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

The denominator is positive: if its two nonnegative summands vanished, $x=0$ and $1-nx=0$ would both hold, which is impossible. Each $f_n$ is therefore continuous, $0\le f_n\le 1$, $f_n(0)=0$, and $f_n(1\mathrm{/}n)=1$.

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

Take $\epsilon =1\mathrm{/}2$. For every $\delta >0$ there is an integer $n\ge 1$ with $1\mathrm{/}n<\delta$, but $\mathrm{\mid }{f}_n(1\mathrm{/}n)-f_n(0)\mathrm{\mid }=1$. Thus no common radius works at $0$, so the family is not equicontinuous.

For every subsequence $(f_{n_j})$, the indices $n_j$ tend to infinity, so step 1.2 still forces any uniform limit to be zero. But $\parallel f_{n_j}{\parallel }_{\mathrm{\infty }}=f_{n_j}(1\mathrm{/}{n}_j)=1$ for every $j$, contradicting uniform convergence to zero. No subsequence is uniformly convergent.

This bounded family is neither equicontinuous nor uniformly sequentially compact.

For every $c$ and all $x,y$, $\mathrm{\mid }{f}_c(x)-f_c(y)\mathrm{\mid }=0$, 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 $\epsilon >0$, the common choice $\delta =\epsilon$ gives $\mathrm{\mid }x-y\mathrm{\mid }<\delta \Rightarrow \mathrm{\mid }{f}_n(x)-f_n(y)\mathrm{\mid }<\epsilon$ for every $n$; this is the ordinary all-metric-space equicontinuity condition used in the refuted claim.

If $\mathrm{\mid }m-n\mathrm{\mid }\ge 2$, then at $x=n$ one has $f_n(n)=1$ and $f_m(n)=0$; consequently $\parallel f_n-f_m{\parallel }_{\mathrm{\infty }}\ge 1$.

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

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

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

The reverse triangle inequality gives $\mathrm{\mid }{d}_a(x)-d_b(x)\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$ for every $x$; evaluating at $x=0$ gives $\parallel d_a-d_b{\parallel }_{\mathrm{\infty }}=\mathrm{\mid }a-b\mathrm{\mid }$.

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 $\max \{\mathrm{\mid }p(-1)+1\mathrm{\mid },\mathrm{\mid }p(1)-1\mathrm{\mid }\}\ge 1$, so $\parallel p-\mathrm{id}{\parallel }_{\mathrm{\infty }}\ge 1$.

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 $p\in A$, $\parallel p-\mathbf{1}{\parallel }_{\mathrm{\infty }}\ge \mathrm{\mid }p(0)-1\mathrm{\mid }=1$. Thus $\mathbf{1}$ is not in the uniform closure of $A$.

Therefore $A$ separates points but is not dense.