# Differentiability of a multidimensional integral

Let $u:\mathbb R^n\to\mathbb R$ be only continuous and let $\Omega\subseteq\mathbb R^n$ be a bounded open set and for $r>0$ fixed, and define $f:\mathbb R^n\to\mathbb R$ via $$f(x) := \int_{B(x,r)}u(y)~dy$$ wherever it makes sense ($B(x,r)\subset\Omega$). Is $f$ differentiable?

Notation: $B(x,r)$ is the ball centered at $x$ with radius $r$.

We were investigating an alternative proof for the fact that the mean value property (in the context of harmonic functions) implies smoothness of a function (the standard one using mollifiers). If $u$ is at least once differentiable, then we can differentiate under the integral to see that $f$ is smooth, but if $u$ is only continuous, can we say that $f$ is differentiable or can we find a counterexample?

I've seen this: Differentiability of multivariable functions represented through integral but I only want to integrate over a ball, not $\mathbb R^n$.

We can try something like \begin{align*} \lim_{h\to0}\frac{f(x+h)-f(x)}{\vert h\vert} = \lim_{h\to0}\frac{\int_{B(x,r)}u(y+h)-u(y)~dy}{\vert h\vert} \end{align*} and then we would see that we need to make $u(y+h)-u(y)$ go to $0$ about as fast as $\vert h\vert$, but it's not clear how to control this, so it seems like it shouldn't be true, but I haven't been able to come up with a counterexample yet.

A start: Let's take $R=1,$ $n=2,$ and let's concentrate on a derivative at $(0,0).$ Let's also assume $u$ is $C^1$ to begin with. Then

$$\tag 1 \frac{f((0,t)-f(0,0)}{t}=\frac{1}{t}\left (\int_{B((0,t),1)} u\,dA - \int_{B((0,0),1)} u\,dA\right )$$ $$= \int_{B((0,0),1)} \frac{u((x,y)+(0,t)) - u(x,y)}{t}\,dA.$$

As $t\to 0,$ the last integrand converges nicely to $\partial u/\partial y(x,y).$ So the limit of $(1)$ equals

$$\tag 2 \int_{B((0,0),1)} \partial u/ \partial y\,dA.$$

If we integrate using Fubini and use the FTC, we see that $(2)$ equals

$$\tag 3\int_{-1}^1 (u(x,(1-x^2)^{1/2}) - u(x,-(1-x^2)^{1/2})\, dx.$$

Now if $u$ is merely continuous, $(3)$ makes sense even if $(1),(2)$ don't.

So there's my guess for $\partial f \partial y(0,0)$ in this situation. That leads to a guess for what $Df(0,0)$ should be. As for dealing with mere continuity alone, I think you could just bang it out with some estimates. Or perhaps approximate $u$ with $v\in C^1$ uniformly and proceed that way.