How to show the linear operator $T_{D,k}[\phi](x)=\int_{\partial D}\frac{|x-y|^{k-1}}{k!}\phi(y) d\sigma(y)$ is uniformly bounded? Let $D$ be a bounded and simply connected domain in $\mathbb{R}^3$ with a continuously differentiable boundary $\partial D$ and let $T_{D,k}$ be the family of linear operators given by
$$
T_{D,k}[\phi](x) = \int_{\partial D} \frac{|x - y|^{k-1}}{k!}\phi(y) d\sigma(y), \quad \quad x \in \partial D.
$$
I read that this family of operators is uniformly bounded with respect to $k$ in the norm $||T_{D,k}||_{\mathcal{L(L^2(\partial D), H^1(\partial D)})}$.
How can this be shown?
 A: Assume $k\geq 2$. Let $\phi\in \mathcal{S}$ a Schwartz function. Since $\partial D$ is continuously differentiable, then without lost of generality we may assume $\Gamma_i=\partial D\cap B(x_i, r) \subset \mathbb{R}^2$, i.e. locally flatten out.  Observe you have
\begin{align}
\bigg\|\frac{1}{k!} \int_{\partial D}|x-y|^{k-1}\phi(y)\ d\sigma(y)\bigg\|_{L^2(\Gamma)}^2 =&\ \frac{1}{k!}\int_\Gamma\left|\int_{\partial D}|x-y|^{k-1}\phi(y)\ d\sigma(y) \right|^2dx\\
\leq&\ \frac{1}{k!} \int_\Gamma \left(\int_{\partial D}|x-y|^{2k-2}\ d\sigma(y) \right)dx\left(\int_{\partial D}|\phi(y)|^2 d\sigma(y)\right)\\
\leq&\ \frac{\operatorname{Area}(\partial D)^2\operatorname{diam}(\partial D)^{2k-2}}{k!}\|\phi\|^2_{L^2(\Gamma)} 
\end{align}
and likewise
\begin{align}
\bigg\|\frac{1}{k!} \int_{\partial D}\nabla_x|x-y|^{k-1}\phi(y)\ d\sigma(y)\bigg\|_{L^2(\Gamma)}^2 =&\ \bigg\|\frac{k-1}{k!} \int_{\partial D}|x-y|^{k-3}(x-y)\phi(y)\ d\sigma(y)\bigg\|_{L^2(\Gamma)}^2\\
\leq&\ \frac{(k-1)\operatorname{Area}(\partial D)^2\operatorname{diam}(\partial D)^{2k-4}}{k!}\|\phi\|_{L^2(\Gamma)}^2.
\end{align}
Hence it follows
\begin{align}
\bigg\|\frac{1}{k!} \int_{\partial D}|x-y|^{k-1}\phi(y)\ d\sigma(y)\bigg\|_{H^1(\partial D)} \leq C\|\phi\|_{L^2(\partial D)}
\end{align}
where $C$ is independent of $k$ since both the above constants are uniformly bounded in $k$. 
