The conjecture on infinitely many nondifferentiable points and semi-algebraicity

Let $f:[0,1]\to\mathbb R$ be continuous and nonnegative, and put

$$C=\{(x,y):0\le x\le1,\ 0\le y\le f(x)\}.$$

We prove the stronger contrapositive: if $C$ is semialgebraic, then $f$ is real-analytic except at finitely many points.

Suppose that $C$ is semialgebraic, and define

$$E=\{(x,y,z)\in\mathbb R^3:(x,y)\in C,\ (x,z)\in C,\ z>y\}.$$

The set $E$ is semialgebraic. By the Tarski–Seidenberg theorem, its coordinate projection $\pi_{x,y}(E)$ is semialgebraic. Therefore

$$\Gamma_f=C\setminus\pi_{x,y}(E) =\{(x,f(x)):0\le x\le1\}$$

is semialgebraic.

Choose a finite Boolean description of $\Gamma_f$ using nonzero polynomials $P_1,\ldots,P_m\in\mathbb R[x,y]$. The graph has empty interior. If all $P_i$ were nonzero at a point of $\Gamma_f$, their signs would remain unchanged throughout a neighborhood, so that neighborhood would also belong to the graph, a contradiction. Consequently the nonzero polynomial

$$P(x,y)=\prod_{i=1}^mP_i(x,y)$$

satisfies $P(x,f(x))=0$ for every $x\in[0,1]$.

Factor $P$ over $\mathbb R[x,y]$. Let $Q(x,y)$ be the product of one copy of each distinct irreducible factor having positive $y$-degree, and let $A(x)$ be the product of all remaining nonconstant, $x$-only factors. There must be at least one positive-$y$-degree factor, since a nonzero polynomial in $x$ alone cannot vanish on the entire graph. Outside the finitely many roots of $A$,

$$Q(x,f(x))=0.$$

Viewed in $\mathbb R(x)[y]$, the polynomial $Q$ is squarefree. Since the characteristic is zero,

$$\gcd_{\mathbb R(x)[y]}(Q,\partial_yQ)=1, \qquad \Delta(x)=\operatorname{Res}_y(Q,\partial_yQ)\ne0.$$

Hence

$$F=\{0,1\}\cup\{x\in[0,1]:A(x)\Delta(x)=0\}$$

is finite. For $x_0\in[0,1]\setminus F$, we have

$$Q(x_0,f(x_0))=0,\qquad \partial_yQ(x_0,f(x_0))\ne0.$$

The real-analytic implicit function theorem gives a unique real-analytic root branch near $(x_0,f(x_0))$. Continuity of $f$ forces $f$ to coincide locally with that branch. Thus $f$ is real-analytic, and therefore differentiable, at every point outside $F$.

It follows that a continuous nonnegative function with infinitely many nondifferentiability points cannot have a semialgebraic subgraph, proving the conjecture.