Stone–Weierstrass in General: Examples and Counterexamples

The reverse triangle inequality derived from [L3] makes $z\mapsto \mathrm{\mid }z\mathrm{\mid }$ continuous, so $D=\{\mathrm{\mid }z\mathrm{\mid }\le 1\}$ is closed; it is bounded because $d_{\mathbb{C}}(z,0)=\mathrm{\mid }z\mathrm{\mid }\le 1$. Thus [L5], [L12], and [L13] make $D$ compact, and [L14] makes its metric topology Hausdorff.

Each $p={\sum }_{j=0}^m{\alpha }_j{z}^j$ in $P$ is continuous: the identity $z^j-w^j=(z-w){\sum }_{k=0}^{j-1}{z}^k{w}^{j-1-k}$ of [L6] together with $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ from [L3] gives $\mathrm{\mid }{z}^j-w^j\mathrm{\mid }\le j\mathrm{\mid }z-w\mathrm{\mid }$ for $z,w\in D$, so $\mathrm{\mid }p(z)-p(w)\mathrm{\mid }\le C_p\mathrm{\mid }z-w\mathrm{\mid }$ with $C_p:={\sum }_{j=1}^{m}j\mathrm{\mid }{\alpha }_j\mathrm{\mid }$, which is continuity for the metric of [L5]; hence [L15] puts every member of $A$ in $C(D,\mathbb{C})$. The set $P$ contains the constants and the coordinate function $z\mapsto z$, which separates points, and is closed under complex linear combinations and products by [L1] and [L4], so $P$ is a unital point-separating complex function algebra and $A\supseteq P$ inherits unitality and point separation. $A$ is a complex vector subspace because approximants add and scale. For products, $\mathrm{\mid }z\mathrm{\mid }\le 1$ and [L3] give $\mathrm{\mid }p(z)\mathrm{\mid }\le M_p:={\sum }_{j=0}^m\mathrm{\mid }{\alpha }_j\mathrm{\mid }$ on $D$ for every $p\in P$, so given $f,g\in A$ and $\eta >0$ one may first fix $b_0\in P$ with $\mathrm{\mid }g-b_0\mathrm{\mid }<1$ everywhere, whence $\mathrm{\mid }g\mathrm{\mid }\le K:=M_{b_0}+1$ on $D$, and then choose $a\in P$ with $\mathrm{\mid }f-a\mathrm{\mid }<\eta \mathrm{/}(2K)$ everywhere and $b\in P$ with $\mathrm{\mid }g-b\mathrm{\mid }<\eta \mathrm{/}(2(M_a+1))$ everywhere; from $ab-fg=a(b-g)+(a-f)g$ and [L3], $\mathrm{\mid }ab-fg\mathrm{\mid }\le M_a\mathrm{\mid }b-g\mathrm{\mid }+K\mathrm{\mid }a-f\mathrm{\mid }<\eta$ pointwise, and $ab\in P$, so $fg\in A$. Finally $A$ is uniformly closed, because a function within $\epsilon \mathrm{/}2$ of a member of $A$ everywhere is within $\epsilon$ of a member of $P$ everywhere.

Suppose for contradiction that $\bar{z}\in A$. Then there is a nonzero polynomial $p(z)={\sum }_{j=0}^m{a}_j{z}^j$ with ${\sup }_{z\in D}\mathrm{\mid }\bar{z}-p(z)\mathrm{\mid }<1$; put $N:=m+2\ge 2$ and $\zeta :=\exp (2\pi i\mathrm{/}N)$.

Repeated use of the addition law [L9], along the induction of [L11] on $k$ with base $\exp 0=1={\zeta }^0$, gives $\exp (2\pi ik\mathrm{/}N)={\zeta }^k$ for every natural $k$; so the list of [L7] is exactly $1,\zeta ,\dots ,{\zeta }^{N-1}$, and these are the $N$th roots of unity. For an integer $r$ with $1\le r<N$ one has ${\zeta }^r=\exp (2\pi ir\mathrm{/}N)$, and [L10] makes this equal to $1=\exp 0$ only when $2\pi ir\mathrm{/}N\in 2\pi i\mathbb{Z}$, that is only when $N$ divides $r$, which fails in that range; hence ${\zeta }^r\mathrm{\ne }1$. The same law gives $({\zeta }^r{)}^N=\exp (2\pi ir)=1$.

For every natural $q\ge 1$ and every $x\in \mathbb{C}$, $(x-1){\sum }_{k=0}^{q-1}{x}^k=x^q-1$. Apply [L11] to the property that this identity holds for $q=n+1$. At $n=0$ the identity reads $(x-1)x^0=x-1$, which is immediate. Assuming it at $n$, adding the term $x^{n+1}$ to the sum changes the left side by $(x-1)x^{n+1}=x^{n+2}-x^{n+1}$, carrying the right side from $x^{n+1}-1$ to $x^{n+2}-1$, which is the identity at $n+1$.

The exponent-one cancellation ${\sum }_{k=0}^{N-1}{\zeta }^k=0$ is [L8]. For $2\le r\le m+1<N$, step 1.5 with $x={\zeta }^r$ and step 1.4 give $({\zeta }^r-1){\sum }_{k=0}^{N-1}{\zeta }^{rk}=0$ with ${\zeta }^r-1\mathrm{\ne }0$, so ${\sum }_{k=0}^{N-1}{\zeta }^{rk}=0$.

Every sampled point lies on the unit circle, so [L3] gives ${\zeta }^k\overline{{\zeta }^k}=1$ and hence $N^{-1}{\sum }_{k=0}^{N-1}{\zeta }^k\overline{{\zeta }^k}=1$.

Expanding $p$ and using step 2.1 for the exponents $j+1\in \{1,\dots ,m+1\}$ gives $N^{-1}{\sum }_{k=0}^{N-1}{\zeta }^{k}p({\zeta }^k)=0$.

Subtracting step 3.1 from step 2.2 and repeatedly applying the triangle inequality in [L3], justified over the finite sum by [L11], yields $1\le N^{-1}{\sum }_{k=0}^{N-1}\mathrm{\mid }\overline{{\zeta }^k}-p({\zeta }^k)\mathrm{\mid }<1$, contradicting step 1.3.

Therefore $\bar{z}\mathrm{\notin }A$. Since the coordinate function $z$ belongs to $A$, the algebra is not self-adjoint by [L1]. The conjugation map is continuous, because $\bar{z}-\bar{w}=\overline{z-w}$ by [L2] and $\mathrm{\mid }\bar{u}{\mathrm{\mid }}^2=\bar{u}u=\mathrm{\mid }u{\mathrm{\mid }}^2$ by [L3], so $\mathrm{\mid }\bar{z}-\bar{w}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$; and $A$ is uniformly closed by step 1.2, so a function uniformly approximable by members of $A$ lies in $A$. Hence $\bar{z}$ is a member of $C(D,\mathbb{C})$ that $A$ cannot approximate uniformly, and $A$ is not dense.

The reverse triangle inequality from [L5] makes $z\mapsto \mathrm{\mid }z\mathrm{\mid }$ continuous, so $\mathbb{T}=\{\mathrm{\mid }z\mathrm{\mid }=1\}$ is closed; it is bounded. Thus [L3], [L8], and [L9] make $\mathbb{T}$ compact, and [L10] makes it Hausdorff.

For $z\in \mathbb{T}$, [L5] gives $z\bar{z}=1$, so uniqueness of the inverse in the field [L6] gives $z^{-1}=\bar{z}$; consequently [L7] gives $z^{-r}=(\bar{z}{)}^r$ for every natural $r$.

Finite Laurent sums are closed under complex linear combinations and products, contain every constant and the coordinate function $z\mapsto z$, and therefore separate points; conjugating such a sum conjugates its coefficients and reverses its exponents by step 1.2, so $T$ is self-adjoint. Every member of $T$ is also continuous, so $T$ is a subalgebra of $C(\mathbb{T},\mathbb{C})$: step 1.2 rewrites a Laurent sum as ${\sum }_{j\ge 0}{a}_j{z}^j+{\sum }_{j<0}{a}_j(\bar{z}{)}^{-j}$ on $\mathbb{T}$; conjugation satisfies $\mathrm{\mid }\bar{z}-\bar{w}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$, because $\bar{z}-\bar{w}=\overline{z-w}$ and $\mathrm{\mid }\bar{u}{\mathrm{\mid }}^2=\bar{u}u=\mathrm{\mid }u{\mathrm{\mid }}^2$ by [L5]; and the identity $u^r-v^r=(u-v){\sum }_{k=0}^{r-1}{u}^k{v}^{r-1-k}$ of [L7] with $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ from [L5] gives $\mathrm{\mid }{u}^r-v^r\mathrm{\mid }\le r\mathrm{\mid }u-v\mathrm{\mid }$ whenever $\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }=1$, so each Laurent sum is Lipschitz for the metric of [L3].

The algebra $T$ is a unital, point-separating, self-adjoint complex function algebra on the compact Hausdorff circle from step 1.1, so [L1] makes it uniformly dense in $C(\mathbb{T},\mathbb{C})$.

By [L4] and [L5], the metric topology makes $X$ a compact Hausdorff space.

For $a,x,y\in X$, the triangle inequality and symmetry in [L3] give $d(a,x)\le d(a,y)+d(x,y)$ and $d(a,y)\le d(a,x)+d(x,y)$; hence $\mathrm{\mid }{d}_a(x)-d_a(y)\mathrm{\mid }\le d(x,y)$, so every $d_a$ is continuous.

The generated lattice $L$ is unital by construction. If $x\mathrm{\ne }y$, then $d_x(x)=0$ while $d_x(y)=d(x,y)\mathrm{\ne }0$ by [L3], so $d_x\in L$ separates $x$ and $y$; thus $L$ is point-separating in the sense of [L2].

Apply [L1] to the unital point-separating real vector sublattice $L$ on the compact Hausdorff space of step 1.1.

If $X=\mathrm{\varnothing }$, then $F^X$ has only the empty function, which is the zero element of $A$. If $X=\{x\}$, every function is constant, so unitality gives $A=F^X$.

Assume $X$ has at least two points and use [L4] to list its points. For a fixed $x\in X$, [L5] supplies, for each listed $y\mathrm{\ne }x$, a nonempty set of open neighbourhoods of $x$ missing $y$; [L3] chooses one from each member of this finite list. Their intersection is the open singleton $\{x\}$. Hence every singleton is open and every function $X\to F$ is continuous.

For every ordered pair $x\mathrm{\ne }y$, point separation in [L1] or [L2] gives $f_{xy}\in A$ with $f_{xy}(x)\mathrm{\ne }{f}_{xy}(y)$; by [L3] and the finite listing in [L4], choose these over the finite list of ordered pairs and define $h_{xy}:=(f_{xy}-f_{xy}(y))\mathrm{/}(f_{xy}(x)-f_{xy}(y))\in A$. Then $h_{xy}(x)=1$ and $h_{xy}(y)=0$.

For each $x\in X$, the finite product $e_x:={\prod }_{y\mathrm{\ne }x}{h}_{xy}$ belongs to $A$, equals $1$ at $x$, and equals $0$ at every other point because the factor indexed by that point vanishes.

For any function $\phi :X\to F$, the finite sum ${\sum }_{x\in X}\phi (x)e_x$ belongs to $A$ and agrees with $\phi$ at every point. By step 1.2 every such $\phi$ is continuous, so $A=C(X,F)=F^X$.

By [L3] and the equivalence in [L4], the subspace $[0,1]$ of $\mathbb{R}$ is a compact topological space. By [L5] and [L6] the line $\mathbb{R}$ is Hausdorff, so [L7] makes the subspace $[0,1]$ Hausdorff.

Endpoint equality is preserved by pointwise sums, real scalar multiples, and products, and every constant has equal endpoint values, so $A$ is a unital real function algebra.

If $g$ is uniformly approximable by members of $A$, then for every $\epsilon >0$ some $f\in A$ satisfies $\mathrm{\mid }g(0)-f(0)\mathrm{\mid }<\epsilon \mathrm{/}2$ and $\mathrm{\mid }g(1)-f(1)\mathrm{\mid }<\epsilon \mathrm{/}2$; since $f(0)=f(1)$, this forces $g(0)=g(1)$. Hence $A$ is uniformly closed.

For $c\in (0,1)$ let $\iota :[0,1]\to \mathbb{R}$ be the inclusion and put $t_c:=\min \{c^{-1}\iota ,(1-c{)}^{-1}(1-\iota )\}$. A constant map is continuous because the preimage of every open set is $\mathrm{\varnothing }$ or all of $[0,1]$, which is the condition in [L10]; $\iota$ is continuous by [L8]; so [L9] makes the two affine maps and their pointwise minimum continuous. For $0\le x\le c$ one has $x\mathrm{/}c\le 1\le (1-x)\mathrm{/}(1-c)$, and for $c\le x\le 1$ the two inequalities reverse, so $t_c(x)=x\mathrm{/}c$ on $[0,c]$ and $t_c(x)=(1-x)\mathrm{/}(1-c)$ on $[c,1]$. Hence $t_c(0)=t_c(1)=0$, so $t_c\in A$; also $t_c(c)=1$, and $t_c(x)>0$ for $0<x<1$, so $t_c$ vanishes only at the two endpoints.

Every member of $A$ identifies $0$ and $1$. Conversely, if $x\mathrm{\ne }y$ and $\{x,y\}\mathrm{\ne }\{0,1\}$, at least one of the two points is interior; choosing that point as $c$ in step 1.4 gives a tent function taking value $1$ there and a value strictly below $1$ at the other point. Thus [L2] says that the only nonsingleton equivalence class is $\{0,1\}$.

Steps 1.1, 1.2, and 1.3 meet the hypotheses of [L1], and step 2.1 identifies its quotient; since $[0,1]$ is nonempty, [L1] gives the isometric conclusion, so descent is an isometric unital algebra isomorphism $A\cong C([0,1]\mathrm{/}\{0,1\},\mathbb{R})$.

By [L4] and the equivalence in [L7], the subspace $[a,b]$ of $\mathbb{R}$ is a compact topological space; by [L5] and [L6] the line $\mathbb{R}$ is Hausdorff, so [L8] makes the subspace $[a,b]$ Hausdorff.

Put $c:=(a+b)\mathrm{/}2$, so $a<c<b$, let $\iota :[a,b]\to \mathbb{R}$ be the inclusion, and put $h:=\mathrm{\mid }\iota -c\mathrm{\mid }$, so that $h(x)=\mathrm{\mid }x-c\mathrm{\mid }$. A constant map is continuous because the preimage of every open set is $\mathrm{\varnothing }$ or all of $[a,b]$, which is the condition in [L11]; $\iota$ is continuous by [L9]; so [L10] makes $\iota -c$ and then $h$ continuous.

The restricted polynomials form a unital real function algebra, and the coordinate polynomial $x\mapsto x$ separates distinct points; hence [L1] makes $P[a,b]$ uniformly dense in $C([a,b],\mathbb{R})$. In particular, [L2] also places the continuous function $h$ from step 1.2 in its uniform closure.

Suppose a real polynomial $p$ agreed with $h$ on $[a,b]$. Then $q(x):=p(x)-(x-c)$ vanishes at every $x\in [c,b]$; if $q$ were nonzero, that nondegenerate interval would contain more distinct roots than the finite bound in [L3], so $q$ is the zero polynomial and $p(x)=x-c$ identically.

Evaluating the identity from step 2.2 at $a$ gives $p(a)=a-c<0$, whereas $h(a)=\mathrm{\mid }a-c\mathrm{\mid }=c-a>0$, a contradiction. Therefore $h\mathrm{\notin }P[a,b]$.

Step 2.1 puts $h$ in the uniform closure and step 3.1 keeps it outside $P[a,b]$, so $P[a,b]$ is not closed; together with the density in step 2.1 this proves the example.

For each $n\in \mathbb{N}$ the function $x\mapsto x^n$ is continuous on $[0,1]$, so $x\mapsto x^{n}f(x)$ is continuous on $[0,1]$ as a product of continuous functions, and is therefore integrable; so each integral in the hypothesis is defined.

$f$ is bounded and integrable on $[0,1]$, so there is a real $M>0$ with $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ for every $x\in [0,1]$; if the bound supplied is $0$, replace it by $1$.

$f^2$ is continuous on $[0,1]$ as a product of continuous functions, hence integrable, and $f(x{)}^2\ge 0$ for every $x\in [0,1]$, so ${\int }_0^1{f}^2\ge 0$.

Let $p(x)=a_0+a_{1}x+\cdots +a_m{x}^m$ be any real polynomial. Each $x\mapsto a_j{x}^{j}f(x)$ is integrable by step 1.1, and applying the linearity identity $m$ times to the finite sum gives ${\int }_0^{1}pf={\sum }_{j\le m}{a}_j{\int }_0^1{x}^{j}f$, every summand of which is $0$ by hypothesis, so ${\int }_0^{1}pf=0$.

Let $\epsilon >0$. Choose a polynomial $p$ with ${\sup }_{x\in [0,1]}\mathrm{\mid }p(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}M$, which is legitimate since $\epsilon \mathrm{/}M>0$ by step 1.2.

The polynomial $p$ chosen in step 2.2 is continuous on $[0,1]$ by [L2] and hence integrable by [L3]; so $f-p$ is integrable by [L4], and both $(f-p)f$ and $pf$ are integrable by [L6]. Since $f^2=(f-p)f+pf$ pointwise on $[0,1]$, [L4] gives ${\int }_0^1{f}^2={\int }_0^1(f-p)f+{\int }_0^{1}pf$, and the second term is $0$ by step 2.1, giving ${\int }_0^1{f}^2={\int }_0^1(f-p)f$.

For every $x\in [0,1]$, $\mathrm{\mid }f(x)-p(x)\mathrm{\mid }\mathrm{\mid }f(x)\mathrm{\mid }<(\epsilon \mathrm{/}M)\cdot M=\epsilon$ by steps 1.2 and 2.2, so $-\epsilon \le (f-p)(x)f(x)\le \epsilon$ on $[0,1]$; since $1-0=1$, the two-sided bound gives ${\int }_0^1(f-p)f\le \epsilon$.

Combining, $0\le {\int }_0^1{f}^2\le \epsilon$.

Step 4.1 holds for every $\epsilon >0$, and the value ${\int }_0^1{f}^2$ does not depend on $\epsilon$; were it positive, taking $\epsilon$ to be half of it would contradict step 4.1, so ${\int }_0^1{f}^2=0$.

$f^2$ is continuous on $[0,1]$, nonnegative there, and has integral $0$ by step 5.1, so $f(x{)}^2=0$ for every $x\in [0,1]$, and hence $f(x)=0$ for every $x\in [0,1]$.