality gives \(|d(x, x_0) - d(y, x_0)| \le d(x, y)\).

\(X\) is unbounded, so no ball \(B(x_0, M)\) contains all of \(X\).

\(d_{x_0}(X)\) is the continuous image of the connected space \(X\), hence connected; the connected subsets of \(\RR\) are exactly the intervals; \(0 = d(x_0, x_0) \in d_{x_0}(X)\).

\(0 \in d_{x_0}(X)\) (<1>3) and \(d_{x_0}\) is unbounded (<1>2), so the interval \(d_{x_0}(X) \supseteq [0, \infty)\); in particular \(r\) lies in the image.

<1>4 says \(r\) has a preimage under \(d_{x_0}\), which is exactly a point at distance \(r\) from \(x_0\).

rationalize: \(\sqrt{n+1} - \sqrt{n} = \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}\).

divide numerator and denominator by \(\sqrt{n}\).

continuity of the square root and \(1 + 1/n \to 1\) (or the standard squeeze \(1 \le \sqrt{1+1/n} \le 1 + 1/(2n)\)-style bound).

<1>2 and <1>3, by the quotient law for limits.

\(a_n = 3, 1, 3, 1, \ldots\) and \(b_n = 1, 3, 1, 3, \ldots\), both positive.

\(a > 0\) since \(a_n > 0\) for all \(n\) (the limit of positive numbers is \(\ge 0\); if \(a = 0\) then \(0 \le \liminf a_n b_n\), and for any subsequence \(a_n b_n\) along it tends to \(0\) because \(a_n \to 0\) and \(b_n\) is bounded below by \(0\), so \(\liminf a_n b_n = 0\)). For \(a > 0\): since \(a_n \to a\), for any \(\eps\) with \(0 < \eps < a/2\) there is \(N\) with \(a - \eps \le a_n \le a + \eps\) for \(n \ge N\).

\(\left|\dfrac{x^n}{1 + n|x|^n}\right| \le |x|^n\) (denominator \(\ge 1\)), and \(\sum |x|^n\) is a convergent geometric series.

every term is positive and \(\dfrac{1}{1+n} \ge \dfrac{1}{2n}\), so the series dominates a divergent harmonic tail.

\(\sum \dfrac{(-1)^n}{1+n}\) is an alternating series whose terms decrease to \(0\); Leibniz’s test applies.

every term is positive and \(\dfrac{x^n}{1 + nx^n} = \dfrac{1}{n + x^{-n}} \ge \dfrac{1}{n+1}\), so the series dominates \(\sum \dfrac{1}{n+1} = \infty\).

\(x^n = (-1)^n|x|^n\); divide numerator and denominator by \(|x|^n\).

\(n + |x|^{-n} \uparrow \infty\); the increment \(\big((n+1) + |x|^{-(n+1)}\big) - \big(n + |x|^{-n}\big) = 1 - |x|^{-n}(1 - 1/|x|) > 0\) since \(|x|^{-n}(1 - 1/|x|) < 1\) for \(|x| > 1\).

alternating series test with <2>2. <2>4. The convergence is not absolute.

\(a_n \ge \dfrac{1}{n+1}\) (by <2>1), so \(\sum a_n\) diverges.

<1>1–<1>5 cover all real \(x\).

\(\sum_{n\ge 0} a_n\) converges, so its terms tend to \(0\).

for \(|x| < 1\), \(|a_n x^n| \le C|x|^n\), and \(\sum_n C|x|^n\) is a convergent geometric series; so \(\sum_n a_n x^n\) converges absolutely for every \(|x| < 1\).

if \(R > 1\), the power series converges absolutely for every \(|x| < R\), in particular at \(x = 1\), so \(\sum_n |a_n| < \infty\), contradicting conditional convergence.

<1>2 and <1>3. <1>5. Q.E.D.

such indices exist at every step: if no index \(\ge n(k-1) + 1\) had \(a_n \le x - s_{k-1} > 0\), then all remaining terms would be \(> x - s_{k-1} > 0\), contradicting \(a_n \to 0\).

each added term satisfies \(a_{n(k)} \le x - s_{k-1}\), so \(s_k \le x\); hence \(s_k \uparrow s^* \le x\).

suppose \(s^* < x\); let \(\eta = x - s^* > 0\).

<1>2 and <1>3 give \(\sum_{k=1}^\infty a_{n(k)} = x\).

we show its partial sums are unbounded, using the divergence of \(\sum a_n\).

\(a_n \le \beta \Rightarrow \beta + a_n \le 2\beta\); \(a_n > \beta \Rightarrow \beta + a_n < 2a_n \Rightarrow \frac{a_n}{\beta + a_n} > \frac{1}{2}\).

<1>2 covers both cases uniformly: for \(a_n \le \beta\) the min is \(a_n/\beta\) and the RHS is \(a_n/(2\beta)\); for \(a_n > \beta\) the min is \(1\) and the RHS is \(1/2\).

let \(I = \{n : a_n \ge \beta\}\).

<1>3 and <1>4 give \(\sum \frac{a_n}{\beta + a_n} \ge \frac12 \sum \min\{a_n/\beta, 1\} = \infty\).

for \(h \neq 0\), \[ \frac{(fg)(x+h) - (fg)(x)}{h} = \frac{f(x+h)g(x+h) - f(x)g(x)}{h} = \frac{f(x+h) - f(x)}{h}\,g(x+h) + f(x)\,\frac{g(x+h) - g(x)}{h}. \] <1>2. \(g\) is continuous at \(x\).

\(g\) is differentiable at \(x\), hence continuous at \(x\): \(g(x+h) \to g(x)\) as \(h \to 0\).

as \(h \to 0\), \(\frac{f(x+h)-f(x)}{h} \to f'(x)\), \(g(x+h) \to g(x)\) (<1>2), and \(\frac{g(x+h)-g(x)}{h} \to g'(x)\); substituting into <1>1, \[ \lim_{h\to 0}\frac{(fg)(x+h)-(fg)(x)}{h} = f'(x)g(x) + f(x)g'(x), \] so \(fg\) is differentiable at \(x\) with derivative \(f'(x)g(x) + f(x)g'(x)\).

Assume that \(f\) and \(g\) are differentiable at \(x\). For \(h\ne0\), \[ \frac{f(x+h)g(x+h)-f(x)g(x)}{h} = f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}. \] Differentiability of \(f\) at \(x\) implies continuity there, so \[ f(x+h)\to f(x). \] Also, \[ \frac{g(x+h)-g(x)}{h}\to g'(x), \qquad \frac{f(x+h)-f(x)}{h}\to f'(x). \] Therefore \[ \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} = f(x)g'(x)+g(x)f'(x). \] Thus \(fg\) is differentiable at \(x\), with \[ \boxed{(fg)'(x)=f'(x)g(x)+f(x)g'(x).} \]

Let the discontinuity points of \(f\) in \((a,b)\) be \[ c_1,\dots,c_N. \] Because \(f\) is bounded, choose \(M>0\) such that \[ |f(x)|\le M \qquad(x\in[a,b]). \] Fix \(\varepsilon>0\).

Choose pairwise disjoint open intervals \(I_j\) around \(c_j\) whose total length is less than \[ \frac{\varepsilon}{4M} \] (if \(M=0\), the conclusion is immediate). The contribution to any upper-minus-lower Darboux sum from subintervals contained in these neighborhoods is at most \[ 2M\sum_{j=1}^N |I_j|<\frac\varepsilon2. \]

Remove the chosen neighborhoods from \([a,b]\). The remaining set is a finite union of compact intervals on each of which \(f\) is continuous, hence uniformly continuous.

Therefore there exists \(\delta>0\) such that whenever \(x,y\) lie in the same remaining compact component and \(|x-y|<\delta\), one has \[ |f(x)-f(y)|<\frac{\varepsilon}{2(b-a)}. \] Choose a partition \(P\) containing all endpoints of the intervals \(I_j\) and subdividing every complementary compact interval into pieces of length less than \(\delta\).

On each complementary partition interval \(J\), the oscillation of \(f\) is less than \(\varepsilon/(2(b-a))\). Hence the total contribution of all such intervals to \[ U(f,P)-L(f,P) \] is at most \[ \frac{\varepsilon}{2(b-a)}(b-a)=\frac\varepsilon2. \] Combining this with step <1>2 gives \[ U(f,P)-L(f,P)<\varepsilon. \] Since \(\varepsilon>0\) was arbitrary, the Darboux criterion for Riemann integrability implies \[ \boxed{f\in\mathscr R([a,b]).} \]

\(a^n f(x) - x^n f(a) = a^n f(x) - a^n f(a) + a^n f(a) - x^n f(a) = a^n(f(x) - f(a)) - f(a)(x^n - a^n)\).

definition of the derivative.

factorization of \(x^n - a^n\); or the derivative of \(x \mapsto x^n\) at \(a\).

set \(h = g - f\); then \(h \ge 0\), \(h(x_0) = 0\), and \(h\) is differentiable with \(h' = g' - f'\).

for \(x > x_0\), \(h(x) \ge 0 = h(x_0)\), so \((h(x) - h(x_0))/(x - x_0) \ge 0\); hence the right derivative \(h'_+(x_0) = \lim_{x \to x_0^+}\frac{h(x)-h(x_0)}{x-x_0} \ge 0\).

for \(x < x_0\), \(x - x_0 < 0\) and \(h(x) - h(x_0) \ge 0\), so \((h(x)-h(x_0))/(x-x_0) \le 0\); hence \(h'_-(x_0) \le 0\).

\(h\) is differentiable at \(x_0\), so \(h'(x_0) = h'_+(x_0) = h'_-(x_0)\); by <1>2 and <1>3 this common value satisfies \(0 \le h'(x_0) \le 0\), so \(h'(x_0) = 0\).

differentiability on \((a,b]\) implies continuity at every point of \((a,b]\), including \(b\).

the Mean Value Theorem applied to \(f\) on \([x, y] \subset (a, b]\) gives \(f(x) - f(y) = f'(\xi)(x - y)\) for some \(\xi \in (x, y)\), and \(|f'(\xi)| \le M\).

by <1>2, \(|f(x) - f(y)| \le M|x - y| < M\delta\); choose \(\delta = \eps/M\).

the Cauchy criterion <1>3 is equivalent to the existence of the limit (pick a sequence \(x_n \to a^+\); <1>3 makes \((f(x_n))\) Cauchy in the complete space \(\RR\), so it converges, and the limit is independent of the sequence by <1>3 again).

by the mean value theorem applied to \(f\) on \([x, x+1]\), for each \(x\) there is \(\xi_x \in (x, x+1)\) with \[ f(x+1) - f(x) = f'(\xi_x) . \] As \(x \to \infty\), the left side tends to \(a - a = 0\) (both \(f(x+1)\) and \(f(x)\) tend to \(a\)). Also \(\xi_x \to \infty\), so \(f'(\xi_x) \to b\) (since \(\lim_{x\to\infty} f'(x) = b\)). Hence \(b = 0\).

\(f(x) = \sin(x^2)/x\) tends to \(0\) as \(x \to \infty\), but \[ f'(x) = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}, \] whose limit as \(x \to \infty\) does not exist (the term \(2\cos(x^2)\) oscillates between \(-2\) and \(2\) while the second term tends to \(0\)). So \(f\) is differentiable, \(\lim f = 0\), yet \(\lim f'\) fails to exist.

\(\frac{f(x) - f(0)}{x - 0} = \frac{xg(x)}{x} = g(x) \to g(0)\) as \(x \to 0\) (continuity of \(g\) at \(0\); also \(f(0) = 0\cdot g(0) = 0\)).

then \(f(x) = xg(x)\) for all \(x\) (for \(x = 0\) both sides are \(0\)).

<1>2 and the definition of the derivative (using \(f(0) = 0\)).

<1>1 and <1>3 give both directions.

continuity at \(0\) with \(\eps = f(0)/2 > 0\) gives \(|f(x) - f(0)| < f(0)/2\) for \(|x| < \delta\), hence \(f(x) > f(0) - f(0)/2 = f(0)/2\).

\(f \ge 0\) on \([0,1]\) gives \(\int_0^1 f \ge \int_0^\delta f\); and on \([0,\delta]\), \(f \ge f(0)/2\) by <1>1, so \(\int_0^\delta f \ge \delta f(0)/2 > 0\).

hypothesis.

possible since there are finitely many of them.

\(K\) is closed and bounded, hence compact, and \(f\) is continuous on \(K\) (all discontinuities lie in the removed open intervals).

uniform continuity on \(K\) (<1>3).

on each such sub-interval the oscillation is \(< \frac{\eps}{2(b-a)}\) by <1>4, and the total length of the good sub-intervals is \(\le b - a\).

on each, the oscillation is \(\le 2M\) (\(f\) bounded by \(M\)), and their total length is the total length of the \(J_i\), \(< \eps/(4M)\).

<2>1 and <2>2 give \(U(P) - L(P) < \eps/2 + \eps/2 = \eps\).

by the Riemann criterion, \(U(P) - L(P) \to 0\) over fine partitions (<1>5 for every \(\eps\)) implies \(f\) is Riemann integrable on \([a,b]\).

\(0 \le f \le M\) on \([a,b]\), so \(\int_a^b f^p \le M^p (b-a)\), and \((b-a)^{1/p} \to 1\).

<1>1 and \((b-a)^{1/p} \to 1\).

since \(f\) is continuous, it attains \(M\) at some \(x_0 \in [a,b]\).

<1>2 and <1>3 (letting \(\eps \to 0\)). <1>5. Q.E.D.

\(f\) is continuous on the compact set \([0,1]\), so it attains its extrema.

\(m \le f(x) \le M\) for all \(x\), and \(x^4 \ge 0\), so \(m x^4 \le x^4 f(x) \le M x^4\); integrating and using \(\int_0^1 x^4\,dx = 1/5\) gives the claim.

by <1>2, \(5\int_0^1 x^4 f \in [m, M]\).

by <1>3, \(5\int_0^1 x^4f = f(\xi)\) for some \(\xi \in [0,1]\).

let \(N(x) \da x\phi'(x) - \phi(x) + \phi(0)\). Differentiating, \[ \frac{d}{dx}\big(x\phi'(x) - \phi(x) + \phi(0)\big) = \phi'(x) + x\phi''(x) - \phi'(x) = x\phi''(x), \] and \(N(0) = 0\cdot\phi'(0) - \phi(0) + \phi(0) = 0\). Hence by the fundamental theorem of calculus, \[ N(x) = \int_0^x y\,\phi''(y)\,dy . \]

by <1>1, \[ |N(x)| \le \int_0^x y\,|\phi''(y)|\,dy \le \norm{\phi''}_\infty \int_0^x y\,dy = \norm{\phi''}_\infty \frac{x^2}{2} . \]

for \(x > 0\), \[ \Big|\cos x \cdot \frac{N(x)}{x^2}\Big| \le \frac{|N(x)|}{x^2} \le \frac{\norm{\phi''}_\infty}{2}, \] using \(|\cos x| \le 1\) and <1>2. (The integrand is bounded as \(x \to 0^+\) by the same estimate, so it extends continuously to \(0\).)

integrating <1>3 over \([0,1]\), \[ \int_0^1 \cos x \,\frac{x\phi'(x) - \phi(x) + \phi(0)}{x^2}\,dx \le \int_0^1 \frac{\norm{\phi''}_\infty}{2}\,dx = \frac{1}{2}\norm{\phi''}_\infty < \frac{3}{2}\norm{\phi''}_\infty . \] (The constant \(3/2\) in the problem is not optimal; \(1/2\) suffices.)

\(x^n \to 0\) for \(0 \le x < 1\) and \(1^n = 1\); \(f\) is continuous, so \(f(x^n) \to f(0)\) for \(x < 1\) and \(f(1^n) = f(1)\).

\(g(1) = f(1) \ne f(0) = \lim_{x \to 1^-}g(x)\) by hypothesis \(f(0) \ne f(1)\).

the uniform limit theorem — a uniform limit of continuous functions (\(f_n\) continuous since \(x \mapsto x^n\) and \(f\) are continuous) is continuous.

the pointwise limit \(g\) of \(f_n\) is unique, and it is discontinuous by <1>2, contradicting <1>3; hence the convergence is not uniform.

\(f_n(x) = \frac{x}{1+nx^2}\), so \[ f_n'(x) = \frac{(1+nx^2) - 2nx^2}{(1+nx^2)^2} = \frac{1 - nx^2}{(1+nx^2)^2}. \]

let \(u = nx^2 \ge 0\); then \(|f_n'(x)| = \frac{|1-u|}{(1+u)^2}\). For \(0 \le u \le 1\): \(\frac{1-u}{(1+u)^2} \le \frac{1}{(1+u)^2} \le 1\). For \(u > 1\): \(\frac{u-1}{(1+u)^2} \le \frac{u}{(1+u)^2} \cdot \frac{u}{u} \le 1\) (indeed \(u-1 < u \le (1+u)^2\)). Hence \(|f_n'| \le 1\) everywhere.

by the mean value theorem and <1>2, for all \(n\) and all \(x, y\), \[ |f_n(x) - f_n(y)| \le \sup |f_n'| \cdot |x - y| \le |x - y|. \] So the family is uniformly \(1\)-Lipschitz: given \(\eps > 0\), take \(\delta = \eps\) independently of \(n\) and \(x\). In particular \(\mathcal{F}\) is equicontinuous on any compact \([a,b]\) (and in fact on all of \(\RR\)).

\(x^n \to 0\) for \(0 \le x < 1\); \(f\) is continuous, so \(f(x^n) \to f(0)\).

<1>1 covers all \(x\) except \(x = 1\), a null set.

\(f\) is continuous on the compact interval \([0,1]\), hence bounded; \(x^n \in [0,1]\).

dominated convergence (<1>2, <1>3) gives \(\lim_n\int_0^1 f(x^n)\,dx = \int_0^1 f(0)\,dx = f(0)\).

mean value theorem — for \(x \ne y\), \(f(x) - f(y) = f'(c)(x - y)\) for some \(c\) between \(x\) and \(y\), and \(|f'(c)| \le M\).

\(M < \infty\) by hypothesis.

<1>1 and <2>1. <2>3. Q.E.D.

\(|f(x)| \le |f(a)| + |f(x) - f(a)| \le M + M|x - a| \le M + M(b - a)\) by <1>1.

\(x^{99}p(x)\) is a finite linear combination of powers \(x^{99}, x^{100}, \ldots\), each of which integrates to \(0\) against \(f\).

Weierstrass approximation theorem.

\(|x^{99}p_k - x^{99}h| \le \|p_k - h\|_\infty \to 0\).

\(\int f\,x^{99}h = \lim_k \int f\,x^{99}p_k = 0\) by uniform convergence under the integral.

<1>2 with \(h = f \in C([0,1])\).

\(x^{99}f(x)^2 \ge 0\) is continuous with integral \(0\), hence identically \(0\); so \(f = 0\) on \((0,1]\), and continuity at \(0\) gives \(f(0) = 0\).

split the difference into the convergence part (uniform) and the endpoint part (small interval).

uniform convergence \(f_n \to f\) means \(\|f_n - f\|_\infty \to 0\); the interval has length \(\le 1\).

\(f\) is continuous on \([0,1]\) (uniform limit of continuous functions), hence bounded, \(|f| \le M\); then \(\left|\int_0^{1/n} f\right| \le M/n \to 0\), so \(\int_{1/n}^1 f = \int_0^1 f - \int_0^{1/n} f \to \int_0^1 f\).

<1>2 and <1>3 combine: \(\int_{1/n}^1 f_n = \int_{1/n}^1 f + o(1) \to \int_0^1 f\).

\(V\) is well-defined and finite because \(f \in \text{BV}[a,b]\) (so \(V(b) < \infty\)), and \(V(a) = 0\).

partitions of \([a,x]\) extend to partitions of \([a,y]\) by adding the point \(y\), so the sup over the larger interval is \(\ge\).

\(V(y) \ge V(x) + |f(y) - f(x)|\): a partition of \([a,x]\) achieving within \(\eps\) of \(V(x)\), extended by \(y\), gives a partition of \([a,y]\) with variation \(\ge V(x) + |f(y)-f(x)| - \eps\); let \(\eps \to 0\).

<2>1 and \(|u| + u \ge 0\).

<2>1 and \(|u| - u \ge 0\).

<1>2 and <1>3 show both \(V + f\) and \(V - f\) are increasing (multiplying by \(1/2\) preserves monotonicity), and the identity \(f = \frac12(V+f) - \frac12(V-f)\) is a direct algebraic rearrangement.

\(\frac{1}{f(x)} - \frac{1}{f(y)} = \frac{f(y) - f(x)}{f(x)f(y)}\); and \(|f(x)f(y)| \ge c^2\) by the hypothesis \(|f| \ge c > 0\).

<1>1 applied term by term; the last inequality is the definition of total variation.

<1>2 bounds every partition sum of \(g\) uniformly, so the total variation of \(g\) is finite.

show the total variation is infinite by evaluating at a sequence of points with large accumulated oscillation.

\(1/t_k = (2k+1)\pi/2\), so \(\sin(1/t_k) = \pm 1\) alternating with \(k\); and \(t_k \downarrow 0\).

\(f(t_k) = t_k \sin((2k+1)\pi/2) = t_k \cdot (-1)^k\).

\(f(t_k) = (-1)^k t_k\) and \(f(t_{k+1}) = (-1)^{k+1} t_{k+1}\) have opposite signs (and \(f(0) = 0\)), so \(|f(t_k) - f(t_{k+1})| = t_k + t_{k+1}\) exactly; write \(t_K = 0\).

the harmonic tail \(\sum \frac{1}{2k+1}\) diverges.

<1>4 and <1>5 give arbitrarily large finite-variation lower bounds; \(f\) is continuous (including at \(0\)) but has unbounded variation.

\(\alpha(x) = a_n\) if \(x = x_n\) and \(0\) otherwise, so \(|\alpha(x)| \le \sup_n |a_n| \le \sum_n |a_n|\), and \(\sum_n |a_n|\) converges absolutely.

let \(0 = t_0 < t_1 < \cdots < t_m = 1\) be any partition.

by <1>2, every partition has variation \(\le 2\sum_n |a_n| < \infty\); taking the supremum, \(V(\alpha) \le 2\sum_n|a_n| < \infty\).

explicit space and subset.

\(E = X\) is the whole space, and the whole space is always closed (its complement \(\varnothing\) is open).

\(E \subseteq B(x_0, 1)\) for any \(x_0 \in X\) (the diameter of \((0,1)\) is \(1\); e.g. \(E \subseteq B(1/2, 1)\)).

the sequence \(x_k = 1/(k+1) \in E\) has no convergent subsequence in \(X\): it converges to \(0 \notin X\), and any subsequence also converges to \(0\), so no subsequence converges to a point of \(X\).

<1>2–<1>4 show closed + bounded \(\not\Rightarrow\) compact in general metric spaces; compactness needs completeness (or the ambient space must be \(\RR^n\)). Alternative example: \(X = \QQ\), \(E = \QQ \cap [0,1]\) (closed and bounded in \(\QQ\), not compact).

if \(f\) is continuous, then \(f^{-1}(a,\infty)\) and \(f^{-1}(-\infty,b)\) are preimages of open sets, hence open.

every open set \(U \subseteq \RR\) is a union of open intervals \((a,b)\); since \(f^{-1}(a,b) = f^{-1}(a,\infty) \cap f^{-1}(-\infty,b)\) is open, \(f^{-1}(U)\) is a union of open sets, hence open.

fix \(a < f(x)\).

suppose \(f\) were not l.s.c.: some \(f^{-1}(a,\infty)\) is not open.

this is Dini’s theorem: a monotone pointwise-convergent sequence of continuous functions on a compact space converges uniformly.

\(f_n\) is continuous, and \(U_n = f_n^{-1}(-\infty, \eps)\) is the preimage of an open set.

\(f_n \ge f_{n+1}\) gives \(U_n \subseteq U_{n+1}\).

\(\{U_n\}\) is an open cover of the compact space \(X\) (<1>2, <1>3), so it has a finite subcover \(U_{n_1} \subseteq \cdots \subseteq U_{n_k}\); since the \(U_n\) are nested, \(X = U_{n_k}\), and taking \(N = n_k\) works.

by <1>4, for \(\eps > 0\) there is \(N\) with \(f_N(x) < \eps\) for all \(x\); monotonicity gives \(0 \le f_n(x) \le f_N(x) < \eps\) for all \(n \ge N\) and all \(x\).

definition of \(\alpha\); the value \(\alpha(0)\) is irrelevant to the Stieltjes integral.

hypothesis.

on intervals where \(\alpha\) is constant, the \(\alpha\)-increment is \(0\), so those terms contribute nothing to \(\sum f(t_i)\Delta\alpha_i\).

the only contributing interval \([x_j, x_{j+1}] \ni 0\) has \(\Delta\alpha = 2\) and oscillation of \(f\) on it \(\le 2\eps\); every other interval has \(\Delta\alpha = 0\).

the Riemann–Stieltjes criterion: \(U(P) - L(P)\) can be made arbitrarily small by refining the partition.

for partitions as in <1>4, \(\sum f(t_i)\Delta\alpha_i = f(t)(\alpha(0+) - \alpha(0-)) = 2f(t)\) for some \(t\) in the interval around \(0\), and \(|f(t) - f(0)| < \eps\), so the sums converge to \(2f(0)\) as the mesh tends to \(0\).

given partitions \(P_1, P_2\) of \([a,c], [c,b]\), their union \(P\) is a partition of \([a,b]\) with \(U(f,P,\alpha) - L(f,P,\alpha) = \big(U(f,P_1,\alpha) - L(f,P_1,\alpha)\big) + \big(U(f,P_2,\alpha) - L(f,P_2,\alpha)\big)\) (the \(\alpha\)-increments on the two subintervals add up to those on \([a,b]\); note \(f\) is bounded on \([a,b]\) by integrability).

integrability on each subinterval.

the upper/lower sums split over the two subintervals as in <1>1 (the boundary point \(c\) contributes its \(\Delta\alpha\) to exactly one of the two sums), so \(U - L = (U_1 - L_1) + (U_2 - L_2) < \eps\).

<2>2 shows \(U - L\) can be made arbitrarily small, i.e. integrability on \([a,b]\) (Cauchy criterion).

for the partitions of <2>2, Riemann–Stieltjes sums satisfy \(S(P) = S(P_1) + S(P_2)\); taking \(P_1, P_2\) with mesh making all three sums converge to the respective integrals gives \(\int_a^b = \int_a^c + \int_c^b\) (both sides are limits of the same telescoping sums).

<1>1, <1>2, <1>3 establish both directions and the additivity formula.

\(\alpha \in C^1[a,b]\), so \(\alpha'\) is continuous on the compact set \([a,b]\), hence uniformly continuous and bounded.

\(f\) is Riemann integrable, hence bounded and continuous a.e.; \(\alpha'\) is continuous, hence Riemann integrable.

let \(P = \{a = x_0 < \cdots < x_n = b\}\) be a partition and pick tags \(\xi_k \in [x_{k-1}, x_k]\).

since \(f\) is bounded (\(|f| \le B\)) and \(\alpha'\) is uniformly continuous, the difference between the sums in <1>3 and the ordinary Riemann sums \(\sum_k f(\xi_k)\alpha'(\xi_k)(x_k-x_{k-1})\) tends to \(0\) as the mesh goes to \(0\): \[ \Big|\sum_k f(\xi_k)\big(\alpha'(\eta_k) - \alpha'(\xi_k)\big)\Delta x_k\Big| \le B \sum_k |\alpha'(\eta_k)-\alpha'(\xi_k)|\Delta x_k \le B\,(b-a)\,\omega_\alpha'(\text{mesh}) \to 0. \] Both sums converge to \(\int_a^b f(x)\alpha'(x)\,dx\) (<1>2), so the Stieltjes sums converge to the same value.

the Riemann–Stieltjes integral \(\int_a^b f\,d\alpha\) exists (as \(f\) is Riemann integrable and \(\alpha\) is of bounded variation, <1>1) and equals the limit of the Stieltjes sums, which by <1>4 is \(\int_a^b f(x)\alpha'(x)\,dx\).

By the root test, since \(\limsup |a_n x^n|^{1/n} = |x| \limsup |a_n|^{1/n}\), the series converges absolutely when \(|x| < R\) and diverges when \(|x| > R\), because then its terms do not tend to \(0\).

For \(|x| < \min\{R_1, R_2\}\) both \(\sum a_n x^n\) and \(\sum b_n x^n\) converge absolutely, and \(|(a_n + b_n)x^n| \le |a_nx^n| + |b_nx^n|\).

Let \(R_1 < |x| < R_2\). Then \(\sum a_n x^n\) diverges and \(\sum b_n x^n\) converges. If \(\sum (a_n + b_n)x^n\) converged, then \(\sum a_n x^n\) would converge as the difference of two convergent series. So \(\sum (a_n + b_n)x^n\) diverges at points with \(|x| > R_1\) arbitrarily close to \(R_1\), and \(R \le R_1\).

The inequality is step <1>1. With \(b_n = a_n\), the coefficients \(2a_n\) give \(R = R_1\). With \(b_n = -a_n\), the series is \(0\) and \(R = \infty\), which exceeds \(R_1\) whenever \(R_1 < \infty\).

If \(R_1 \neq R_2\), steps <1>1 and <1>2, with the roles of the series exchanged if \(R_2 < R_1\), give \(R = \min\{R_1, R_2\}\). Step <1>3 is the case \(R_1 = R_2\).

The set \(\theset{uv : 0 \le u < R_1,\ 0 \le v < R_2}\) is the interval \([0, R_1R_2)\), with \(R_1R_2 = \infty\) if either radius is infinite and the other positive.

\(\sum a_n u^n\) and \(\sum b_n v^n\) converge, since \(u < R_1\) and \(v < R_2\), so their terms tend to \(0\).

By step <2>2, \(|a_n b_n x^n| = (|a_n|u^n)(|b_n|v^n)\left|\frac{x}{uv}\right|^n \le \left|\frac{x}{uv}\right|^n\) for large \(n\), and \(\sum_n |x/(uv)|^n\) is a convergent geometric series by step <2>1.

Step <1>1.

\(\limsup |a_n|^{1/n} = 1\) along the even indices and \(\limsup|b_n|^{1/n} = 1\) along the odd indices, so the Cauchy–Hadamard formula gives \(R_1 = R_2 = 1\). Every \(a_nb_n = 0\), so \(\sum a_nb_nx^n\) is the zero series and \(R = \infty > 1 = R_1R_2\).

Step <1>2 is the inequality and step <1>3 shows that it can be strict.

\(2^{-nx} = (2^{-x})^n\).

\(h'(x) = e^{-x\ln 2}(1 - x\ln 2)\) vanishes only at \(x = 1/\ln 2\), where \(h\) changes from increasing to decreasing; \(h(0) = 0\) and \(h \to 0\) as \(x \to \infty\).

\(h(1/\ln 2) = \frac{1}{\ln 2}e^{-1} = \frac{1}{e\ln 2}\), and \(e\ln 2 > 1\).

on \([0,B]\), \(x2^{-x} \le \max(h(B), h(1/\ln 2)) < 1\) by <1>2, so the Weierstrass M-test applies: \(\sum (x2^{-x})^n \le \sum r^n\) with \(r < 1\).

<1>2 gives a uniform ratio bound \(x2^{-x} \le \frac{1}{e\ln 2} < 1\) on all of \([0,\infty)\), so the M-test applies globally: \(\left|\sum_{n>N}(x2^{-x})^n\right| \le \frac{r^{N+1}}{1-r}\) with \(r = \frac{1}{e\ln 2}\), which \(\to 0\) as \(N \to \infty\) independent of \(x\).

the M-test requires \(\sum_n M_n < \infty\) where \(M_n = \sup_x |f_n(x)|\).

the intervals \((2^{-(n+1)}, 2^{-n}]\) for \(n = 1, 2, \ldots\) are disjoint (they partition \((0, 1/2]\)), and \(f_n = 0\) elsewhere.

by <1>2, at most one term is nonzero at each \(x\), so the series has at most one nonzero summand and converges.

for the tail, \(S(x) - S_N(x) = \sum_{n>N} f_n(x)\), which by <1>2 is either \(0\) (if the support containing \(x\) has index \(\le N\)) or \(f_n(x) = 1/n \le 1/(N+1)\) (if its index is \(n > N\)). Hence \[ \sup_{x\in\RR}\big|S(x) - S_N(x)\big| \le \frac{1}{N+1} \to 0, \] so the series converges uniformly.

for \(x \in [0,1)\), \(0 \le \frac{x^k}{1+x^k} \le x^k\), and \(\sum_k x^k\) is a convergent geometric series; hence \(\sum_k \frac{x^k}{1+x^k}\) converges, defining \(f(x)\).

for \(x \le a\), \(\frac{x^k}{1+x^k} \le a^k\), and \(\sum_k a^k\) converges (geometric, \(a < 1\)); by the Weierstrass \(M\)-test the convergence is uniform on \([0,a]\).

each term \(g_k(x) = \frac{x^k}{1+x^k}\) is differentiable with \(g_k'(x) = \frac{kx^{k-1}}{(1+x^k)^2}\).

Since \(f\) is continuous on \([0,1]\), let \[ M:=\max_{[0,1]}|f|<\infty. \] The derivative bound gives \(|f'|\le kM\), so \(f\) is Lipschitz and hence absolutely continuous. Therefore, using \(f(0)=0\), \[ |f(x)| =\left|\int_0^x f'(t)\,dt\right| \le k\int_0^x |f(t)|\,dt \qquad(0\le x\le1). \]

We claim that for every integer \(n\ge1\), \[ |f(x)|\le M\frac{(kx)^n}{n!} \qquad(0\le x\le1). \] For \(n=1\), Step 1 and \(|f(t)|\le M\) give \[ |f(x)|\le kMx. \] If the estimate holds for \(n\), then Step 1 gives \[ |f(x)| \le k\int_0^x M\frac{(kt)^n}{n!}\,dt =M\frac{(kx)^{n+1}}{(n+1)!}. \] Thus the claim follows by induction.

For each fixed \(x\in[0,1]\), \[ 0\le |f(x)|\le M\frac{(kx)^n}{n!}\longrightarrow0 \qquad(n\to\infty). \] Hence \(f(x)=0\). Since \(x\) was arbitrary, \[ \boxed{f\equiv0\text{ on }[0,1].} \]