3.3.3 Regularity of ball averages

Write $v_ d=|B(0,1)|=s_{d-1}/d$. For an open set $\Omega \subset \mathbb R^ d$ and $r>0$, put $\Omega _ r=\{ x\in \Omega :\operatorname {dist}(x,\Omega ^ c)>r\} $ and

Theorem 3.3.4 (Ball-average regularity).F549D908E93A If $f\in L^1_{\mathrm{loc}}(\Omega )$, then $A_ rf$ is continuous on $\Omega _ r$. If $f\in C^ k(\Omega )$, where $k\geq 0$ is an integer, then $A_ rf\in C^{k+1}(\Omega _ r)$. For continuous $f$ and any $\eta \in \mathbb R^ d$,

6CEBA4319BDE 3.3.5

\begin{equation} D_\eta A_ rf(x)=\frac1{v_ dr}\int _{S^{d-1}} f(x+r\theta )(\eta \cdot \theta )\, dS(\theta ). \end{equation}

Proof. Fix $x_0\in \Omega _ r$ and choose $\varepsilon >0$ so that $\overline{B(x_0,r+\varepsilon )}\subset \Omega $. For $|x-x_0|<\varepsilon $,

\[ |A_ rf(x)-A_ rf(x_0)|\leq \frac1{v_ dr^ d} \int _{B(x,r)\, \triangle \, B(x_0,r)}|f(y)|\, dy. \]

The symmetric difference lies in $\{ y:r-|x-x_0|\leq |y-x_0|\leq r+|x-x_0|\} $, whose volume tends to zero. Absolute continuity of the integral of $|f|$ on the fixed compact ball therefore proves continuity.

Suppose now that $f$ is continuous. We prove the formula for a coordinate direction $e_ i$ by slicing the ball perpendicular to $e_ i$. For $w\in e_ i^\perp $, $|w|<r$, set $a(w)=\sqrt{r^2-|w|^2}$. Fubini and translation in the $i$th coordinate give, for small nonzero $h$,

\begin{align*} v_ dr^ d\bigl (A_ rf(x+he_ i)-A_ rf(x)\bigr ) =\int _{|w|<r}\biggl (& \int _{a(w)}^{a(w)+h}f(x+w+se_ i)\, ds\\ & -\int _{-a(w)}^{-a(w)+h}f(x+w+se_ i)\, ds\biggr )\, dw. \end{align*}

The formula uses signed integrals and holds for either sign of $h$. On a fixed compact neighborhood of the closed ball, $f$ is uniformly continuous and bounded. Divide by $h$ and let $h\to 0$; dominated convergence on the $(d-1)$-dimensional disk yields

\[ v_ dr^ d\partial _ iA_ rf(x)=\int _{|w|<r} \bigl (f(x+w+a(w)e_ i)-f(x+w-a(w)e_ i)\bigr )\, dw. \]

The graph area formula on the two hemispheres gives $\nu _ i\, dS_ r=dw$ on the upper graph and $\nu _ i\, dS_ r=-dw$ on the lower graph, where $\nu $ is the outward unit normal of $\partial B(0,r)$. The equator has surface measure zero. Consequently the last integral is $\int _{\partial B(0,r)} f(x+y)\nu _ i(y)\, dS_ r(y)$. Substitute $y=r\theta $: the surface element supplies $r^{d-1}$ and $\nu _ i(r\theta )=\theta _ i$. Dividing by $v_ dr^ d$ proves (3.3.5) for $\eta =e_ i$. The right side is continuous in $x$, by uniform continuity on compact sets; hence all coordinate derivatives are continuous and $A_ rf\in C^1$. Linearity in $\eta $ gives the stated directional formula.

If $f\in C^ k$, repeated differentiation under this compact sphere integral, justified by the same uniform convergence argument as in Lemma 3.3.2, shows that each $\partial _ i A_ rf$ has continuous derivatives of order at most $k$. Thus $A_ rf\in C^{k+1}$. The argument includes $k=0$ and $f=0$. $\square$

Formula (3.3.5) includes the surface dilation factor missing in the printed proof: the coefficient of the unit-sphere integral is $1/(v_ dr)$, rather than $1/(v_ dr^ d)$. For $f(x)=x_ i$ it gives $\partial _ iA_ rf=1$ for every radius, as required.

3.3.3 Regularity of ball averages`AABF9B330CC1`

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3.3.3 Regularity of ball averagesAABF9B330CC1

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3.3.3 Regularity of ball averages`AABF9B330CC1`