On a heat equation-like PDE

Question

For $f: \mathbb R^n \to \mathbb R$ a locally integrable function, $\varepsilon \in (0, \infty)$, and $x \in \mathbb R^n$, define $I(f, \varepsilon, x)$ to be the averaged integral of $f$ over $B_{\varepsilon} (x)$, the ball of radius $\varepsilon$ around $x$.

Define $$ K(f, \varepsilon, x) := \begin{cases} 1 & \text{if }\; I(f, \varepsilon, x) > f(x),\\ -1 & \text{if }\; I(f, \varepsilon, x) < f(x),\\ 0, &\text{if }\; I(f, \varepsilon, x) = f(x).\\ \end{cases} $$

Finally, let $$ H(f, \varepsilon, x) = \dfrac{1}{\varepsilon} \int\limits_{(0, \varepsilon]} K(f, s, x) ds $$

Intuitively, H is the weighted average amount of time a function spends greater than (resp. less than) its value at a point, in an infinitesimal neighbourhood of said point.

Questions

1. Is it true that any $C^2$ function $f$ satisfies $\limsup_{\varepsilon \to 0} H(f, \varepsilon, x) = \liminf_{\varepsilon \to 0} H(f, \varepsilon, x)$ for almost all $x \in \mathbb R$?

2. Consider the PDE $\partial_t u(x, t)$ = $\limsup_{\varepsilon \to 0} H(u(x, t), \varepsilon, x)$.
   If (1) is true, then the $\limsup$ may be replaced by a limit, so that no arbitrary choice between limsup and liminf must be made.
   The PDE is meant in a strong sense, to be solved over functions $u: \mathbb R \times [0, \infty) \to \mathbb R$; denoted $u(x, t)$ that are $C^2$ in $x$ for each fixed $t$, and $C^1$ in $t$ for each fixed $x$; with initial condition $u(x, 0) = f(x)$ for arbitrary $f \in C^2$. Do strong solutions exist? Are they unique?

3. The PDE in (2) is solvable by $u(x, t) := f(x)$ if the initial condition $f$ is a harmonic function, since harmonic functions satisfy the mean value property. Suppose the PDE in (ii) is uniquely solvable for some initial condition $f \in C^2$. Denoting by $u$ the solution, is it true that the functions $u(., t)$ converge pointwise to a harmonic function $u_\infty$ as $t \to \infty$?

Answer 1

I am puzzled by this question, so, at the risk of ridicule (which doesn't kill as we all know), I'll write what I understand from what is written.

• It seems that if $f$ is affine,then $K(f,e,x)=0.$ This is in fact true of all functions with the mean value property, that is, harmonic functions.

• If $f=x\cdot Ax$ where $A$ is a symmetric matrix, then $K=1$ when $A$ is positive definite (it is convex), $K=-1$ when $A$ is negative definite (it is concave). After a rotation, and a diagonalization, it appears that $K=\text{sgn}(\text{trace}(A)))$.

• Even though $K$ isn't linear in $f$, it is a fact that if $K(f,e,x)=0$, then $K(f+g,e,x)=K(g,e,x)$, therefore a Taylor expansion at $x$ shows that if $f\in C^2(\mathbb R)$ then $$ \lim_{e\to0} H(f,e,x)= \text{sgn}(\text{trace}(D^2f(x))))= \text{sgn}(\Delta f) $$ except where $D^2f=0$, but $Df$ not constant (thank you Nate River for pointing it out). Luckily this is a set of zero measure by Sard's Theorem. Thus the answer to 1. is yes indeed.

Moving on the question 3. as noted above, when $f$ is harmonic, then $\Delta f=0$, and in turn $\lim_{e\to0} H(f,e,x)=0$. So a solution to $u(0,x)=f$, and $\partial_t u = \text{sgn}(\Delta u)$ is simply $u=f$ for all $t\geq0$. Now take $f=x^2$ (when $n=1$). Then the pde is $\partial_t u = 1$, a solution would be $u=x^2+t$ for all $t \geq0$ and all $x$, so the answer to question 3 is no.

Finally 2. The pde is $$ \begin{cases} \partial_t u &= \text{sgn}(\Delta u) \text{ on }\mathbb R^n \times [0,\infty), \\ u(0,x)&=f(x). \end{cases} $$ Apart from the case discussed in 3, the question of existence / uniqueness of solutions isn't obvious (to me). Taking for example $f=x^3$ (when $n=1$), a solution which would be continuous in time would be, for $x\neq0$, $u(x,t)=x^3 + t\frac{x}{|x|}$, but that cannot be made $C^2$ in $x$..