Sequences of functions
A sequence \(f_n\colon X\to\RR\) converges pointwise to \(f\) if for every \(x\) and \(\varepsilon>0\) there is \(N\), depending on \(x\) and \(\varepsilon\), with \(\abs{f_n(x)-f(x)}<\varepsilon\) for \(n\geq N\); it converges uniformly if \(N\) can be chosen independently of \(x\). A uniform limit of continuous functions is continuous, and on a space of finite measure a uniform limit of integrable functions has integral equal to the limit of the integrals.
Limit superior of a function at a point (Tag D-S2YWR)
Let \(A\subseteq\RR^n\), let \(f\colon A\to \RR\), and let \(y\in\RR^n\) be a limit point of \(A\). The limit superior of \(f\) at \(y\) is \[ \limsup_{x\to y} f(x) \coloneqq \lim_{\varepsilon\to 0^+} \sup f\qty{A \cap B_\varepsilon(y) \setminus\theset{y}} \in [-\infty,\infty], \] where \(B_\varepsilon(y)\) is the open ball of radius \(\varepsilon\) about \(y\); the limit exists because the supremum is nonincreasing as \(\varepsilon\) decreases.
(A continuous sum of a series that does not converge uniformly)
On \((0,1]\), let \[ g(x) \coloneqq \sum_{n=1}^{\infty} {1 \over 1 + n^2 x}. \] For every \(a>0\) the series converges uniformly on \([a,1]\) by comparison with \(\sum_{n\ge1}(n^2a)^{-1}\), so \(g\) is continuous on \((0,1]\). The convergence is not uniform on \((0,1]\): the \(n\)th summand at \(x=1/n^2\) equals \(1/2\), so the summands do not converge uniformly to zero.
Sequences of numbers
\(\limsup_n a_n\) is the largest limit of a convergent subsequence of \((a_n)\) in \([-\infty,\infty]\), and \(\liminf_n a_n\) is the smallest.
The Cauchy condensation test (Tag PR-4EVYE)
Let \((a_k)_{k\geq1}\) be a nonincreasing sequence of nonnegative real numbers. Then \[ \sum_{k\geq 1} a_k < \infty \iff \sum_{k\geq 1} 2^k a_{2^k}<\infty . \]
Group the terms in dyadic blocks \(2^k\le j<2^{k+1}\). Since \((a_j)\) is nonincreasing, each block satisfies \[ 2^k a_{2^{k+1}}\le\sum_{j=2^k}^{2^{k+1}-1}a_j\le 2^k a_{2^k}. \] Summing over \(0\le k\le K\) gives \[ \frac12\sum_{k=1}^{K+1}2^{k}a_{2^{k}}\le\sum_{j=1}^{2^{K+1}-1}a_j\le\sum_{k=0}^{K}2^k a_{2^k}. \] Both series have nonnegative terms, so their partial sums are nondecreasing, and these bounds show that one sequence of partial sums is bounded if and only if the other is; the term \(a_1\) omitted from \(\sum_{k\geq1}2^ka_{2^k}\) does not affect convergence.
(Cauchy condensation test)
Let \(a_1\geq a_2\geq\cdots\geq0\). The block \(2^k\leq n<2^{k+1}\) has \(2^k\) terms, each at most \(a_{2^k}\), so \(\sum_{n\geq1}a_n\leq\sum_{k\geq0}2^ka_{2^k}\). The block \(2^{k-1}<n\leq2^k\) has \(2^{k-1}\) terms, each at least \(a_{2^k}\), so \(2^{k-1}a_{2^k}\leq\sum_{2^{k-1}<n\leq2^k}a_n\) and \[ \sum_{k\geq0}2^ka_{2^k} \leq a_1 + 2\sum_{n\geq2}a_n. \] Hence \(\sum_n a_n\) and \(\sum_k 2^ka_{2^k}\) converge or diverge together.
Growth rates of common functions (Tag FF-QLRXX)
For real sequences \(a(n)\), \(b(n)\) that are positive for all sufficiently large \(n\), write \(a\prec b\) if \(a(n)/b(n)\to 0\) as \(n\to\infty\). For every real \(c>1\), \[ 1 \prec \log(\log n) \prec \log n \prec n \prec n\log n \prec n^{c} \prec c^n \prec n!. \]
Limit definition of exponential function (Tag FD-D2QPH)
The exponential function \(\RR\to\RR\), \(x\mapsto e^x\), is given by \[ e^x \coloneqq \lim_{n \to \infty} \qty{1 + \frac{x}{n}}^n, \] the limit existing for every \(x\in\RR\).
Series
A series \(\sum_n a_n\) of real numbers converges if and only if for every \(\varepsilon>0\) there is \(N\) with \(\abs{\sum_{n=M}^{M'}a_n}<\varepsilon\) for all \(M'\geq M\geq N\) (the Cauchy criterion). For \(a_n\geq0\), the comparison test gives convergence of \(\sum_n a_n\) from \(a_n\leq b_n\) with \(\sum_n b_n<\infty\), and the \(p\)-test gives convergence of \(\sum_n n^{-p}\) exactly for \(p>1\); a Taylor series of \(f\) converges to \(f\) at \(x\) exactly when the remainder in Taylor’s theorem tends to \(0\) at \(x\). For series of functions, the Cauchy criterion in the norm \(\norm{\wait}_\infty\) characterizes uniform convergence.
Comparison test for series with nonnegative terms (Tag PR-P6NHI)
Let \((a_n)_{n\geq1}\) and \((b_n)_{n\geq1}\) be real sequences with \(0\leq a_n \leq b_n\) for all \(n\).
If \(\sum_n b_n < \infty\), then \(\sum_n a_n < \infty\).
If \(\sum_n a_n = \infty\), then \(\sum_n b_n = \infty\).
The terms of a uniformly convergent series tend to zero uniformly (Tag PR-UJ64S)
Let \(S\) be a set and let \(f_n\colon S\to\CC\) for \(n\geq1\).
If \(\sum_{n\geq1}f_n(x)\) converges for every \(x\in S\), then \(f_n\to0\) pointwise on \(S\).
If \(\sum_{n\geq1}f_n\) converges uniformly on \(S\), then \(f_n\to0\) uniformly on \(S\).
Pointwise convergence of the series does not give uniform convergence of the terms: on \(S\coloneqq[0,1)\), \(\sum_{n\geq0}x^n\) converges for every \(x\in S\), but \(\sup_{x\in S}\abs{x^n}=1\) for all \(n\).
Uniform limits of series of continuous functions are continuous (Tag C-3S4XS)
Let \((X,d)\) be a metric space and let \(f_n\colon X\to\CC\) be continuous for every \(n\geq 1\). If the partial sums \(S_N\coloneqq\sum_{n=1}^N f_n\) converge uniformly on \(X\) to a function \(f\colon X\to\CC\), then \(f\) is continuous.
Each \(S_N\) is a finite sum of continuous functions, hence continuous. Fix \(x_0\in X\) and \(\varepsilon>0\). Choose \(N\) with \(\abs{S_N(x)-f(x)}<\varepsilon/3\) for all \(x\in X\), and then \(\delta>0\) with \(\abs{S_N(x)-S_N(x_0)}<\varepsilon/3\) whenever \(d(x,x_0)<\delta\). For such \(x\), \[ \abs{f(x)-f(x_0)}\leq\abs{f(x)-S_N(x)}+\abs{S_N(x)-S_N(x_0)}+\abs{S_N(x_0)-f(x_0)}<\varepsilon. \]
Uniform Cauchy criterion for sequences and series of functions (Tag PR-GT5RS)
Let \(S\) be a set, let \(f_n\colon S\to\CC\) for \(n\geq1\), and for \(h\colon S\to\CC\) put \(\norm{h}_\infty\coloneqq\sup_{x\in S}\abs{h(x)}\). The sequence \((f_n)\) converges uniformly on \(S\) to some function \(f\colon S\to\CC\) if and only if it is uniformly Cauchy: \[ \forall\varepsilon>0\ \exists N\ \forall m,n\geq N:\quad \norm{f_n-f_m}_\infty<\varepsilon . \] In particular, a series \(\sum_{n\geq1}f_n\) converges uniformly on \(S\) if and only if \[ \forall\varepsilon>0\ \exists N\ \forall n>m\geq N:\quad \sup_{x\in S}\abs{\sum_{k=m+1}^{n}f_k(x)}<\varepsilon . \]
A Taylor series converges to $f$ when $\abs{f^{(n)}}\leq M^n$ (Tag PR-H4CYN)
Let \(c\in\RR\), \(\varepsilon>0\), and \(f\in C^\infty((c-\varepsilon,c+\varepsilon))\) with values in \(\RR\). Suppose there exists \(M\geq0\) such that \[ \abs{f^{(n)}(x)} \leq M^n \quad\text{for all } n\geq0 \text{ and all } x\in(c-\varepsilon,c+\varepsilon). \] Then for every \(x\in(c-\varepsilon,c+\varepsilon)\), \[ f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n . \]
$p$-tests for series and integrals (Tag PR-6OHTJ)
Let \(p\in\RR\), let \(\varepsilon>0\), let \(n\geq1\), and let \(B \coloneqq \theset{x\in \RR^n \suchthat \abs{x} \leq 1}\) be the closed unit ball for the Euclidean norm. Then \[ \begin{aligned} \sum_{k=1}^\infty \frac 1 {k^p} < \infty &\iff p>1, \\ \int_\varepsilon^\infty \frac {dx} {x^p} < \infty &\iff p>1, \\ \int_0^1 \frac {dx} {x^p} < \infty &\iff p<1, \\ \int_B \frac{dx}{\abs{x}^p} < \infty &\iff p < n, \\ \int_{\RR^n\setminus B} \frac{dx}{\abs{x}^p} < \infty &\iff p > n . \end{aligned} \]
The terms of a uniformly convergent series tend to zero uniformly (Tag PR-LEDI3)
Let \(S\) be a set and let \(f_n\colon S\to\CC\) for \(n\geq1\). If the series \(\sum_{n\geq1}f_n\) converges uniformly on \(S\), then \[ \sup_{x\in S}\abs{f_n(x)}\convergesto{n\to\infty}0 . \]
Uniform limits of series of continuous functions are continuous (Tag C-VSE32)
Let \((X,d)\) be a metric space, let \(f_n\colon X\to\CC\) be continuous for every \(n\geq 1\), and let \(F\colon X\to\CC\). If \[ \norm{\sum_{n=1}^N f_n - F}_\infty \coloneqq \sup_{x\in X}\abs{\sum_{n=1}^N f_n(x) - F(x)} \xrightarrow{N\to\infty} 0, \] then \(F\) is continuous.
The partial sums \(S_N\coloneqq\sum_{n=1}^N f_n\) are continuous and converge uniformly to \(F\). Fix \(x_0\in X\) and \(\varepsilon>0\), choose \(N\) with \(\norm{S_N-F}_\infty<\varepsilon/3\), and choose \(\delta>0\) with \(\abs{S_N(x)-S_N(x_0)}<\varepsilon/3\) whenever \(d(x,x_0)<\delta\). For such \(x\), the triangle inequality gives \(\abs{F(x)-F(x_0)}<\varepsilon\).
Taylor remainders in Lagrange, Cauchy, and integral form (Tag T-2R7PC)
Let \(n\geq1\), \(p\in\RR\), \(\delta>0\), and let \(f\colon(p-\delta,p+\delta)\to\RR\) be \(n\) times differentiable. For \(x\in(p-\delta,p+\delta)\) put \[ R_n(x)\coloneqq f(x)-\sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k . \] Let \(x\in(p-\delta,p+\delta)\) with \(x\neq p\).
(Lagrange form) There exists \(\xi\) strictly between \(p\) and \(x\) such that \[ R_n(x)=\frac{f^{(n)}(\xi)}{n!}(x-p)^n . \]
(Cauchy form) There exists \(\eta\) strictly between \(p\) and \(x\) such that \[ R_n(x)=\frac{f^{(n)}(\eta)}{(n-1)!}(x-\eta)^{n-1}(x-p) . \]
(Integral form) If moreover \(f^{(n)}\) is continuous on \((p-\delta,p+\delta)\), then \[ R_n(x)=\int_p^x \frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}\,dt . \]
Part 1 is in [@Rud76].
Uniform convergence
For bounded \(f_n - f\), \[ f_n\to f \text{ uniformly} \quad\Longleftrightarrow\quad \norm{f_n-f}_\infty\to 0. \] Uniform convergence is proved by bounding \(\sup_x\abs{f_n(x)-f(x)}\) by a quantity independent of \(x\) tending to \(0\), and disproved by exhibiting \(x_n\) and \(\varepsilon>0\) with \(\abs{f_n(x_n)-f(x_n)}\geq\varepsilon\) for infinitely many \(n\). For a series, the Weierstrass \(M\)-test reduces uniform convergence to convergence of a numerical series.
Sup-norm test for uniform convergence (Tag PR-WUZSG)
Let \(S\) be a set and let \(f_n,f\colon S\to\CC\) for \(n\geq1\). Then \(f_n\to f\) uniformly on \(S\) if and only if there are real numbers \(M_n\) with \[ \sup_{x\in S}\abs{f_n(x) - f(x)} \leq M_n \text{ for all } n \quad\text{and}\quad M_n\convergesto{n\to\infty}0 . \]
Implications among uniform, pointwise, almost everywhere, and $L^p$ convergence (Tag FF-I6VGK)
Let \((X,\mcm,\mu)\) be a measure space, let \(1\leq p<\infty\), and let \(f_n, f\colon X\to\CC\) be measurable.
If \(f_n\to f\) uniformly on \(X\), then \(f_n\to f\) pointwise on \(X\); if \(f_n\to f\) pointwise on \(X\), then \(f_n\to f\) almost everywhere.
If \(\mu(X)<\infty\), \(f_n, f\in L^p(\mu)\), and \(f_n\to f\) uniformly on \(X\), then \(f_n\to f\) in \(L^p(\mu)\).
If \(f_n\to f\) in \(L^p(\mu)\), then some subsequence \((f_{n_k})\) converges to \(f\) almost everywhere.
Without \(\mu(X)<\infty\), uniform convergence does not imply \(L^1\) convergence: \(\frac1n\chi_{[0,n]}\to 0\) uniformly on \(\RR\) with \(\int_\RR\frac1n\chi_{[0,n]} = 1\) (FE-3RPMC). Almost everywhere convergence does not imply \(L^1\) convergence: \(n\chi_{[0,1/n]}\to 0\) almost everywhere on \(\RR\) with integral \(1\) (FE-VT5N3).
\(L^1\) convergence does not imply almost everywhere convergence. Write each \(n\geq 1\) uniquely as \(n = 2^k + j\) with \(k\geq 0\) and \(0\leq j<2^k\), and let \(f_n\coloneqq\chi_{[j2^{-k},\,(j+1)2^{-k}]}\) on \([0,1]\). Then \(\norm{f_n}_1 = 2^{-k}\to 0\), but for every \(x\in[0,1]\) and every \(k\) some \(f_n\) with \(2^k\leq n<2^{k+1}\) has \(f_n(x) = 1\), and for \(k\geq 2\) another has \(f_n(x) = 0\), so \((f_n(x))\) diverges at every \(x\in[0,1]\).
Showing uniform convergence of a sequence of functions (Tag FS-FZL2X)
Let \(S\) be a set, and let \(f_n\colon S\to\CC\) for \(n\geq1\) and \(f\colon S\to\CC\) be functions. Find constants \(M_n\), independent of \(x\), such that \[ \abs{f_n(x) - f(x)} \leq M_n \quad\text{for all } x\in S \text{ and } n\geq1, \qquad M_n \to 0. \] Then \((f_n)\) converges uniformly to \(f\) on \(S\).
Negating uniform convergence of a sequence of functions (Tag FS-THMMW)
Let \(S\) be a set, and let \(f_n\colon S\to\CC\) for \(n\geq1\) and \(f\colon S\to\CC\) be functions. To show that \((f_n)\) does not converge uniformly to \(f\) on \(S\), exhibit \(\varepsilon>0\) and, for infinitely many \(n\), a point \(x_n\in S\) with \(\abs{f_n(x_n) - f(x_n)} \geq \varepsilon\).
On \(S=(0,\infty)\), the functions \(f_n(x)=\frac{1}{1+nx}\) converge pointwise to \(f=0\), but not uniformly: with \(\varepsilon=\frac12\) and \(x_n=\frac1n\), \(\abs{f_n(x_n)-f(x_n)}=\frac{1}{2}\) for every \(n\).
Weierstrass $M$-test for uniform convergence (Tag FF-IAUQG)
Let \(A\) be a set and let \(f_n\colon A\to\CC\) for \(n\geq 0\). Put \(\norm{f_n}_{\infty, A}\coloneqq\sup_{x\in A}\abs{f_n(x)}\). If \(\sum_{n\geq 0}\norm{f_n}_{\infty, A} < \infty\), then for every \(x\in A\) the series \(\sum_{n\geq 0} f_n(x)\) converges absolutely, and its partial sums converge uniformly on \(A\) to \(\sum_{n\geq 0} f_n\).
Equivalently, the conclusion holds whenever there are real numbers \(M_n\) with \(\abs{f_n(x)} \leq M_n\) for all \(x\in A\) and \(n\geq 0\) and \(\sum_{n\geq 0} M_n < \infty\).
For each \(x\), \(\sum_n\abs{f_n(x)}\leq\sum_n M_n<\infty\), so \(S(x)\coloneqq\sum_{n\geq 0}f_n(x)\) exists. For \(x\in A\) and \(N\geq 0\), \(\abs{S(x) - \sum_{n=0}^{N}f_n(x)}\leq\sum_{n>N}M_n\), which tends to \(0\) as \(N\to\infty\) independently of \(x\). The first form is the case \(M_n = \norm{f_n}_{\infty,A}\), and conversely \(\norm{f_n}_{\infty,A}\leq M_n\).
Conversely, if \(\sum_n f_n\) converges uniformly on \(A\), then \(\sup_{x\in A}\abs{f_n(x)}\to0\); this necessary condition does not imply \(\sum_n\sup_{x\in A}\abs{f_n(x)}<\infty\), as the constant functions \(f_n\coloneqq(-1)^n/n\) show.
Showing uniform convergence of a series of functions (Tag FS-5FKPD)
Use the Weierstrass \(M\)-test. Let \(S\) be a set and \(f_n\colon S\to\CC\) for \(n\geq1\). Find constants \(M_n\geq0\), independent of \(x\), such that \(\sup_{x\in S}\abs{f_n(x)} \le M_n\) for every \(n\) and \(\sum_{n\geq1} M_n < \infty\). Then \(\sum_{n\geq1} f_n\) converges absolutely at every point of \(S\), and its partial sums converge uniformly on \(S\), since for \(N>N'\) and all \(x\in S\), \[ \abs{\sum_{n=N'+1}^{N} f_n(x)} \le \sum_{n>N'} M_n \xrightarrow{N'\to\infty} 0. \]
(Negating uniform convergence)
\(f_n\not\to0\) uniformly if and only if there are \(\varepsilon>0\), infinitely many \(n\), and points \(x_n\) with \(\abs{f_n(x_n)}\geq\varepsilon\). \((f_n)\) is not uniformly Cauchy if and only if there are \(\varepsilon>0\) and, for every \(N\), indices \(m,n\geq N\) and a point \(x\) with \(\abs{f_n(x)-f_m(x)}\geq\varepsilon\).
$C([0,1])$ with the sup norm is complete (Tag PR-RWROV)
Let \(C([0,1])\) be the space of continuous functions \(f\colon[0,1]\to\RR\) with the norm \[ \norm{f}_\infty \coloneqq \sup_{x\in [0, 1]} \abs{f(x)} . \] Then \(C([0,1])\) with the metric \(d(f,g)\coloneqq\norm{f-g}_\infty\) is a complete metric space.
Let \((f_k)\) be Cauchy in \(C([0,1])\) with the norm \(\norm{\wait}_\infty\). For each \(x\), \(\abs{f_k(x) - f_j(x)} \leq \norm{f_k - f_j}_\infty\), so \((f_k(x))\) is Cauchy in \(\RR\); let \(f(x) \coloneqq \lim_k f_k(x)\).
Given \(\varepsilon>0\), choose \(N\) with \(\norm{f_k-f_j}_\infty<\varepsilon\) for \(j,k\geq N\). For fixed \(k\geq N\) and every \(x\), letting \(j\to\infty\) in \(\abs{f_k(x)-f_j(x)}<\varepsilon\) gives \(\abs{f_k(x)-f(x)}\leq\varepsilon\), so \(\norm{f_k-f}_\infty\leq\varepsilon\). Thus \(f_k\to f\) uniformly, and \(f\) is continuous as a uniform limit of continuous functions, so \(f\in C([0,1])\).
The same argument proves completeness of other spaces of functions with the norm \(\norm{\wait}_\infty\), such as the bounded functions on a set or \(C_b(X)\) for a metric space \(X\), provided the uniform limit is shown to lie in the space.
Weierstrass approximation theorem (Tag T-3QNBQ)
Let \(a<b\) and let \(f\colon[a,b]\to\CC\) be continuous. For every \(\varepsilon>0\) there exists a polynomial \(p\) such that \[ \sup_{x\in[a,b]}\abs{f(x)-p(x)}<\varepsilon . \] If \(f\) is real-valued, \(p\) may be taken with real coefficients. Equivalently, the polynomials are dense in the Banach space \(C([a,b])\) with the norm \(\norm{g}_\infty\coloneqq\sup_{x\in[a,b]}\abs{g(x)}\).