Let $H:\mathbb{R^3}\to \mathbb{C^3}$ such that H is continuously differentiable in $\mathbb{R^3}$. Denote the sphere of radius $R$ centered around origin by $\Omega_{R}$. Considering $\mathbf{r}$ to be the radial coordinate and $\hat{r}$ to be the unit vector in the radial direction. How to show that the closed surface integral
$$\int_{\Omega_{R}} \frac{H(\mathbf{r})}{|\mathbf{r}|^3} \times \hat{r} \ dS$$
converges to the zero vector as $R\to 0$?
It is sufficient to show that
$$\left\Vert \int_{\Omega_{R}} \frac{H(\mathbf{r})}{|\mathbf{r}|^3} \times \hat{r} \ dS \right\Vert \to 0 $$ as $R$ goes to $0$. But then I do not know how to proceed from there.
Another way would be to show that each component goes to zero. For instance, if $H(\mathbf{r})= ( H_x(\mathbf{r}) , H_y(\mathbf{r}) , H_z(\mathbf{r}) )$ and $\hat{r} = (r_x,r_y,r_z)$, then the integrand for the x-component of the integral is $H_y(\mathbf{r})r_z - H_z(\mathbf{r})r_y $. If we assume that $H_y(\mathbf{r}) = H_y(\mathbf{0}) + \varepsilon_y(\mathbf{r})$ and $H_z(\mathbf{r}) = H_z(\mathbf{0}) + \varepsilon_z(\mathbf{r})$, then for $\mathbf{r} \in \Omega_R $
$$ \left|\frac{H_y(\mathbf{r})r_z - H_z(\mathbf{r})r_y}{|\mathbf{r}|^3} - \frac{H_y(\mathbf{r_0})r_z - H_z(\mathbf{r_0})r_y}{|\mathbf{r}|^3} \right|= \left|\frac{\varepsilon_y(\mathbf{r})r_z - \varepsilon_z(\mathbf{r})r_y}{|\mathbf{r}|^3} \right| \leq \frac{|\varepsilon_y(\mathbf{r})r_z| + |\varepsilon_z(\mathbf{r})r_y|}{|\mathbf{r}|^3} < \frac{|r_z| + |r_y|}{|\mathbf{r}|^3} \varepsilon $$
where $\varepsilon$ is the maximum of the supremum (over $\Omega_R$) of $|\varepsilon_y(\mathbf{r})|$ and $|\varepsilon_z(\mathbf{r})|$. The above bound cannot be used to show that the integral goes to zero because after integration we will be left with $|\mathbf{r}|$ in the denominator.
The claim is wrong. Consider the field $${\bf H}({\bf r}):=(y,-x,0)\ .$$ Using the parametrization $${\bf r}(\phi,\theta):=R(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$$ of $\Omega_R$, whereby $\phi$ and $\theta$ are geographical coordinates, we have $$\hat r=(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta),\qquad {\rm d}\omega=R^2\cos\theta\>{\rm d}(\phi,\theta)\ ,$$ and $${\bf H}\bigl({\bf r}(\phi,\theta)\bigr)\times\hat r=R(-\cos\theta\sin\theta\cos\phi,\ -\cos\theta\sin\theta\sin\phi,\ \cos^2\theta)\ .$$ The first two components of the integral $$\int_{\Omega_R}{{\bf H}({\bf r})\times\hat r\over|{\bf r}^3|}\>{\rm d}\omega$$ then vanish trivially, and for the third component we obtain $${1\over R^3}\cdot 2\pi R\cdot \int_{-\pi/2}^{\pi/2}\cos^2\theta\cdot R^2\cos\theta\>d\theta={8\pi\over3}\ ,$$ and this does not converge to $0$ when $R\to0+$.