Show that $|\int_{S_r} {\langle\nu,\varphi\rangle f} \, d\sigma| \rightarrow 0$ when $r\to0$. Let $f\in C^1(\mathbb{R}^n\setminus\{0\})$ and assume that $f\in L^2_\text{loc}(\mathbb{R}^n)$ and $\nabla f\in L^2_\text{loc}(\mathbb{R}^n)$. Show that $$\left|\int_{S_r} {\langle\nu,\varphi\rangle f} \, d\sigma\right| \rightarrow 0,\qquad r\to0, \varphi\in C^{\infty}_{0}(\mathbb{R}^n)$$
where $S_r=\{x\in\mathbb{R}^n: |x|=r\}$ and $\sigma$ is the area on the sphere $S_r$, $\nu$ is the unit normal vector fields on $S_r$ pointing at $o$.
My approach: I suppose that
$$\left|\int_{S_r} {\langle\nu,\varphi\rangle f} \, d\sigma\right|\leq \int_{S_r} {|\langle\nu,\varphi\rangle| |f|} \, d\sigma\leq c\int_{S_r}{|f|d\sigma}$$
From cauchy inequality $|\langle \nu,\varphi\rangle|\leq |\nu||\varphi|\leq c$, where $c$ is a constant.
I define $h(r)=\left(\int_{S_r} |f| \, d\sigma \right)^2$, then by Cauchy-Scharwz inequality I get,
$$h(r)=\left(\int_{S_r}{|f| d\sigma}\right)^2\leq \left(\int_{S_r}{d\sigma}\right)\left(\int_{S_r}{|f|^{2} d\sigma}\right)\leq\sigma(S_r)\int_{S_r}{|f|^{2} d\sigma}=\omega_{n}r^{n-1}\int_{S_r}{|f|^{2} d\sigma}$$
Where $\omega_{n}$ is the area of the $(n-1)-$sphere on $\mathbb{R}^n$. Since $f\in L^{2}_{loc}(\mathbb{R}^n)$, we have $$\frac{h(r)}{\omega_{n}r^{n-1}}\leq \int_{S_r}{|f|^{2} d\sigma}<\infty$$
But $f\in L^2_\text{loc}(\mathbb{R}^n)$, so the integral is finite then, $F(r)<\infty$, and now If I take $r\to 0$, how can prove that $F(r)\to 0$. Any hint will be appreciated. Thanks!
 A: In the new version of the question, the integral is the flux of the vector field $f\mathbf{\varphi}$ ($f$ real-valued with $f\in L^2_{\textrm{loc}}$ and $\nabla f\in L^2_{\textrm{loc}}$, $\mathbf{\varphi}$ vector-valued and smooth) through the surface $S_r$. The Divergence Theorem for this case shows that
\begin{equation}
\begin{split}
\int_{S_r}f\mathbf{\varphi}\cdot\mathbf{\nu}d\sigma
&=~
\int_{B_r}\mathsf{div}(f\mathbf{\varphi})dV\\
&=~ \int_{B_r}\left[\nabla f\cdot\mathbf{\varphi} + f\mathsf{div}(\mathbf{\varphi})\right]dV,
\end{split}
\end{equation}
where $B_r$ is the ball of radius $r$. We will use the local $L^2$ properties of $f$ and $\nabla f$ to make bounds that depend on $r$.
\begin{equation}
\begin{split}
\left|\int_{B_r}\nabla f\cdot\mathbf{\varphi}dV\right|
&\leq~
\int_{B_r}\left|\nabla f\cdot\mathbf{\varphi}\right|dV\\
&\leq~ \left[\int_{B_r}\left\Arrowvert\nabla f\right\Arrowvert^2dV\right]^{1/2}\left[\int_{B_r}\left\Arrowvert\mathbf{\varphi}\right\Arrowvert^2dV\right]^{1/2}
\end{split}
\end{equation}
by Cauchy-Schwarz.
Since $\left\Arrowvert\nabla f\right\Arrowvert^2\geq 0$, the integral over $B_r$ is less than or equal to the integral over $B_R$ for $r < R$:
\begin{equation}
\left[\int_{B_r}\left\Arrowvert\nabla f\right\Arrowvert^2dV\right]^{1/2}~
~\leq~
\left[\int_{B_R}\left\Arrowvert\nabla f\right\Arrowvert^2dV\right]^{1/2}
~<~ \infty.
\end{equation}
Since $\mathbf{\varphi}$ is smooth, $\left\Arrowvert\mathbf{\varphi}\right\Arrowvert$ achieves a maximum on each ball, in particular on $B_R$. Because of this, we can write, for $r < R$,
\begin{equation}
\begin{split}
\left[\int_{B_r}\left\Arrowvert\mathbf{\varphi}\right\Arrowvert^2dV\right]^{1/2}
&\leq~ \max_{x\in B_r}\left\Arrowvert\mathbf{\varphi}(x)\right\Arrowvert\times\textrm{vol}(B_r).
\end{split}
\end{equation}
The volume of $B_r$ shrinks to 0 as $r\to 0$, and
\begin{equation}
\begin{split}
\left|\int_{B_r}\nabla f\cdot\mathbf{\varphi}dV\right|
&\leq~ \left[\int_{B_R}\left\Arrowvert\nabla f\right\Arrowvert^2dV\right]^{1/2}\times\max_{x\in B_r}\left\Arrowvert\mathbf{\varphi}(x)\right\Arrowvert\times\textrm{vol}(B_r)\longrightarrow 0~\textrm{as}~r\to 0.
\end{split}
\end{equation}
We apply almost identical arguments to $f\mathsf{div}(\mathbf{\varphi})$, starting with Cauchy-Schwarz:
\begin{equation}
\begin{split}
\left|\int_{B_r}f\mathsf{div}(\mathbf{\varphi})dV\right|
&\leq~ \left[\int_{B_r}|f|^2dV\right]^{1/2}\left[\int_{B_r}\left\Arrowvert\mathsf{div}(\mathbf{\varphi})\right\Arrowvert^2dV\right]^{1/2}.
\end{split}
\end{equation}
Since $f\in L^2_{\textrm{loc}}$, the integral of $|f|^2$ on $B_r$ is bounded by its (finite) integral on $B_R$ for $r < R$. Since $\mathbf{\varphi}$ is smooth, $\left\Arrowvert\mathsf{div}(\mathbf{\varphi})\right\Arrowvert^2$ achieves a maximum on $B_R$, and its integral over $B_r$ ($r < R$) is bounded by this maximum times the volume of $B_r$.  The volume of $B_r$ shrinks to 0 as $r\to 0$.
