Approximation and Compactness in $C(K)$: Examples and Counterexamples

Each $f_n$ is continuous, $0\le f_n\le1$, $f_n(0)=0$, and $f_n(1/n)=1$.

For every fixed $x>0$, $f_n(x)\to0$, while $f_n(0)=0$. Thus any uniformly convergent subsequence would have the zero function as its pointwise limit, consistently with [L1].

The common values at $0$ and $1/n\to0$ contradict equicontinuity at $0$.

Uniform convergence to zero is impossible because $\lVert f_n\rVert_\infty\ge f_n(1/n)=1$.

This bounded family is neither equicontinuous nor uniformly sequentially compact.

For every $c$ and all $x,y$, $|f_c(x)-f_c(y)|=0$, so any positive $\delta$ works simultaneously for every member of $\mathcal F$.

At each fixed point $a$, $\{f_c(a):c\in\mathbb R\}=\mathbb R$, which is unbounded.

Hence this family is equicontinuous but not pointwise bounded.

The triangular bump is $1$-Lipschitz and takes values in $[0,1]$. Every translate $f_n$ has the same properties. Thus the family is uniformly bounded and, for every $\varepsilon>0$, the common choice $\delta=\varepsilon$ gives $|x-y|<\delta\Rightarrow|f_n(x)-f_n(y)|<\varepsilon$ for every $n$; this is the ordinary all-metric-space equicontinuity condition used in the refuted claim.

If $|m-n|\ge2$, then at $x=n$ one has $f_n(n)=1$ and $f_m(n)=0$; consequently $\lVert f_n-f_m\rVert_\infty\ge1$.

Every infinite subsequence contains two indices separated by at least $2$, so no subsequence is uniformly Cauchy.

By [L1], no subsequence can converge uniformly.

For $n\ge1$, substitute $f(k/n)=(k/n)^2$ in the definition and use [L1] to obtain $B_n(f)=x^2+x(1-x)/n$.

Since $0\le x(1-x)\le1/4$ with equality at $x=1/2$, the stated supremum error is $1/(4n)$.

The separate definition at degree zero gives $B_0(f)=f(0)=0$.

The reverse triangle inequality gives $|d_a(x)-d_b(x)|\le|a-b|$ for every $x$; evaluating at $x=0$ gives $\lVert d_a-d_b\rVert_\infty=|a-b|$.

Given a sequence $d_{a_n}$ in $\mathcal D$, use [L1] to choose $a_{n_j}\to a\in[0,1]$. Step 1.1 then gives $d_{a_{n_j}}\to d_a$ uniformly.

Thus $\mathcal D$ is sequentially compact, and it is compact by [L2].

The constants belong to $A$, and sums and products of even polynomials are even, so $A$ is a unital algebra.

Every $p\in A$ has $p(-1)=p(1)$; therefore $A$ does not separate these two points.

If $p\in A$, then $\max\{|p(-1)+1|,|p(1)-1|\}\ge1$, so $\lVert p-\operatorname{id}\rVert_\infty\ge1$.

Hence the identity function is not in the closure of $A$, and $A$ is not dense.

The set $A$ is an algebra, and the function $x\mapsto x$ in $A$ separates every distinct pair of points of $[0,1]$.

Every $p\in A$ vanishes at $0$, so $A$ contains no constant-one function and is not unital.

For every $p\in A$, $\lVert p-\mathbf1\rVert_\infty\ge|p(0)-1|=1$. Thus $\mathbf1$ is not in the uniform closure of $A$.

Therefore $A$ separates points but is not dense.