# Second derivatives of magnetic potential

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1. Feb 23, 2016

### DavideGenoa

Hello, friends! I have been told that, if $\mathbf{J}$ is of class $C^2$ and $V\subset \mathbb{R}^3$ is a $\mu$-measurable and bounded set, where $\mu$ is the ordinary Lebesgue measure on $\mathbb{R}^3$, then, for all $\mathbf{x}\in\mathbb{R}^3$, $$\mathbf{A}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$where $\mu_0$ is a constant defines a function of class $C^2(\mathbb{R})$. If $\mathbf{J}$ is the density of current, the physical interpretation of $\mathbf{A}$ is that of the magnetic potential.
By using this argument I have not found it difficult to prove, by deriving under the integral sign, that $\mathbf{A}\in C^1(\mathbb{R}^3)$.

How can we prove that the second derivatives of $\mathbf{A}$ belong to $C(\mathbb{R}^3)$?

I have thought that, if $\mathbf{J}\in C^2(\mathbb{R}^3)$ and its support is contained in $V$, we might write $$\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{l}-\mathbf{x}\|}d\mu_{\mathbf{l}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y}+\mathbf{x})}{\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$and try to derive under the integral sign by using the corollary to the dominated convergence theorem that I quote here, but I cannot find a $\varphi$ (notation of the linked post) to use in the desired inequality...
I $\infty$-ly thank you!

2. Feb 28, 2016

### Greg Bernhardt

By using this result, with $f:\mathbf{y}\mapsto \|\mathbf{y}\|^{-1}$ and $g$ as each component of $\frac{\mu_0}{4\pi}\mathbf{J}$, we can see that, if $\mathbf{J}\in C^k(\mathbb{R}^3)$ is compactly supported, then, if $k_1+k_2+k_3=k$, $$\frac{\partial^k\mathbf{A}(\mathbf{x})}{\partial x_1^{k_1}\partial x_2^{k_2}\partial x_3^{k_3}}=\frac{\mu_0}{2\pi}\int_{\mathbb{R}^3}\frac{1}{\|\mathbf{x}-\mathbf{y}\|} \frac{\partial^k\mathbf{J}(\mathbf{y})}{\partial y_1^{k_1}\partial y_2^{k_2}\partial y_3^{k_3}}\, d\mu_{\mathbf{y}}$$and $\mathbf{A}\in C^k(\mathbb{R}^3)$. If I am not wrong...