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While studying on texts of physics I notice that differentiation under the integral sign is usually introduced without any comment on the conditions permitting to do so. In that case, I take care of thinking about what the author is assuming and the usual assumption made in physics that all the functions are of class $C^\infty$, at least piecewise on compact subsets, often is enough to guarantee the liceity of freely commutating the derivative and integral signs.

While studying the derivation of Ampère's law from the Biot-Savart law, someting has surprised me in this proof which seems to be ubiquitous on line and in cartaceous texts. In fact the magnetic field in a point $\mathbf{x}$ is $$\mathbf{B}(\mathbf{x}):=\frac{\mu_0}{4\pi}\iiint_V\mathbf{J}(\mathbf{l})\times\frac{\mathbf{x}-\mathbf{l}}{\|\mathbf{x}-\mathbf{l}\|^3}d^3l=\frac{\mu_0}{4\pi}\iiint_V\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]d^3l$$where I would prove the identity of the integrands at both members by considering the derivatives as... well, ordinary derivatives. I keep Wikipedia's notation except for $\mathbf{x}$, which is more common as a variable, and the norm sign, for which I have always seen $\|\cdot\|$ elsewhere. Then we can notice that the proof uses a differentiation under the integral sign (at $(1)$ below): since $\nabla_x\times\left[\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]\right]=\nabla_x\left[\nabla_x\cdot\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]\right]-\nabla_x^2\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]=\nabla_x\left[\mathbf{J}(\mathbf{l})\cdot\nabla_x\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\right]$ $-\nabla^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})$, where I would calculate the derivatives as ordinarily understood, again, we have that$$\nabla_x\times\mathbf{B}(\mathbf{x})=\nabla_x\times\left[\frac{\mu_0}{4\pi}\iiint_V\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]d^3l\right]$$$$=\frac{\mu_0}{4\pi}\iiint_V\nabla_x\times\left[\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{x}-\mathbf{l}\|}\right]\right]d^3l\quad(1)$$$$=\frac{\mu_0}{4\pi}\iiint_V\nabla_x\left[\mathbf{J}(\mathbf{l})\cdot\nabla_x\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\right]-\nabla_x^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d^3l$$and then the integral is split as licit for Riemann, and Lebesgue, integrals when both integrands are integrable, and the gradient and integral signs are commutated in the first of the two resulting integrals to get$$\frac{\mu_0}{4\pi}\nabla_x\iiint_V\mathbf{J}(\mathbf{l})\cdot\nabla_x\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]d^3l-\frac{\mu_0}{4\pi}\iiint_V\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d^3l$$where the first addend is $\mathbf{0}$ (I do not understand how it is calculated, but that is not the main focus of my question) and where the identity $\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]=-4\pi\delta(\mathbf{x}-\mathbf{l})$, where the derivatives are this time intended as derivatives of a distribution, is used to get$$-\frac{\mu_0}{4\pi}\iiint_V\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d^3l=\mu_0\mathbf{J}(\mathbf{x}).$$ Everything of my reasoning seemed to me to work by assuming $V\subset\mathbb{R}^3$ to be compact and such that $\mathbf{x}\notin V$ and intending the integral $\iiint...d^3l$ to be a Riemann (or Lebesgue, which, in that case, I think to be the same) integral, but at this last step I see that it was not what I thought.

What are, then, the integrals appearing in such calculations? They cannot be Riemann integrals, as far as I understand, because then it must be $\mathbf{x}\notin V$ and then $\iiint_V\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d^3l=\mathbf{0}$, and they cannot be Lebesgue integrals, because, even with $\mathbf{x}\in V$, then $\iiint_V\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d^3l$ $=\int_{V\setminus\{\mathbf{x}\}}\nabla_l^2\left[\frac{1}{\|\mathbf{x}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})\,d\mu_{\mathbf{l}}$ $=\mathbf{0}$, even if $\mathbf{J}(\mathbf{x})$ is not null.

What else if not Riemann or Lebesgue integrals? Why is the commutation of the integral and differential operators licit and what do the derivatives mean in such a context? If we intend them to represent functionals as in the context of functional analysis (which is the only one that I know of where Dirac's $\delta$ is defined), which function ($\varphi$, to use the notation used here) is the argument of the functional and what does the functional maps it to?

What are the derivatives expressed by $\nabla$? Since theorems such as Stokes' are usually applied when integrating $\nabla\times\mathbf{B}$, I would believe that they are the ordinary derivatives of elementary multivariate calculus, but then the $\delta$, which is a tool of the theory of distributions, pops up in the outline of proof, and in the theory of distributions there exist derivatives of distributions which are a very different thing, but they are taken, as far as I know, with respect to the variables written as "variables of integration" in the distribution integral notation, while we start with $\nabla_r\times \mathbf{B}$ with $r$ , while the integral appears with $d^3l$...

Or is that one of those cases, whose set I have been told not to be empty, where physics methods, at least at the didactic level, are not as rigourous as mathematics would require? I admit that I was rather inclined to think so until a user of PSE told me, without explaining how to interpretate the integrals and justify the steps, that the quoted proof is rigourous. I heartily thank any answerer.

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    Start by looking at the Helmholtz Theorem for vector fields. https://en.wikipedia.org/wiki/Helmholtz_decomposition . You can use this in conjunction with $d^2=0$ types of identities to do most things in a classical way. Though a delta function is nice to learn about, the machinery to argue properly often isn't worth it, especially when classical approaches are available that work generally for most of what you'll ever want to do. – Disintegrating By Parts Jan 29 '16 at 23:25
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    Take a closer look at this argument to see what is really meant by such expressions: http://math.stackexchange.com/questions/1625017/diracs-delta-in-3-dimensions-proof-of-nabla2-boldsymbolx-boldsymbolx/1625172#1625172 . That's why I left what I posted there. :) – Disintegrating By Parts Jan 30 '16 at 13:55
  • @TrialAndError Thanks again! I think I understood your proof, which I think to be essentially the same as Daniel's, but my doubts remains: which part of the integrands is $\varphi$ in the above derivation of Ampère's law, why does the differential operators commute with the integral signs, what are those integrals? The (physics) resources I've had access to don't explain that and I haven't been able to receive explanations in physics fora, so that I'm beginning to suspect that it might even be a sort of artifice compatible with experimental results but not so grounded in mathematical rigour... – Self-teaching worker Jan 30 '16 at 14:25
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    What I gave you shows a left inverse for $\nabla^2$. And I never interchanged derivatives and the integral. That's how Helmholtz Theorem is proved: by a left inverse that starts with $\nabla\cdot \vec{F}$ and $\nabla\times \vec{F}$ to reconstruct $\vec{F}$. – Disintegrating By Parts Jan 30 '16 at 14:30
  • @TrialAndError I meant "why does the differential operators commute with the integral signs" in the derivation of derivation of Ampère's law in the OP. As to what you mean, forgive me if sometimes it's hard for me to understand comments, which aren't by their nature exhaustive... Are you saying that the use of the $\delta$ in proofs such as that of Ampère's law may be avoided, since Ampère's law can be proved by using the Helmholtz decomposition, which can in turn be proved by using the ordinary Riemann or Lebesgue integrals only? [...] – Self-teaching worker Jan 30 '16 at 15:06
  • [...] If the Helmholtz decomposition can be proved in such a way, I think it would be worth a separate question in MSE... Moreover, are you suggesting that the $\delta$'s use may be avoided because the derivation that I wrote in the OP isn't mathematically rigourous (as I've hypothesised)? – Self-teaching worker Jan 30 '16 at 15:06
  • Physicists are the Mathematical Devils. – Count Iblis Jan 31 '16 at 17:54

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The difference between the way that Mathematicians and Physicists handle such things is based in an underlying assumption of Physics that solutions of Physical equations exist. The Mathematician would say that, if you have a $C^2$ scalar field $F$ in a region of space, then you can reconstruct $F$ if you know $\nabla^2F$. First, for $r \ne r'$, $$ \frac{1}{|r-r'|}\nabla'^2F(r')=\frac{1}{|r-r'|}\nabla'^2F(r')-F(r')\nabla'^2\frac{1}{|r-r'|}\\ = \nabla'\cdot\left[\frac{1}{|r-r'|}\nabla'F(r')-F(r')\nabla'\frac{1}{|r-r'|}\right] $$ The way this is done is with the integration rules of Calculus: $$ \frac{1}{4\pi}\int_{V}\nabla^2 F(r')\frac{1}{|r-r'|}dV(r') \\ =\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\nabla^2F(r')\frac{1}{|r-r'|}dV(r')\\ = \lim_{\epsilon\downarrow 0}\left[\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon}(r))}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r') -\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon})}F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r')\right] \\ = -F(r)+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}(r')dS(r')-\frac{1}{4\pi}\int_{\partial V}F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r') $$ Therefore, the scalar field can be reconstructed if $\nabla^2F$ and $\frac{\partial F}{\partial n}$ are know on some nice region $V$: \begin{align} F(r)& =\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r'-r|}\frac{\partial F}{\partial n}(r')dS-\frac{1}{4\pi}\int_{\partial V}F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r') \\ & -\frac{1}{4\pi}\int_{V}\frac{1}{|r'-r|}\nabla^2F(r')dV(r').\;\;\;\; (*) \end{align} But the Physicist has more information. The Physicist assumes that they're dealing with a scalar field with known volume and surface density functions $\rho$ and $\sigma$. And therefore it must be that $$ \nabla^2\left[F(r)=\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r'-r|}\sigma(s')dS+\frac{1}{4\pi}\int_{V}\frac{1}{|r'-r|}\rho(r')dV(r')+\cdots\right] \\ =\nabla^2\left[\frac{1}{4\pi}\int_{V}\frac{1}{|r'-r|}\rho(r')dV(r')\right]=-\rho(r) $$ So the Mathematician has a left inverse $L\nabla^2F = F$ from which the Physicist obtains a right inverse: $\nabla^2 L\nabla^2F = \nabla^2F$. This is a full right inverse because, based on Physical grounds, every reasonable density $\rho$ can be written as $\rho=-\nabla^2F$. Hence $\nabla^2 L\rho = -\rho$. Another way of saying it: If $\rho$ is a nice density function (continuous or piecewise continuous, for example) for which there exists some $F$ such that $\nabla^2F=-\rho$, then $$ \nabla^2 \left[\frac{1}{4\pi}\int_{V}\frac{\rho(x')}{|x-x'|}dV(x')\right]=-\rho(x) \;\;\;\; (\dagger) $$ Why? Because $F$ has the above integral representation (*) involving $\nabla^2F$ and the values of $F$, $\frac{\partial F}{\partial n}$ on the boundary, which forces $(\dagger)$ by the argument given above. It's not so hard to construct some such $F$, for example, if $\rho$ is smooth: extend $\rho$ to a cube and use Fourier techniques. All you need to know is there exists some $F$, and the precise boundary information does not matter in order to argue that $(\dagger)$ must hold based on the representation $(*)$ for $F$.

The Mathematician would insist that you prove the existence of such a solution, while the Physicist knows how the world works and doesn't have to ask if nature knows how to solve the equations. :) A solution exists--let's go find it. The Physicst's approach is a productive one; you just need to be aware of the assumptions.

The Helmholtz Theorem of Mathematics is this: every smooth vector field $\vec{F}$ on a nice region can be reconstructed from $\nabla\cdot\vec{F}$ and $\nabla\times\vec{F}$: $$ \begin{align} \vec{F}(x,y,z)=&-\nabla\left[\int_{V}\frac{\nabla'\cdot \vec{F}(\vec{x}')}{4\pi|\vec{x}-\vec{x}'|}dV'-\oint_{S}\frac{\vec{F}(\vec{x}')\cdot\hat{n}}{4\pi|\vec{x}-\vec{x}'|}dS'\right] \\ &+\nabla\times\left[\int_{V}\frac{\nabla'\times \vec{F}(\vec{x}')}{4\pi|\vec{x}-\vec{x}'|}dV'+\oint_{S}\frac{\vec{F}(\vec{x}')\times\hat{n}}{4\pi|\vec{x}-\vec{x}'|}dS'\right] \end{align} $$ I'll let you construct your own right inverses from the left inverse given above. The end result is that it looks like you're interchanging differentiation operators with the integrals, but you're really not. You're assuming a solution of an equation and deriving the required form from the vector identities, knowing that you have a solution. You automatically know that when you apply the differential operators to the solution, you get back what you started with. And, of course, you're using operator identities such as $\nabla\cdot\nabla\times \vec{F}=0$ and $\nabla\times \nabla f=0$ to eliminate terms from the above when you apply differential operators.

These powerful techniques generalize to Differential Geometry. But the roots of this subject lie in Heavside's vectorization of Maxwell's Equations. There were a lot of geniuses involved in this work.

Disintegrating By Parts
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  • Wow! I think it will take some time for me to understand everything you say, but I will try to do so because you present tremendously interesting topics! Forgive me if I don't understand a thing in particular: I completely miss the relation of what you say and the problems I have raised concerning the derivation of Ampère's law given in Wikipedia. Are you suggesting an alternative way to prove it? (in that case who is who in the equations you wrote?) [...] – Self-teaching worker Jan 30 '16 at 16:59
  • [...] Is Wikipedia's proof rigourous? If it is, what are those integrals, as I ask in the OP? (I don't think they are Lebesgue/Riemann integrals... are they?) Thank you very much again!!! – Self-teaching worker Jan 30 '16 at 17:00
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    @Self-teachingDavide : Start with observations about $\nabla\times H$ and $\nabla\cdot B$, and see how that affects the representation. – Disintegrating By Parts Jan 30 '16 at 18:45
  • I'll proceed with order trying to understand the equalities you wrote, which, if I correctly understand, are necessary to find the answers to my questions in the OP. The first problem I find in understanding them is this. Thank you in advance both if you can answer me there or if you cannot! – Self-teaching worker Jan 30 '16 at 21:08
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    @Self-teachingDavide : For some reason I am inclined to always drop this one particular integral term. I've added it back in for you. Helmholtz Theorem I copied from a text; so I didn't forget terms there. – Disintegrating By Parts Jan 30 '16 at 21:40
  • I don't understand the minus sign in $\nabla^2 L\rho=-\rho$, but I don't think that it's relevant to answer the questions I pose in the OP... The main problem I have at this stage of my study of your answer is the Helmholtz decomposition, about which I ask here. I heartily thank you if you can answer me there! – Self-teaching worker Jan 31 '16 at 14:26
  • @Self-teachingDavide : Don't worry so much about the $L$. That part was a way to discussing how, if you know $\nabla^2 F$ (and boundary information) then you can reconstruct $F$ using various integrals, which is why when you start with integrals of $\rho$ (where $\rho=\nabla^2F$) then you end up with a function $G$ such that $\nabla^2G = \rho$, regardless of the boundary integrals. And that explains how to approach your problem to get the end result you want, without some interchange of differentiation and integration using a $\delta$ function. – Disintegrating By Parts Jan 31 '16 at 16:27
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    @Self-teachingDavide : I added more about the point of a left inverse. – Disintegrating By Parts Jan 31 '16 at 16:45
  • $\infty$ thanks and forgive me again, I'm study these things by myself and know nothing of many things you says (such as Fourier analysis outside $\mathbb{R}$)... Why does the term $\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F (r')}{\partial n'}$ $-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS'$ of $F(r)=\frac{1}{4\pi}\int_{V}\frac{\rho(r')}{|r-r'|}dV'$ $+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F(r')}{\partial n'}$ $-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS'$ disappear at $(\dagger)$? – Self-teaching worker Jan 31 '16 at 17:12
  • As to the Helmholtz decomposition, which I cannot prove to myself, is it necessary to answer my OP questions? Thanks again! – Self-teaching worker Jan 31 '16 at 17:12
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    @Self-teachingDavide : For the boundary integrals, you can easily interchange differentiation in $x$ with the integral, provided $x$ is inside the volume. There are no $\delta$ issues there. As for the Helmholtz Theorem, you need to find a reference because the proof uses many vector identities, and you end up with integrals where normals are crossed with vectors; it's not a simple proof, but definitely worth knowing for Physics. It shows you what to expect of Physics representations of magnetic field, especially. – Disintegrating By Parts Jan 31 '16 at 17:47
  • Wow, very interesting, thank you so much! Then, if $F$ and $V$ are as described here, then $\nabla^2\left[\frac{1}{4\pi}\int_V\frac{\nabla'^2F(r')}{|r-r'|}]dV'\right]=- \nabla^2 F(r)$ (where the integral is intended as a Lebesgue integral, or as a Riemann $\lim_{\epsilon\to0}\int_{V\setminus B_\epsilon(r)}$) which, although I don't think it to be enough to prove Ampère's law or justify the steps of Wikipedia's proof, is a tremendously interesting result! – Self-teaching worker Jan 31 '16 at 19:32
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    @Self-teachingDavide : Now look at the Helmholtz Theorem, which allows you to work with gradient and curl. By the way, the Laplace result is needed to boostrap up to the Helmholtz Theorem. $\nabla\cdot B=0$. – Disintegrating By Parts Jan 31 '16 at 19:38
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    If $F\in C^2(\mathbb{R}^3)$ is compactly supported, I think (hope) I have been able to prove the Helmholtz theorem here. If $F$ is not compactly supported, I have no idea how to move the $\nabla$ under the integral signs... Do you know a text where it is rigourously proved with the explanation of the various steps? – Self-teaching worker Mar 09 '16 at 16:00
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    @Self-teachingworker : Once you prove the representation for the Laplacian, you can get it by applying it to vector functions. I've basically explained how to get the representation using the Laplacian. Then you can consult a good Engineering text that derives the full case from that one. The Engineers can help you with you with this. You're not going to find the $\mathbb{R}^3$ case as a special case of manifolds or higher dimensions, even though this result is also known as "The Fundamental Theorem of Vector Calculus." https://en.wikipedia.org/wiki/Helmholtz_decomposition – Disintegrating By Parts Mar 09 '16 at 17:30
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    Thank you for your comment! The problem is that all "proofs" aimed at physics/engineering students I've seen nonchalantly commute $\nabla$ and $\int$ without a word to explain what theorem allows that and what the integral sign means (Lebesgue, distribution or other), in a way that I've verified to lead not few students to believe that such commutation is always allowed and Dirac's $\delta$ can be introduced to get almost any result one wants to derive. As rigourous as dividing by $0$... – Self-teaching worker Mar 09 '16 at 18:19
  • @Self-teachingworker : I gave you a way to deal with the $\nabla^2$ case using an improper Riemann integral, assuming you have enough smoothness to invoke classical Stokes/Divergence Theorems. Everything can be done in that context, and I recommend working there. With the $\nabla^2$ result, you can can get the general result from Engineering texts, ignoring their use of $\nabla^2 \cdots = \delta$ and replacing with what I gave you instead. The result of the Engineering details are rigorous after that, provided you choose a decent Engineering text. Dist. Theory here is excessive IMHO. – Disintegrating By Parts Mar 09 '16 at 18:48
  • By applying some vector identities, commuting $\int_{\mathbb{R}^3\setminus V}$ and $\nabla_r$, which can be done thanks to Leibniz's rule for a compact domain outside of which $\vec{F}$ is 0, and the divergence theorem, I think I've been able to prove (here) the Helmholtz theorem in the form you quote for te particular case of $\vec{F}\in C^2(\mathbb{R}^3)$ compactly supported, and $r\in\mathring{V}$. Is it intended to hold for more general cases? Thank you so much both if you can answer me or if you cannot! – Self-teaching worker Mar 15 '16 at 17:31
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    @Self-teachingworker : You should be able to work with a specific decay rate on $\vec{F}$ if you want to extend to free space. I seem to recall that it turns out the Physics-motivated estimates are good enough to be able to take a limit in order to get rid of the boundary integral terms. – Disintegrating By Parts Mar 15 '16 at 18:04
  • The problem is that, although I do see that, given $G:\mathring{A}\subset\mathbb{R}^3\to \mathbb{R}^3$, if $\bar{V}\subset\mathring{A}$ satisfies the assumptions of the divergence theorem, $x\in\mathring{V}$ and $\exists F\in C^2(\mathring{A}):G=\nabla^2 F$, then $$G(x)=\frac{1}{4\pi}\nabla\times\left(\nabla\times\int_V\frac{G(x')}{|x-x'|}d\mu'\right)-\frac{1}{4\pi}\nabla\left(\nabla\cdot\int_V\frac{G(x')}{|x-x'|}d\mu'\right)$$which already is an interesting version of the Helmholtz decomposition, [...] – Self-teaching worker Apr 03 '16 at 14:43
  • [...] I have no idea of what can allow to "transform" $\nabla\times\int_E|x-x'|^{-1}$ into $\int_E|x-x'|^{-1}\nabla'\times$ and $\nabla\cdot\int_E|x-x'|^{-1}$ into $\int_E|x-x'|^{-1}\nabla'\cdot$... – Self-teaching worker Apr 03 '16 at 14:44
  • @Self-teachingworker : Are you following the Wikipedia proof? – Disintegrating By Parts Apr 03 '16 at 15:37
  • Thank you for the comment! No, I am not following the Wikipedia proof at all, in particular where it commutes $\nabla$ and $\int_V$ at the 6-th line... – Self-teaching worker Apr 03 '16 at 15:50
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    @Self-teachingworker : The derivatives are in $L^1$ and the function $F$ is in $L^{\infty}$. I believe that this is true when you're looking at the curl. – Disintegrating By Parts Apr 03 '16 at 16:57
  • I think (I hope) I've found a proof. If you were interested in answering me with a confirmation, if it's correct, or a correction, if it's wrong, I would accept your answer confirming or correcting the proof I've written here... – Self-teaching worker Apr 06 '16 at 21:56
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    @Self-teachingworker : I'm having trouble following the links. I think some of the links take me to the wrong place. Is the solution you're looking start by discussing $F=\nabla^2G$? If so ... note that because all integrals can be localized, you only need a local solution of this equation, which is easily achieved with a Fourier series, for example. That means Fourier conditions of Holder continuity or some other mild condition should be enough. I have my doubts about continuous $F$ only (even though I see such claims,) but any little bit beyond that (as mentioned in this comment) is enough. – Disintegrating By Parts Apr 06 '16 at 23:17
  • As to $F\in C^k(A)$ for some $k$ as a sufficient condition for the decomposition to hold, well, although I see a lot of enunciates assuming some "smoothness" with no further specifications, but then, when there is a proof, it appears to be an informal non-proof, like Wikipedia's, with a lot of unexplained commutations $\int/\nabla$, the introduction of $\delta$ with no explanation of why we can introduce it... a horrible way of "proving" typical, sadly, of not-so-few textbooks of physics, even graduate ones... – Self-teaching worker Apr 08 '16 at 16:21
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    Yes, it's the answer beginning with "Thanks to a conversation I have had with user...": http://math.stackexchange.com/questions/1634520/helmholtz-theorem/1730859#1730859 . I've corrected the links. I think that, by locality, you mean that $\exists G\in C^3(A)$ with $A$ open such that $\forall x\in A\quad$ $F(x)=\nabla^2 G(x)$, as I wrote in the linked answer: am I right? – Self-teaching worker Apr 08 '16 at 16:21
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I hope I have been able, thanks to Daniel Fischer's proof of this very interesting result, to find a proof of Ampère's law from the Biot-Savart law based on the interpretation of the integral representing the magnetic potential$$\frac{\mu_0}{4\pi}\iiint_Vd^3l\frac{\mathbf{J}(\mathbf{r})}{|\mathbf{r}-\mathbf{l}|}$$ as a Lebesgue integral $$\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}(\mathbf{r})}{\|\mathbf{r}-\mathbf{l}\|}\,d\mu_{\mathbf{l}}$$where $\mu_{\mathbf{l}}$ is the usual tridimensional Lebesgue measure defined on $\mathbb{R}^3$.

Following this reasoning, I have reached an interpretation of the integrals and differennt signs of Wikipedia's (and Jackson's Classical Electrodynamics's) outline of proof as Lebesgue integrals and ordinary derivatives, except for the integral where $\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$ appears: I suppose that $$\iiint_V d^3l\mathbf{J}(\mathbf{l})\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$$ is to be interpretated, differently from all the other integrals in that outline of proof, as a symbolic notation for the linear operator whose components are the Laplacians $\nabla^2 T_f$ of the linear functional defined by, $f:\mathbf{l}\mapsto\|\mathbf{r}-\mathbf{l}\|^{-1}$, i.e. $$T_f:J_i\mapsto\int_V\frac{J_i(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}.$$Such Laplacian is such that, for $J_i$ compactly supported within $V$, $$\nabla^2T_f(J_i)=\int_V\frac{\nabla_l^2J_i(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}=-4\pi J(\mathbf{r})=\int\delta(\mathbf{x}-\mathbf{r})J_i(\mathbf{x})$$which, as never stressed enough, is not at all, in general, the same as $$\int_V\nabla_r^2\left(\frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\mathbf{J}(\mathbf{l})d\mu_{\mathbf{l}}\equiv\mathbf{0}.$$

I admit that requiring the reader to realise read them as Lebesgue integrals except for the integral where $\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$ appears.

That would be a tremendously confusing notation for me (even more with that performing an integration by parts), and, I suspect, for many other readers, in particular students. I am not sure that the authors using that outline of proof require such a care to distinguish different meanings of the same integral notations, used in the very same equality, from their readers.

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