A $C^1$ map from ${\mathbb{R}}^2$ to $\mathbb{R}$ with an interval of critical values

Put $p=2\mathrm{/}5$, start with $K_0=[0,1]$, and replace each level-$n$ closed interval $[a,b]$ by its left and right closed subintervals of relative length $p$; let $K$ be the intersection of the resulting finite unions. Each stage is closed, so $K$ is closed. Every binary word ${ϵ}_1\dots {ϵ}_n$ determines a closed interval of length $p^n$. For each infinite binary sequence these intervals are nested and their lengths tend to zero by [F5], so [F4] gives a unique point in their intersection. Conversely the two children of every interval are separated by a positive gap, so each point of $K$ determines a unique branch at every level. The code map is therefore a bijection from binary sequences to $K$; the level-$n$ interval for a code has left endpoint ${\sum }_{j=1}^n(1-p){ϵ}_j{p}^{j-1}$ and length $p^n$, so the point differs from that partial sum by at most $p^n\to 0$ and hence is $x={\sum }_{j=1}^{\mathrm{\infty }}(1-p){ϵ}_j{p}^{j-1}$. By [F6], matching these digits defines a bijection $h:K\to C$ by $h(x)={\sum }_{j=1}^{\mathrm{\infty }}2{ϵ}_j\mathrm{/}{3}^j$. If distinct $x,y\in K$ first differ after a common prefix of length $k$, their source intervals are separated by at least $(1-2p)p^k=\frac{1}{5}(\frac{2}{5}{)}^k$, while their images lie in one target interval of length $3^{-k}$. Hence $\mathrm{\mid }h(x)-h(y)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }\le 5(5\mathrm{/}6{)}^k$. For every $N$, if $k\le N$ the source gap is at least $(1-2p)p^N$, so $\mathrm{\mid }x-y\mathrm{\mid }<(1-2p)p^N$ forces $k>N$; thus the ratio tends to zero uniformly on $K$.

For each $a\in K$ prescribe the constant polynomial $P_a(t)=h(a)$. For order $r=1$, the Whitney compatibility condition for $\mathrm{\mid }\alpha \mathrm{\mid }=0$ is exactly the estimate from step 1.1; for $\mathrm{\mid }\alpha \mathrm{\mid }=1$ both derivatives of these polynomials are zero, which is the required $o(1)$ condition. Therefore [F2] supplies a map $g\in C^1(\mathbb{R},\mathbb{R})$ with $g(a)=h(a)$ and $g^{\mathrm{\prime }}(a)=0$ for every $a\in K$.

Define $F(x,y)=g(x)+g(y)$. For increments $(u,v)$, subtracting the candidate linear map $L_{(x,y)}(u,v):=g^{\mathrm{\prime }}(x)u+g^{\mathrm{\prime }}(y)v$ leaves the sum of the one-variable remainders $g(x+u)-g(x)-g^{\mathrm{\prime }}(x)u=o(\mathrm{\mid }u\mathrm{\mid })$ and $g(y+v)-g(y)-g^{\mathrm{\prime }}(y)v=o(\mathrm{\mid }v\mathrm{\mid })$. Since $\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }\le \sqrt{2}\sqrt{u^2+v^2}$, the total remainder is $o(\sqrt{u^2+v^2})$; hence $D{F}_{(x,y)}=L_{(x,y)}$, and this derivative is continuous because $g^{\mathrm{\prime }}$ is continuous. Thus $F\in C^1({\mathbb{R}}^2,\mathbb{R})$. On $K\times K$ its derivative is the zero linear map ${\mathbb{R}}^2\to \mathbb{R}$, which is not onto, so [F1] makes every point there critical. Since $h[K]=C$ by step 1.1, $F(K\times K)=C+C$.

Fix $u\in [0,1]$, set $r_0=u$, and recursively define digits $d_j\in \{0,1,2\}$ and remainders $r_j\in [0,1]$: if $0\le r_{j-1}\le 1\mathrm{/}3$, take $(d_j,r_j)=(0,3r_{j-1})$; if $1\mathrm{/}3<r_{j-1}\le 2\mathrm{/}3$, take $(1,3r_{j-1}-1)$; if $2\mathrm{/}3<r_{j-1}\le 1$, take $(2,3r_{j-1}-2)$. These exhaustive cases keep $r_j$ in $[0,1]$ and give $r_{j-1}=(d_j+r_j)\mathrm{/}3$. At $n=0$, $u=r_0$; substituting this recurrence proves by induction that $u={\sum }_{j=1}^n{d}_j\mathrm{/}{3}^j+r_n\mathrm{/}{3}^n$ for every $n$. Since $0\le r_n\le 1$ and $(1\mathrm{/}3{)}^n\to 0$ by [F5], the partial sums converge to $u$, hence represent it by [F7]. Apply this to $u=t\mathrm{/}2$ for any $t\in [0,2]$, and choose $({ϵ}_j,{\delta }_j)=(0,0),(1,0),(1,1)$ when $d_j=0,1,2$, respectively. By [F6], $a={\sum }_{j=1}^{\mathrm{\infty }}2{ϵ}_j\mathrm{/}{3}^j$ and $b={\sum }_{j=1}^{\mathrm{\infty }}2{\delta }_j\mathrm{/}{3}^j$ lie in $C$, and [F7] gives $a+b={\sum }_{j=1}^{\mathrm{\infty }}2d_j\mathrm{/}{3}^j=2u=t$. Thus $[0,2]\subseteq C+C$; the reverse inclusion follows from $C\subseteq [0,1]$ in [F6]. Step 3.1 now shows $[0,2]\subseteq {\mathrm{CV}}_{C^1}(F)$.

If the critical-value set were Euclidean null, [F3] would make it null for the single-chart atlas of $\mathbb{R}$, and [L1] would make its complement dense. Step 4.1 shows that its complement misses the nonempty open interval $(0,2)$, a contradiction. Thus the critical-value set is not null.