Sequences and series of functions

A sequence $f_n\colon E\to\RR$ converges uniformly to $f$ if $\sup_{x\in E}\abs{f_n(x)-f(x)}\to0$. A uniform limit of continuous functions is continuous, and if Riemann integrable $f_n$ converge uniformly to $f$ on $[a,b]$, then $f$ is Riemann integrable and $\int_a^bf_n\to\int_a^bf$. If the $f_n$ are differentiable on $[a,b]$, $(f_n(x_0))$ converges for some $x_0\in[a,b]$, and $(f_n')$ converges uniformly, then $(f_n)$ converges uniformly to a differentiable $f$ with $f'=\lim_nf_n'$. The Weierstrass M-test: if $\abs{f_n}\le M_n$ on $E$ and $\sum_nM_n<\infty$, then $\sum_nf_n$ converges uniformly on $E$. The Arzelà–Ascoli theorem: for a compact metric space $K$, a subset of $C(K)$ has compact closure if and only if it is bounded and equicontinuous. The Stone–Weierstrass theorem: a subalgebra of $C(K,\RR)$ that contains the constants and separates points is dense for the supremum norm. On $[0,1]$, $x^n$ converges pointwise and not uniformly; $n\chi_{(0,1/n)}$ converges to $0$ at every point and not in $L^1$; and the typewriter sequence of indicator functions of dyadic intervals converges to $0$ in $L^1$ and at no point.

Pointwise and uniform convergence of function sequences and series (Tag D-RA-WORKSHOP-D7-CONVERGENCE)

Let $E$ be a set, and let $f_n\colon E\to\RR$ for $n\geq 1$ and $f\colon E\to\RR$ be functions.

$f_n\to f$ pointwise on $E$ if $\lim_{n\to\infty}f_n(x)=f(x)$ for each $x\in E$.
$f_n\to f$ uniformly on $E$ if for every $\varepsilon>0$ there exists $N\in\NN$ such that $\abs{f_n(x)-f(x)}<\varepsilon$ for all $n\geq N$ and all $x\in E$.
$\sum_{n=1}^{\infty}f_n=f$ pointwise on $E$ if the partial sums $\sum_{n=1}^{N}f_n$ converge pointwise to $f$ as $N\to\infty$.
$\sum_{n=1}^{\infty}f_n=f$ uniformly on $E$ if the partial sums $\sum_{n=1}^{N}f_n$ converge uniformly to $f$ as $N\to\infty$.

With the supremum norm $\norm{g}_\infty\coloneqq\sup_{x\in E}\abs{g(x)}$ for $g\colon E\to\RR$, condition (2) holds if and only if $\norm{f_n-f}_\infty\to 0$.

Uniform limits of continuous functions are continuous (Tag T-RA-WORKSHOP-D7-6-1)

Uniform convergence preserves continuity.Suggest a slogan

Let $E$ be a metric space, and let $f_n\colon E\to\RR$ for $n\geq1$ and $f\colon E\to\RR$ be functions. If each $f_n$ is continuous on $E$ and $f_n\to f$ uniformly on $E$, then $f$ is continuous on $E$. [@Rud76].

Uniform limit theorem (Tag T-ERNLN)

Uniform convergence preserves continuity.Suggest a slogan

Let $X$ be a topological space, and let $f_n\colon X\to\CC$ for $n\geq1$ and $f\colon X\to\CC$ be functions. If each $f_n$ is continuous and $f_n\to f$ uniformly on $X$, then $f$ is continuous.

Weierstrass M-test (Tag T-RA-WORKSHOP-D7-6-2)

A uniformly summable numerical majorant forces uniform convergence of the function series.Suggest a slogan

Let $E$ be a set, let $f_n\colon E\to\RR$ for $n\geq1$, and let $(M_n)_{n\geq1}$ be a sequence of real numbers such that $\abs{f_n(x)}\leq M_n$ for all $n\geq1$ and all $x\in E$. If $\sum_{n=1}^{\infty}M_n$ converges, then $\sum_{n=1}^{\infty}f_n$ converges uniformly on $E$. [@Rud76].

Weierstrass $M$-test for uniform convergence (Tag FF-IAUQG)

A summable uniform majorant forces absolute and uniform convergence of a function series.Suggest a slogan

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.

What is the $M{\hbox{-}}$test?

Uniform limits of Riemann--Stieltjes integrable functions (Tag T-RA-WORKSHOP-D7-6-3)

Uniform convergence preserves Riemann--Stieltjes integrability and commutes with the integral.Suggest a slogan

Let $a<b$, let $\alpha\colon[a,b]\to\RR$ be monotonically increasing, and let $\mathcal R(\alpha)$ denote the set of bounded functions $[a,b]\to\RR$ that are Riemann–Stieltjes integrable with respect to $\alpha$ on $[a,b]$. Let $f_n\in\mathcal R(\alpha)$ for $n\geq1$, and let $f\colon[a,b]\to\RR$. If $f_n\to f$ uniformly on $[a,b]$, then $f\in\mathcal R(\alpha)$, the limit $\lim_{n\to\infty}\int_a^b f_n\,d\alpha$ exists, and \[ \lim_{n\to\infty}\int_a^b f_n\,d\alpha=\int_a^b f\,d\alpha. \] [@Rud76].

Uniform convergence of derivatives and differentiability of the limit (Tag T-RA-WORKSHOP-D7-6-4)

If $(f_n^\prime)$ converges uniformly and $(f_n(x_0))$ converges, then $f_n\to f$ uniformly, $f$ is differentiable, and $f^\prime=\lim f_n^\prime$.Suggest a slogan

Let $a<b$, and let $f_n\colon[a,b]\to\RR$ for $n\geq1$ be differentiable on $[a,b]$. Suppose that $\{f_n(x_0)\}$ converges for some $x_0\in[a,b]$ and that $\{f_n'\}$ converges uniformly on $[a,b]$. Then $\{f_n\}$ converges uniformly on $[a,b]$ to a function $f\colon[a,b]\to\RR$, $f$ is differentiable on $[a,b]$, and \[ f'(x)=\lim_{n\to\infty}f_n'(x)\qquad\text{for all }x\in[a,b]. \] [@Rud76].

Term-by-term differentiation of a series of functions (Tag T-TR526)

Uniform convergence of the derivative series plus convergence at one point permits term-by-term differentiation.Suggest a slogan

Let $a<b$, let $A\coloneqq[a,b]$, and let $f_n\colon A\to\RR$ for $n\geq1$ be differentiable on $A$. Suppose that $\sum_{n=1}^\infty f_n'$ converges uniformly on $A$ to a function $g\colon A\to\RR$, and that $\sum_{n=1}^\infty f_n(x_0)$ converges for some $x_0\in A$. Then $\sum_{n=1}^\infty f_n$ converges uniformly on $A$ to a function $f\colon A\to\RR$, $f$ is differentiable on $A$, and $f'=g$ on $A$ [@Abb16].

Equicontinuity at a point (Tag D-2CCDB)

Let $X$ be a topological space, let $x\in X$, and let $\mathcal F\subseteq C(X)$ be a family of continuous functions $X\to\CC$. The family $\mathcal F$ is equicontinuous at $x$ if \[ \forall \varepsilon > 0 \quad \exists \text{ an open set } U \ni x \quad \forall y\in U \quad \forall f\in \mathcal{F}: \quad \abs{f(y) - f(x)} < \varepsilon. \]

Arzelà--Ascoli theorem for $C(K,\RR^m)$ (Tag T-RA-WORKSHOP-D7-6-5)

In $C(K,\RR^m)$, compactness is exactly closedness, boundedness, and equicontinuity.Suggest a slogan

Let $n,m\geq1$, let $K\subseteq\RR^n$ be compact, and equip $C(K,\RR^m)$ with the metric $d(f,g)\coloneqq\sup_{x\in K}\norm{f(x)-g(x)}$. A subset $\mathcal F\subseteq C(K,\RR^m)$ is compact if and only if $\mathcal F$ is closed, bounded, and equicontinuous at every point of $K$, that is, \[ \forall x\in K\quad\forall\varepsilon>0\quad\exists\delta>0\quad\forall y\in K\quad\forall f\in\mathcal F:\quad\norm{x-y}<\delta\implies\norm{f(x)-f(y)}<\varepsilon. \] [@Mun00].

Arzelà--Ascoli theorem (Tag FF-XZGIY)

For $K$ compact metric, a family in $C(K)$ has compact closure in the sup norm exactly when it is uniformly bounded and equicontinuous.Suggest a slogan

Let $K$ be a compact metric space, let $C(K)$ be the space of continuous functions $K\to\CC$ with the norm $\norm{f}_\infty\coloneqq\sup_{x\in K}\abs{f(x)}$, and let $\mcf\subseteq C(K)$. Then the closure of $\mcf$ in $C(K)$ is compact if and only if $\mcf$ is uniformly bounded, that is, $\sup_{f\in\mcf}\norm{f}_\infty<\infty$, and equicontinuous at every point of $K$.

In particular, every sequence $(f_n)$ in $C(K)$ that is uniformly bounded and equicontinuous at every point of $K$ has a subsequence that converges uniformly on $K$.

Since $K$ is compact, a family $\mcf\subseteq C(K)$ that is equicontinuous at every point of $K$ is uniformly equicontinuous.

What is the Arzela-Ascoli theorem?

State the Arzela-Ascoli theorem.

Weierstrass approximation and Stone--Weierstrass theorems (Tag T-RA-WORKSHOP-D7-6-7)

Polynomials are uniformly dense in $C([a,b])$; a self-adjoint subalgebra of $C(K)$ that separates points and vanishes at no point is uniformly dense in $C(K)$.Suggest a slogan

Let $a<b$ and let $f\colon[a,b]\to\CC$ be continuous. Then there exist polynomials $P_n$ with complex coefficients such that $P_n\to f$ uniformly on $[a,b]$. If $f$ is real-valued, the $P_n$ may be taken with real coefficients.
Let $K$ be a compact metric space, let $C(K)$ be the algebra of continuous functions $K\to\CC$, and let $\mathcal A\subseteq C(K)$ be a subalgebra, that is, a complex vector subspace closed under pointwise multiplication. Suppose that

$\mathcal A$ is self-adjoint: $\bar g\in\mathcal A$ for every $g\in\mathcal A$;
$\mathcal A$ separates points of $K$: for all $x\neq y$ in $K$ there exists $g\in\mathcal A$ with $g(x)\neq g(y)$;
$\mathcal A$ vanishes at no point of $K$: for every $x\in K$ there exists $g\in\mathcal A$ with $g(x)\neq0$.

Then for every $f\in C(K)$ there exist $f_n\in\mathcal A$ such that $f_n\to f$ uniformly on $K$. [@Rud76].

Weierstrass approximation theorem (Tag T-3QNBQ)

Polynomials are uniformly dense in the continuous functions on a compact interval.Suggest a slogan

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 $\norm{g}_\infty\coloneqq\sup_{x\in[a,b]}\abs{g(x)}$.

State the Weierstrass approximation theorem.

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