Real analysis

The triangle inequality applied to \(x = (x - y) + y\) gives \(|x| \le |x - y| + |y|\).

The triangle inequality applied to \(y = (y - x) + x\) gives \(|y| \le |y - x| + |x|\), and \(|y - x| = |x - y|\).

Steps <1>1 and <1>2 bound both \(|x| - |y|\) and its negative by \(|x - y|\).

Suppose no \(n \in \NN\) exceeds \(x\). Then \(\NN\) is bounded above, so by completeness it has a least upper bound \(s\). Since \(s - 1\) is not an upper bound, some \(n \in \NN\) has \(n > s - 1\), so \(n + 1 > s\) with \(n + 1 \in \NN\), contradicting that \(s\) is an upper bound.

\(\abs{x-y} > 0\), and a common point \(z\) would give \(\abs{x-y} \le \abs{x-z} + \abs{z-y} < 2\abs{x-y}/3\).

The balls \(U_y\), \(y \in K\), form an open cover of the compact set \(K\).

\(V = \bigcap_{i=1}^k V_{y_i}\) is an open ball about \(x\). A point \(z \in V \cap K\) would lie in some \(U_{y_i}\) by step <2>2 and in \(V_{y_i}\), contradicting step <2>1. So every point of \(\RR^d \setminus K\) has a neighborhood disjoint from \(K\), and \(\RR^d \setminus K\) is open.

For \(\eps > 0\), all but finitely many terms lie within \(\eps\) of \(x\), and so do all but finitely many terms of any subsequence. Limits in \(\RR\) are unique.

\(x^n \to 0\) for \(x \in [0,1)\) and \(1^n = 1\), so the pointwise limit is \(\chi_{\theset1}\), but \(\sup_{[0,1]}|x^n - \chi_{\theset1}(x)| = 1\) for every \(n\).

Call \(n\) a peak if \(a_n \ge a_m\) for all \(m \ge n\). If there are infinitely many peaks \(n_1 < n_2 < \cdots\), then \(a_{n_1} \ge a_{n_2} \ge \cdots\). Otherwise let \(n_1\) exceed every peak. Since \(n_k\) is not a peak, there is \(n_{k+1} > n_k\) with \(a_{n_{k+1}} > a_{n_k}\), which gives an increasing subsequence.

If \(s = \sup_n a_n < \infty\) and \(\eps > 0\), then \(s - \eps\) is not an upper bound, so \(a_N > s - \eps\) for some \(N\), and \(s - \eps < a_n \le s\) for \(n \ge N\). If \((a_n)\) is not bounded above, for every \(M\) some \(a_N > M\), and \(a_n \ge a_N > M\) for \(n \ge N\). Decreasing sequences are handled by passing to \((-a_n)\).

By step <1>3 it has a monotone subsequence, which is bounded and therefore converges by step <1>4.

\(\sum_{n\ge 1}a_n\) converges iff the sequence of partial sums \(s_N = \sum_{n\le N}a_n\) converges (to a finite limit). Practical checks: (i) necessary: \(a_n \to 0\); (ii) Cauchy criterion: for every \(\eps > 0\) there is \(N\) with \(|\sum_{n=M+1}^{K}a_n| < \eps\) for all \(K > M \ge N\); (iii) tests: comparison, ratio, root, integral, alternating series, and absolute convergence.

\(\sum f_n\) converges uniformly on \(E\) iff the partial sums \(S_N = \sum_{n\le N}f_n\) converge uniformly on \(E\): \(\sup_{x\in E}|S_N(x) - S(x)| \to 0\). To show it does NOT converge uniformly: exhibit \(\eps > 0\) and points \(x_N \in E\) with \(|S_N(x_N) - S(x_N)| \ge \eps\) for infinitely many \(N\) (or: the tail \(\sup_{x\in E}|\sum_{n>N}f_n(x)|\) fails to tend to \(0\)).

\(a_n = s_n - s_{n-1} \to s - s = 0\).

\(\sum_{n\ge1}a_n\) converges iff the tails \(T_N = \sum_{n>N}a_n\) satisfy \(T_N \to 0\) as \(N \to \infty\) (this is just \(s_N \to s\), with \(T_N = s - s_N\)). So yes, it is a necessary and sufficient condition for convergence (of the partial sums themselves); note this concerns series, and it is equivalent to the Cauchy criterion.

if \(a_n \ne 0\) eventually and \(\limsup |a_{n+1}/a_n| < 1\), then \(\sum a_n\) converges absolutely; if \(\liminf |a_{n+1}/a_n| > 1\), it diverges. Proof of the first: pick \(r\) with \(\limsup < r < 1\); then \(|a_n| \le C r^n\) eventually, and the geometric series \(\sum r^n\) converges.

if \(\limsup \sqrt[n]{|a_n|} < 1\), then \(\sum a_n\) converges absolutely (choose \(r\) with the limsup \(< r < 1\); eventually \(\sqrt[n]{|a_n|} \le r\), i.e. \(|a_n| \le r^n\)); if \(\limsup\sqrt[n]{|a_n|} > 1\), the terms do not tend to \(0\) and the series diverges.

if \(f \ge 0\) is decreasing on \([1,\infty)\), then \(\sum_{n=1}^N f(n)\) and \(\int_1^N f\) converge or diverge together, since \(\int_1^{N+1}f \le \sum_{n=1}^N f(n) \le f(1) + \int_1^N f\) (compare the sum to the integral over unit intervals).

if \(a_n \ge 0\) decreases to \(0\), then \(\sum (-1)^{n+1}a_n\) converges: the even and odd partial sums are monotone and bounded (interleaved), so both converge to a common limit, and the error after \(N\) terms is bounded by \(a_{N+1}\).

group terms: \(1 + \tfrac12 + (\tfrac13+\tfrac14) + (\tfrac15+\cdots+\tfrac18) + \cdots \ge 1 + \tfrac12 + \tfrac12 + \tfrac12 + \cdots = \infty\); or use the integral test with \(f(x) = 1/x\) (\(\int_1^\infty dx/x = \infty\)).

\(s_N = \sum_{n=0}^{N-1} x^n\) satisfies \((1-x)s_N = 1 - x^N\), so \(s_N = \frac{1-x^N}{1-x}\) for \(x \ne 1\); if \(|x| < 1\), \(s_N \to \frac{1}{1-x}\). If \(|x| \ge 1\) and \(x \ne 1\), the terms do not tend to \(0\), so it diverges; at \(x = 1\), \(s_N = N \to \infty\).

\(\sum_{n\ge1} n^{-p}\) converges iff \(p > 1\). For \(p > 1\) use the integral test: \(\int_1^\infty x^{-p}dx = \frac{1}{p-1} < \infty\); for \(p \le 1\): \(\sum n^{-p} \ge \sum n^{-1} = \infty\) (or integral test for \(p < 1\)).

\(\sum a_n\) converges absolutely iff \(\sum |a_n| < \infty\), which implies convergence (by the Cauchy criterion: \(|\sum_{M+1}^K a_n| \le \sum_{M+1}^K |a_n|\)). Conditional convergence means \(\sum a_n\) converges but \(\sum|a_n| = \infty\). Example: \(\sum (-1)^{n+1}/n\) converges (alternating series test, <1>8) but \(\sum 1/n\) diverges (<1>9).

Given \(\eps > 0\), the set \(\theset{p/q : |p/q - x| < 1,\ 1/q \ge \eps}\) is finite and does not contain \(x\). Choose \(\delta < 1\) smaller than the distance from \(x\) to each of its points. Then \(|f(y)| < \eps = \eps + |f(x)|\) for \(|y - x| < \delta\).

\(f(p/q) = 1/q > 0\), while irrationals \(y \to p/q\) have \(f(y) = 0\).

Steps <2>1 and <2>2.

Let \(\omega_f(x) = \inf_{\delta > 0}\sup\theset{|f(y) - f(z)| : y, z \in B(x,\delta)}\) be the oscillation of \(f\) at \(x\). The set \(\theset{\omega_f < 1/n}\) is open, so \(\theset{\omega_f \ge 1/n}\) is closed, and \(f\) is discontinuous at \(x\) exactly when \(\omega_f(x) \geq 1/n\) for some \(n\).

If \(\RR\setminus\QQ = \bigcup_n F_n\) with \(F_n\) closed, then \(\QQ = \bigcap_n U_n\) with \(U_n = \RR \setminus F_n\) open and dense, since \(U_n \supseteq \QQ\). The sets \(U_n\) and \(\RR \setminus \theset q\), \(q \in \QQ\), are countably many open dense sets with empty intersection, contradicting the Baire category theorem.

Steps <2>1 and <2>2.

\(f_n(0) = 0\), and for \(x > 0\), \(0 \le f_n(x) \le \frac{nx}{n^2 x^2} = \frac{1}{nx} \to 0\). But \(f_n(1/n) = \frac12\) for every \(n\), so \(\sup_{[0,1]}|f_n - 0| \ge \frac12\).

\(\frac{f(h) - f(0)}{h} = h\sin(1/h)\) and \(|h\sin(1/h)| \leq |h| \to 0\). For \(x \neq 0\), \(f\) is a product and composition of differentiable functions near \(x\), and the product and chain rules give the formula.

At \(x_k = 1/(2k\pi)\), \(f'(x_k) = -1\) for every \(k\), while \(x_k \to 0\) and \(f'(0) = 0\).

Steps <2>1 and <2>2.

The first is uniform convergence; the second is continuity of \(f_n\) at \(x_0\).

For \(x \in E\) with \(|x - x_0| < \delta\), step <2>1 and the triangle inequality give \(|f(x) - f(x_0)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \eps\).

Choose \(N\) with \(\|f - f_n\|_\infty \le 1\) for \(n \geq N\). Then \(\|f\|_\infty \le \|f_N\|_\infty + 1\), and \(\|f_n\|_\infty \leq \|f\|_\infty + 1\) for \(n \geq N\), so \(\sup_n \|f_n\|_\infty \leq \max(\|f_1\|_\infty, \ldots, \|f_{N-1}\|_\infty, \|f\|_\infty + 1)\).

\(\sup_{x\ge1}|f_n(x) - x| = 1/n \to 0\), and \(g_n\) is constant in \(n\). But \(h_n(x) = x^2 + \frac{x}{n} \to x^2\) pointwise, and \(\sup_{x\ge1}|h_n(x) - x^2| = \sup_{x\ge1}\frac{x}{n} = \infty\) for every \(n\).

By step <1>3, \(g\) is bounded and there is \(M\) with \(\|f_n\|_\infty \le M\) for all \(n\) and \(\|g\|_\infty \le M\). Then \(|f_n g_n - fg| \le |f_n|\,|g_n - g| + |g|\,|f_n - f| \le M\|g_n - g\|_\infty + M\|f_n - f\|_\infty \to 0\) uniformly on \(E\).

Steps <2>1 and <2>2.

\(|f_n| \leq 1/n\), so \(f_n \to 0\) uniformly and \(\frac{d}{dx}\lim_n f_n(x) = 0\). But \(f_n'(x) = \cos(nx)\), so \(\lim_n f_n'(0) = 1 \ne 0\).

\(0 \le \sqrt{x^2 + 1/n} - |x| = \frac{1/n}{\sqrt{x^2 + 1/n} + |x|} \le \frac{1/n}{\sqrt{1/n}} = \frac{1}{\sqrt n} \to 0\).

\(x \mapsto \sqrt{x^2 + c}\) is smooth for \(c > 0\), since \(x^2 + c > 0\). The function \(|x|\) has left derivative \(-1\) and right derivative \(1\) at \(0\).

Steps <2>1 and <2>2.

\(C = \bigcap_n C_n\), where \(C_n\) is the union of the \(2^n\) closed intervals of length \(3^{-n}\) remaining after \(n\) stages. Each \(C_n\) is closed, so \(C\) is closed, and closed sets are Borel.

\(m(C) \le m(C_n) = (2/3)^n\) for all \(n\), so \(m(C) = 0\). A nonempty open interval has positive measure, so it is not contained in \(C\).

Assume \(f\in C([0,1])\) and \(\int_0^1 f(x)x^n\,dx=0\) for every \(n\ge0\). Then \(\int_0^1 f p=0\) for every polynomial \(p\). By Weierstrass, choose polynomials \(p_k\to f\) uniformly. Since \(f\in L^1\), \[ \int_0^1 f(x)^2\,dx =\lim_{k\to\infty}\int_0^1 f(x)p_k(x)\,dx=0. \] Hence \(f=0\) almost everywhere, and continuity gives \(f\equiv0\).

Taylor’s theorem applied from \(x\) to \(x+2h\) gives \[ f(x+2h)=f(x)+2hf'(x)+2h^2f''(\xi) \] for some \(\xi\) between \(x\) and \(x+2h\), hence \[ f'(x)=\frac{f(x+2h)-f(x)}{2h}-h f''(\xi). \] If \(f(x)\to0\) as \(x\to\infty\) and \(|f''|\le M\), fix \(\varepsilon>0\) and choose \(h>0\) so that \(Mh<\varepsilon/2\). For \(x\) large, both \(|f(x)|\) and \(|f(x+2h)|\) are \(<h\varepsilon/2\), so the first term has absolute value \(<\varepsilon/2\). Thus \(|f'(x)|<\varepsilon\) eventually, proving \(f'(x)\to0\).

Away from the origin the function is continuous. In polar coordinates, \[ \left|\frac{xy}{(x^2+y^2)^2}\right| =\frac{|\sin\theta\cos\theta|}{r^2}. \] On any small disk around the origin, \[ \iint |f|\,dx\,dy \ge c\int_0^\delta \frac{dr}{r}=\infty \] for a positive angular constant \(c\). Hence \(f\notin L^1([-1,1]^2)\); in particular it is not Lebesgue integrable.

If \(f\) is continuous at \(x_0\) and \(x_n\to x_0\), the epsilon-delta definition immediately gives \(f(x_n)\to f(x_0)\). Conversely, if continuity fails, there is \(\varepsilon_0>0\) such that for every \(n\) one can choose \(x_n\) with \(d(x_n,x_0)<1/n\) but \(\rho(f(x_n),f(x_0))\ge\varepsilon_0\). Then \(x_n\to x_0\) while \(f(x_n)\not\to f(x_0)\), contradicting the sequential condition.

On \((1,\infty)\), \[ \left|\frac1x-\frac1y\right|=\frac{|x-y|}{xy}\le |x-y|, \] so \(1/x\) is Lipschitz and therefore uniformly continuous. On \((0,\infty)\) take \(x_n=1/n\) and \(y_n=1/(n+1)\). Then \(|x_n-y_n|\to0\), but \[ \left|\frac1{x_n}-\frac1{y_n}\right|=1, \] so uniform continuity fails.

The supports are pairwise disjoint, so at each \(x\) at most one term is nonzero. Therefore the tail satisfies \[ \sup_x\left|\sum_{n>N}f_n(x)\right|\le\frac1{N+1}\to0, \] so the series converges uniformly. But \(\|f_n\|_\infty=1/n\) and \(\sum_n1/n=\infty\), so the M-test does not apply.

For fixed \(x<1\), \[ 0\le \frac{x^k}{1+x^k}\le x^k, \] so the series converges. On every \([0,a]\) with \(a<1\), the bound by \(a^k\) gives uniform convergence by the M-test. Moreover \[ \left(\frac{x^k}{1+x^k}\right)'=\frac{kx^{k-1}}{(1+x^k)^2}, \] and on \([0,a]\) this is bounded by \(ka^{k-1}\), whose series converges. Hence termwise differentiation is valid locally uniformly, and the sum is differentiable on \((0,1)\).

Given \(\varepsilon>0\), choose \(N\) with \(1/N<\varepsilon\). Only the finitely many points \(r_1,\dots,r_N\) can have value at least \(\varepsilon\). For a fixed \(c\), choose a punctured neighborhood of \(c\) avoiding those finitely many points other than possibly \(c\) itself. Then \(|f(x)|<\varepsilon\) throughout that punctured neighborhood, so \[ \lim_{x\to c}f(x)=0. \] Thus \(f\) is continuous exactly at the irrational points and discontinuous exactly at the rational points.

Uniform convergence on \((0,\infty)\) fails because a necessary condition would be \[ \sup_{x>0}\frac1{1+n^2x}\to0, \] but the supremum is \(1\) for every \(n\). On every \([a,\infty)\), \(a>0\), \[ \frac1{1+n^2x}\le\frac1{n^2a}, \] so the series converges uniformly there and \(g\) is continuous. Also \[ g'(x)=-\sum_{n\ge1}\frac{n^2}{(1+n^2x)^2}. \] On \([a,\infty)\) the derivative terms are bounded by \(1/(n^2a^2)\), so the derivative series converges uniformly. Hence \(g\) is differentiable on \((0,\infty)\) and \(g'\) is continuous there.

For \(n\ge1\), differentiating shows the maximum on \([0,\infty)\) occurs at \(x=1/n\), with \[ \|h_n\|_\infty=\frac{n^n}{(n+1)^{n+1}}\le\frac1{n+1}\to0. \] Thus \(h_n\to0\) uniformly. For \(x>0\), \[ \sum_{n=0}^\infty h_n(x) =\frac{x}{1+x}\sum_{n=0}^\infty(1+x)^{-n}=1, \] while at \(x=0\) the sum is \(0\). Hence the series cannot converge uniformly on \([0,\infty)\) because its pointwise sum is discontinuous. On \([a,\infty)\) with \(a>0\), the tail beginning at \(N\) equals \((1+x)^{-N}\), so its supremum is \((1+a)^{-N}\to0\); therefore convergence is uniform there.

Repeated differentiation on \(x>0\) gives \(f^{(n)}(x)=P_n(1/x)e^{-1/x^2}\) for a polynomial \(P_n\). Since \(t^m e^{-t^2}\to0\) as \(t\to\infty\) for every \(m\), all right derivatives tend to \(0\) at the origin, matching the left derivatives. Hence \(f\in C^\infty\) and \(f^{(n)}(0)=0\) for every \(n\). Its Taylor series at \(0\) is therefore identically zero, so it does not equal \(f\) for \(x>0\). Finally \(g(x)=f(x)f(1-x)\) is smooth, nonzero on \((0,1)\), and vanishes outside \((0,1)\).

If \(|f(x)-f(y)|\le C|x-y|\), then for pairwise disjoint intervals \((a_j,b_j)\), \[ \sum_j|f(b_j)-f(a_j)|\le C\sum_j(b_j-a_j), \] so \(f\) is absolutely continuous. If \(f\) is absolutely continuous, then \[ f(y)-f(x)=\int_x^y f'(t)\,dt. \] Thus \(f'\in L^\infty\) implies \[ |f(y)-f(x)|\le\|f'\|_\infty|y-x|, \] so \(f\) is Lipschitz. Conversely, a Lipschitz function is differentiable a.e. and \(|f'|\) is bounded a.e. by its Lipschitz constant.

For nonnegative integrable \(f\), \[ f(x)^{1/n}\to\mathbf1_{\{f>0\}}(x) \] pointwise. Also \(t^{1/n}\le1+t\) for \(t\ge0\), so \(f^{1/n}\le1+f\in L^1([0,1])\). Dominated convergence yields \[ \lim_{n\to\infty}\int_0^1 f(x)^{1/n}\,dx =m(\{x:f(x)>0\}). \]

1. If \(s_n\to s\), then \(\sigma_n\to s\) by the Cesaro theorem, proved directly by splitting the average into a fixed initial block and a tail where \(|s_k-s|\) is small.

2. Take \(s_n=(-1)^n\). Then \(s_n\) does not converge, while \(\sigma_n\to0\).

3. Let \(s_{2^k}=k\) and \(s_n=0\) otherwise. Then \(s_n\ge0\) and \(\limsup s_n=\infty\), but for \(2^m\le n<2^{m+1}\), \[ 0\le\sigma_n\le\frac{1+2+\cdots+m}{2^m}\to0. \]

4. With \(a_n=s_n-s_{n-1}\), \[ \sum_{k=1}^nka_k =ns_n-\sum_{k=0}^{n-1}s_k =(n+1)(s_n-\sigma_n), \] so \[ s_n-\sigma_n=\frac1{n+1}\sum_{k=1}^nka_k. \] If \(na_n\to0\) and \(\sigma_n\) converges, the right side tends to \(0\) by Cesaro averaging, hence \(s_n\) converges to the same limit as \(\sigma_n\).

Let \(f\in C([0,1])\). Since \(\int_0^1kx^{k-1}\,dx=1\), \[ k\int_0^1x^{k-1}f(x)\,dx-f(1) =k\int_0^1x^{k-1}(f(x)-f(1))\,dx. \] Given \(\varepsilon>0\), choose \(a<1\) so that \(|f(x)-f(1)|<\varepsilon\) on \([a,1]\). The contribution from \([a,1]\) is at most \(\varepsilon\). On \([0,a]\) it is at most \(2\|f\|_\infty a^k\). Hence the expression tends to \(0\), proving the limit is \(f(1)\).

For each fixed \(x\in(0,1]\), exponential decay of \((1-x)^n\) dominates the factor \(n\), while \(f_n(0)=0\), so \(f_n\to0\) pointwise. The maximum occurs at \(x=1/(n+1)\) and equals \[ \frac{n}{n+1}\left(\frac{n}{n+1}\right)^n\to e^{-1}, \] so convergence is not uniform. Finally, since \(0\le\sin x\le x\) on \([0,1]\), \[ 0\le \int_0^1 n(1-x)^n\sin x\,dx \le n\int_0^1x(1-x)^n\,dx =\frac{n}{(n+1)(n+2)}\to0. \]

Let \((x_n)\) be decreasing, positive, and suppose \(\sum_nx_n<\infty\). Then \[ \sum_{k=\lfloor n/2\rfloor+1}^{n}x_k \ge \bigl(n-\lfloor n/2\rfloor\bigr)x_n \ge \frac n2x_n. \] The left side is a tail block of a convergent positive series and therefore tends to \(0\). Hence \(nx_n\to0\).