Approximation and Compactness in $C(K)$

The balls $B(a,r_a\mathrm{/}2)$ for $a\in K$ cover $K$, so choose finitely many centres $a_0,\dots ,a_N$ whose balls cover $K$.

Let $\delta$ be the least of the finitely many positive radii $r_{a_i}\mathrm{/}2$. If $d(x,y)<\delta$, choose $i$ with $x\in B(a_i,r_{a_i}\mathrm{/}2)$; then both $x$ and $y$ lie in $B(a_i,r_{a_i})$.

The two estimates from [L1] and the triangle inequality give $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon$ for every $f\in \mathcal{F}$.

The $\delta$-balls cover $K$; compactness supplies finitely many centres $a_0,\dots ,a_N$ whose $\delta$-balls cover it.

By pointwise boundedness, choose $M_i\ge 0$ with $\mathrm{\mid }f(a_i)\mathrm{\mid }\le M_i$ for every $f\in \mathcal{F}$, and let $M:=1+{\max }_i{M}_i$.

For $x\in K$, choose $i$ with $d(x,a_i)<\delta$; then $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }f(x)-f(a_i)\mathrm{\mid }+\mathrm{\mid }f(a_i)\mathrm{\mid }<1+M_i\le M$.

Thus $M$ uniformly bounds $\mathcal{F}$.

Choose a finite $\delta$-net $E=\{a_0,\dots ,a_N\}$ in $K$ from the uniform equicontinuity radius for $\epsilon \mathrm{/}3$.

By [L2], every vector $(f(a_0),\dots ,f(a_N))$ lies in one bounded box in ${\mathbb{R}}^{N+1}$; cover that box by finitely many coordinate cubes of side less than $\epsilon \mathrm{/}3$.

Choose one member of $\mathcal{F}$ from each nonempty inverse image of such a cube. Every $f\in \mathcal{F}$ and its chosen representative differ by less than $\epsilon \mathrm{/}3$ on $E$.

For any $x\in K$, choose $a_i\in E$ with $d(x,a_i)<\delta$ and use equicontinuity for both functions and step 2.1 to obtain $\mathrm{\mid }f(x)-g(x)\mathrm{\mid }<\epsilon$.

The finitely many representatives form an $\epsilon$-net, so $\mathcal{F}$ is totally bounded.

Choose $N$ such that $\mathrm{\mid }{f}_n(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}3$ for every $n\ge N$ and $x\in K$.

By [L1], choose a neighbourhood of $a$ on which $\mathrm{\mid }f(x)-f(a)\mathrm{\mid }<\epsilon \mathrm{/}3$; by continuity of the finitely many $f_n$ with $n<N$, shrink it so the same inequality with $\epsilon$ holds for all of them.

On that neighbourhood, the triangle inequality gives $\mathrm{\mid }{f}_n(x)-f_n(a)\mathrm{\mid }<\epsilon$ for $n\ge N$, while step 1.2 covers $f$ and $n<N$.

Thus the displayed family is equicontinuous at arbitrary $a$, hence equicontinuous.

Suppose $\mathcal{F}$ is equicontinuous and pointwise bounded. By [L1] it is totally bounded, and its closure is totally bounded as well.

Conversely suppose the closure is compact. For a positive $\epsilon$, choose a finite $\epsilon \mathrm{/}3$-net $g_0,\dots ,g_N$ in the closure; by [L5], a common positive radius makes every $g_i$ vary by less than $\epsilon \mathrm{/}3$.

The closure is closed in the complete space of [L2], hence complete by [L3]. Therefore its closure is compact by [L3].

For $f\in \mathcal{F}$, choose $g_i$ within $\epsilon \mathrm{/}3$ in supremum distance. The two uniform-distance bounds and step 1.2 give $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon$ whenever $d(x,y)$ is below the common radius.

The same finite net bounds $\mathrm{\mid }f(a)\mathrm{\mid }$ at each fixed $a\in K$, so $\mathcal{F}$ is pointwise bounded.

Steps 2.2 and 2.3 give equicontinuity and pointwise boundedness, completing the converse.

The sequence lies in its compact closure by [L1].

By [L2], it has a subsequence converging in the supremum metric to a point of that closure.

Supremum-metric convergence is uniform convergence, so the claimed subsequence converges uniformly.

Apply [L1] to $u=x$ and $v=1-x$ to obtain the zeroth identity.

Reindex the terms $k\left(\genfrac{}{}{0px}{}{n}{k}\right)=n\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$ and apply [L1] at exponent $n-1$ to obtain the first identity.

Apply the same reindexing to $k(k-1)\left(\genfrac{}{}{0px}{}{n}{k}\right)=n(n-1)\left(\genfrac{}{}{0px}{}{n-2}{k-2}\right)$ when $n\ge 2$, combine it with step 1.2, and expand the square.

For $n=1$ the displayed centred identity is checked directly; together with step 2.1 this proves it for all $n\ge 1$.

Choose $\delta >0$ such that $\mathrm{\mid }f(s)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever $\mathrm{\mid }s-x\mathrm{\mid }<\delta$, and choose $M$ with $\mathrm{\mid }f\mathrm{\mid }\le M$.

On the far part, $(k\mathrm{/}n-x{)}^2\ge {\delta }^2$; hence its total basis weight is at most $x(1-x)\mathrm{/}(n{\delta }^2)\le 1\mathrm{/}(4n{\delta }^2)$ by [L2].

Split the Bernstein sum into $\mathrm{\mid }k\mathrm{/}n-x\mathrm{\mid }<\delta$ and its complement. The near part is at most $\epsilon \mathrm{/}2$ by the zeroth identity.

Choose $n$ so large that $2M\mathrm{/}(4n{\delta }^2)<\epsilon \mathrm{/}2$. The far part is then below $\epsilon \mathrm{/}2$, uniformly in $x$.

The near and far estimates give $\mathrm{\mid }{B}_n(f)(x)-f(x)\mathrm{\mid }<\epsilon$ for every $x$ and all sufficiently large $n$.

Choose $n$ with ${\sup }_x\mathrm{\mid }{B}_n(f)(x)-f(x)\mathrm{\mid }<\epsilon$.

The finite defining sum for $B_n(f)$ is a polynomial in $x$, so $p:=B_n(f)$ has the required property.

If $a=b$, the constant polynomial $q(x)=f(a)$ agrees with $f$ on the singleton interval.

Now suppose $a<b$ and define $g(t)=f(a+(b-a)t)$ on $[0,1]$.

Choose a polynomial $p$ with $\mathrm{\mid }p(t)-g(t)\mathrm{\mid }<\epsilon$ on $[0,1]$ by [L1].

Then $q(x)=p((x-a)\mathrm{/}(b-a))$ is a polynomial and satisfies $\mathrm{\mid }q(x)-f(x)\mathrm{\mid }<\epsilon$ on $[a,b]$. Together with step 1.1 this proves both cases.

Choose $a_n\in A$ converging uniformly to $u$. Their ranges, together with that of $u$, lie in one bounded interval.

By [L2], choose polynomials $p_j$ with $p_j(0)=0$ converging uniformly to $\mathrm{\mid }t\mathrm{\mid }$ on that interval. Then $p_j(a_n)\in A$ by [L1].

A diagonal choice of $j,n$ makes $p_j(a_n)$ uniformly converge to $\mathrm{\mid }u\mathrm{\mid }$, so $\mathrm{\mid }u\mathrm{\mid }\in \bar{A}$.

The identities $u\vee v=(u+v+\mathrm{\mid }u-v\mathrm{\mid })\mathrm{/}2$ and $u\wedge v=(u+v-\mathrm{\mid }u-v\mathrm{\mid })\mathrm{/}2$ and [L1] give the remaining closure.

Choose $h$ from [L1] and put $g(z)=r+(s-r)(h(z)-h(x))\mathrm{/}(h(y)-h(x))$.

The denominator is nonzero, $g\in A$, and direct substitution gives $g(x)=r$ and $g(y)=s$.

For fixed $x,y\in K$, choose $u_{x,y}\in A$ with $u_{x,y}(x)=f(x)$ and $u_{x,y}(y)=f(y)$, using $f(x)\mathbf{1}$ when $x=y$ and [L2] otherwise.

For fixed $x$, the open sets $U_{x,y}=\{z:u_{x,y}(z)>f(z)-\epsilon \}$ cover $K$, since $y\in U_{x,y}$. Select $y_1,\dots ,y_r$ whose sets cover $K$ by [L3].

Put $g_x={\max }_j{u}_{x,y_j}\in \bar{A}$. Then $g_x(x)=f(x)$ and $g_x>f-\epsilon$ on $K$.

The open sets $V_x=\{z:g_x(z)<f(z)+\epsilon \}$ cover $K$. By [L3], choose $x_1,\dots ,x_s$ whose $V_{x_i}$ cover $K$.

The function $h={\min }_i{g}_{x_i}$ belongs to $\bar{A}$ by [L1], and $f-\epsilon <h<f+\epsilon$ pointwise. Thus $\parallel h-f{\parallel }_{\mathrm{\infty }}<\epsilon$.

Since $\epsilon$ was arbitrary and every $h$ of step 5.1 lies in the closed set $\bar{A}$, the function $f$ lies in $\bar{A}$. Thus $\bar{A}=C(K,\mathbb{R})$ and $A$ is dense.

If $X$ is empty the assertion is vacuous. Otherwise choose an open ball $B_0$ whose closure lies in $O$; this is possible because $O$ is open.

Given a nonempty open ball $B_{n-1}$, its intersection with $X\setminus F_n$ is nonempty because $F_n$ has empty interior. Choose an open ball $B_n$ with nonempty closure, $\overline{B_n}\subseteq B_{n-1}\setminus F_n$, and radius below $2^{-n}$.

Dependent choice gives balls $B_n$ satisfying step 1.2 for every $n$. Choose centres $x_n\in B_n$.

The nesting and the radius bound make $(x_n)$ Cauchy: for $m>n$, both $x_m$ and $x_n$ lie in $\overline{B_n}$, so their distance is at most twice the radius of $B_n$, which tends to zero.

Let $x={\lim }_n{x}_n$, supplied by completeness [L1]. For every $n$, the tail lies in the closed set $\overline{B_n}$, so $x\in \overline{B_n}\subseteq B_{n-1}\setminus F_n$.

Thus $x\in O\setminus {\bigcup }_n{F}_n$. Every nonempty open $O$ meets this complement, proving both stated formulations.

For each $n$, choose a witness $a_n\in [0,1]$ for $f_n\in E_{p,q}$. Pass to a subsequence with $a_n\to a\in [0,1]$ using [L2].

By [L1], $f_n\to f$ uniformly, and [L3] confirms that $f\in C([0,1],\mathbb{R})$.

Fix $t\in [0,1]$ with $\mathrm{\mid }t-a\mathrm{\mid }<1\mathrm{/}q$. For all sufficiently large $n$, $\mathrm{\mid }t-a_n\mathrm{\mid }<1\mathrm{/}q$, hence $\mathrm{\mid }{f}_n(t)-f_n(a_n)\mathrm{\mid }\le p\mathrm{\mid }t-a_n\mathrm{\mid }$.

Letting $n$ tend to infinity in step 2.1, uniform convergence and continuity of $f$ give $\mathrm{\mid }f(t)-f(a)\mathrm{\mid }\le p\mathrm{\mid }t-a\mathrm{\mid }$.

The point $a$ witnesses $f\in E_{p,q}$; therefore $E_{p,q}$ is sequentially closed, hence closed in this metric space.

By [L1], choose a finite partition $0=x_0<\cdots <x_r=1$ so that $\mathrm{\mid }f(s)-f(t)\mathrm{\mid }<\epsilon \mathrm{/}4$ whenever $s,t$ are in one partition interval. Let $g$ be the affine interpolant through $(x_i,f(x_i))$.

The affine-interpolation formula makes $\mathrm{\mid }g(t)-f(t)\mathrm{\mid }<\epsilon \mathrm{/}4$ on every partition interval. Let $S$ be the maximum of the finitely many absolute slopes of $g$.

Choose $0<\eta <\epsilon \mathrm{/}4$. By [L2], subdivide every partition interval evenly enough to support a continuous triangular sawtooth $w$, zero at the old vertices, with $\parallel w{\parallel }_{\mathrm{\infty }}\le \eta$ and every nonvertex slope of absolute value greater than $S+M$.

Put $h=g+w$. On each new affine piece, the reverse triangle inequality gives $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }\ge \mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }-\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }>M$, while $\parallel h-f{\parallel }_{\mathrm{\infty }}<\epsilon \mathrm{/}2<\epsilon$.

This $h$ has the required finite polygonal structure, approximation, and slope bound.

Each $E_{p,q}$ has empty interior. Indeed, every supremum ball contains by [L2] a polygonal $h$ all of whose nonvertex slopes have absolute value greater than $p$; at any point, including a vertex or endpoint, a sufficiently nearby point on an adjacent affine piece violates the defining $p$-bound.

If $f\in G$ had a finite derivative at $a$, [L4] would bound its difference quotients by some integer $p$ on a sufficiently small radius $1\mathrm{/}q$, placing $f$ in $E_{p,q}$; this contradicts $f\in G$.

Hence every $C([0,1])\setminus E_{p,q}$ is open and dense by [L1].

Apply [L5] to the complete space of [L3]. The intersection $G={\bigcap }_{p,q\ge 1}(C([0,1])\setminus E_{p,q})$ is dense.

Thus $G$ is a dense subset of the continuous nowhere-differentiable functions, which proves the statement.

Since $0\le \varphi \le 1\mathrm{/}2$, the $n$th summand is bounded by $2^{-n-1}$; its majorant series converges. Thus $T$ converges uniformly by [L1].

For $x\in [0,1)$ and each $N$, let $u_N\le x<v_N$ be the adjacent dyadic rationals of order $N$; at $x=1$ use the left adjacent interval. Every summand of index at least $N$ vanishes at both endpoints, while each earlier summand is affine on that interval with slope ${\epsilon }_k\in \{-1,1\}$.

Every summand is continuous, so $T$ is continuous by [L2].

Hence $(T(v_N)-T(u_N))\mathrm{/}(v_N-u_N)={\sum }_{k<N}{\epsilon }_k$. These secant slopes cannot converge to a finite real number, because consecutive partial sums differ by ${\epsilon }_N$ of absolute value one.

If a finite derivative existed at an interior point, both endpoint difference quotients and therefore their secant combination would tend to it; at a dyadic point or endpoint the same argument uses the nested one-sided dyadic intervals. This contradicts step 2.2 and [L3].

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.