On the continuity of a function given by evaluating compact subsets of smooth functions

Question

Let $M$ be a compact connected smooth manifold. Write $C^{\infty}(M)$ for the Frechet space of the smooth real-valued functions on $M$ equipped with the usual $C^{\infty}$-topology.

Given a compact subset $K$ of $C^{\infty}(M)$, define a non-negative function on $M$ via

$$F_K(x)=\mu(\mathrm{Ev}_x(K)), \quad x \in M,$$

where $\mathrm{Ev}_x: C^{\infty}(M) \rightarrow \mathbb{R}$ is the map that evaluates a function at the point $x \in M$ and $\mu$ is the Lebesgue measure on $\mathbb{R}$.

My question: Is the function $F_K$ continuous for any compact $K$?

Few remarks:

1. I asked a similar question recently that didn't get answered but according to this partial answer $F_K$ is upper semicontinuous.

2. Let $\mathcal{K}(\mathbb{R})$ be the set of the compact subsets of $\mathbb{R}$ equipped with the Hausdorff metric. I was able to show, using the equicontinuity of $K$, that the map $$ M \ni x \mapsto \mathrm{Ev}_x(K) \in \mathcal{K}(\mathbb{R})$$ is continuous. Since the Lebesgue measure is known to be upper semicontinuous with respect to the Hausdorff metric (see this article and also this), this gives another proof of the upper semicontinuity of $F_K$.

Answer 1

The maps $F_K$ are certainly upper semicontinuous, because, by the regularity property of the Lebesgue measure, for every $x\in M$ and for $\epsilon>$ there exists $\delta>0$ such that the uniform $\delta$-neighbourhood $N_\delta$ of the compact set $\text{Ev}_x(K)$ has measure $|N_\delta|< | \text{Ev}_x(K)|+\epsilon$; since all $f$ in $K$ have a bounded Lipschitz constant $k$, for all $y\in M$ with $d(x,y)<\delta/k$ one has $\text{Ev}_y(K)\subset N_\delta$, thus $| \text{Ev}_y(K)|< | \text{Ev}_x(K)|+\epsilon.$ The same is also true in the $C^0$ case (then all $f$ in the compact $K$ are equicontinuous, and the conclusion remains true).

However $F_K$ is not in general continuous. In fact, the characteristic function of every closed subset $C$ of a smooth compact manifold $M$ can be obtained as an $F_K$ for a suitable $K$:

Let $\eta\in C^\infty(M)$ such that $\eta(x)=1/2$ for all $x\in C$ and $0\le \eta(x)<1/2$ for $x\in M\setminus C$. For $\epsilon \in \{0,1\}^{\mathbb N}$, define $f_\epsilon(x)=\sum_{k=1}^\infty\epsilon_k x^k$, and $K:=\big\{f_\epsilon\circ \eta:\epsilon\in \{0,1\}^{\mathbb N}\big\}\subset C^\infty(M).$

Since $K$ is a closed bounded subset of $C^\infty(M)$, it is compact. For $x\in C$ the set $\text{Ev}_x(K)$ is the whole interval $[0,1]$ so $F_K(1/2)=1$; for $x\in M\setminus C$ it is a null Cantor set, and $F_K(x)=0.$ So $F_K=\chi_C.$

The example also holds in the $C^0$ setting.

Answer 2

I will answer the $\mathrm{C}^0$ case on a closed ball as in the other post. Let $f \colon \mathbb{R} \to \mathbb{R}$ be the continuous function $$ f(x) = \begin{cases} \frac{1}{2} &\text{if } x < \frac{1}{2} \\ 1-x &\text{if } \frac{1}{2} \leq x \leq 1 \\ 0 &\text{if } x > 1 \end{cases}. $$ Then consider the map $\psi \colon \{0,1\}^{\mathbb{N}} \to \mathrm{C}^0([0, 1])$ given by $\psi_c = \sum_{i=0}^{\infty} \frac{c_i}{2^i} f(2^i x)$. It is continuous, since for an open ball of radius $\epsilon$ around a function, arbitrary changes to $c$ after a certain index do not make $\psi_c$ leave this ball. Moreover, $\{0,1\}^\mathbb{N}$ is compact, so the image is as well. However, the functions "branch" in such a way that $\mu\{\psi_c(x) \mid c \in \{0,1\}^{\mathbb{N}}\}$ is zero for $x \neq 0$ (here, this set is finite) and $1$ for $x = 0$ (any number has a binary expansion).