Specifically, I was wondering the following:
Let $I$ stand for the closed interval $[0,1]$. Let $f:I^n \rightarrow \mathbb{R}$ be a continuous function. Take an open subset $U \subset I^n$, and define a function $g: I^n \rightarrow \mathbb{R}$ where $g(x) = \inf_{y \in I^n - U} ||x - y||$, and $||.||$ is the euclidean norm (though, I will accept a solution using a different norm as well).
Now, construct the function $h: I^n \rightarrow \mathbb{R}$ the following way:
$h(x) = f(x)$ when $g(x) = 0$
and
$h(x) = \frac{1}{\mu(B_{g(x)}(x))}\int_{B_{g(x)}(x)}f(x)d\mu$ otherwise,
Where $\mu$ is the standard measure on $I^n$, and $B_{g(x)}(x)$ is the set of all elements $y$ such that $||x-y|| < g(x)$.
With all that setup, the question is the following: is $h$ differentiable at all points where $g(x)$ is not zero? My intuition tells me yes, as $h$ kind of just "averages out" certain parts of the original function, but I do not know for certain.
