Continuity of a certain vector field

Question

Let us define $$\boldsymbol{E}(\boldsymbol{x}):=\lim_{\varepsilon\to 0}\int_{D\setminus B(\boldsymbol{x},\varepsilon_)}\frac{k\rho(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}(\boldsymbol{x}-\boldsymbol{y})\text{d}y_1\text{d}y_2\text{d}y_3$$where $(y_1,y_2,y_3)=\boldsymbol{y}$, where $\rho:D\subset\mathbb{R}^3\to\mathbb{R}$ is a bounded piecewise continuous function, $k$ a positive constant and $D\subset\mathbb{R}^3$ is bounded. If $k$ is Coulomb's constant, we can consider $\boldsymbol{E}$ as an expression of the electrical field in $\boldsymbol{x}$ generated by the charge having density $\rho$ on $D$.

I would like to prove that, under these assumptions, $\boldsymbol{E}:\mathbb{R}^3\to \mathbb{R}^3$ is continuous. Such function has been proved to be defined on all $\mathbb{R}^3$ here (at the same link there is a proof, in the addendum to the answer, that under such assumptions $\boldsymbol{E}\in C(\mathbb{R}^3)$, but I have to follow another path because I really do not understand the calculations and inequalities in that addendum).

I have included a trial of proof by me, hoping that somebody more mathematically cultivated can correct my trial, confirm its (improbable) correctness or provide an alternative answer (using tools that I can understand - of course - such as multivariate calculus, while I have no knowledge of Lebesgue integration in $\mathbb{R}^3$). I thank any answerer very much.

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Proof (trial of): Let us define$$\boldsymbol{E}_n(\boldsymbol{x}):=\int_{D\setminus B(\boldsymbol{x},\varepsilon_n)}\frac{k\rho(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}(\boldsymbol{x}-\boldsymbol{y})\text{d}y_1\text{d}y_2\text{d}y_3$$It can be easily shown, by using polar coordinates around $\boldsymbol{x}$, that the sequence $\{\boldsymbol{E}_n\}$ uniformly converges to $\boldsymbol{E}$ if $\lim_{n}\varepsilon_n=0$. Morever, the limit of a uniformly convergent sequence of continuous functions is a continuous function, which implies that, if $\boldsymbol{E}_n$ is continuous for any $n$ and $\varepsilon_n\to 0$, then $\boldsymbol{E}\in C(\mathbb{R}^3)$. To prove that $\boldsymbol{E}_n$ is continuous on $\mathbb{R}^3$, let us introduce the function $$\boldsymbol{F}_{n,\boldsymbol{x}_0}(\boldsymbol{x}):=\int_{D\setminus B(\boldsymbol{x}_0,\varepsilon_n)}\frac{k\rho(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}(\boldsymbol{x}-\boldsymbol{y})\text{d}y_1\text{d}y_2\text{d}y_3$$defined on $B(\boldsymbol{x}_0,\varepsilon_n/2)$, where $\boldsymbol{x}_0\in\mathbb{R}^3$ is a given point. Let us notice the domain of integration, different in general from that of $\boldsymbol{E}_n$, although $\boldsymbol{E}_n(\boldsymbol{x}_0)=\boldsymbol{F}_{n,\boldsymbol{x}_0}(\boldsymbol{x}_0)$. The function $\boldsymbol{f}:\overline{B\left(\boldsymbol{x}_0,\frac{\varepsilon_n}{2}\right)}\times\overline{D\setminus B(\boldsymbol{x}_0,\varepsilon_n)}\to\mathbb{R}^3$, where the over line means closure, defined by $\boldsymbol{f(\boldsymbol{x},\boldsymbol{y})}:=\frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}$ can be straightforwardly proved to be continuous (if I am not wrong) and therefore uniformly continuous on its compact domain. Therefore, for all $\varepsilon>0$ there is a $\delta$ such that, if $\|(\boldsymbol{x},\boldsymbol{y})-(\boldsymbol{x}_0,\boldsymbol{y})\|<\delta$, then $$\|\boldsymbol{F}_{n,\boldsymbol{x}_0}(\boldsymbol{x})-\boldsymbol{E}_n(\boldsymbol{x}_0)\|\leq k\sup_{D}|\rho|\int_{D\setminus B(\boldsymbol{x}_0,\varepsilon_n)}\left\|\frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}} - \frac{\boldsymbol{x}_0-\boldsymbol{y}}{\|\boldsymbol{x}_0-\boldsymbol{y}\|^{3}} \right\|\,\text{d}y_1\text{d}y_2\text{d}y_3$$ $$\leq k\sup_{D}|\rho|\,\varepsilon\,\mu(D\setminus B(\boldsymbol{x}_0,\varepsilon_n))$$ where $\mu$ is volume.

Moreover, I would say that, if we can prove that, as $\boldsymbol{x}\to \boldsymbol{x}_0$, $\boldsymbol{E}_n(\boldsymbol{x})$ approaches $\boldsymbol{F}_{n,\boldsymbol{x}_0}(\boldsymbol{x})$, we have proved that $\boldsymbol{E}_n$ is continuous in the arbitrarily chosen point $\boldsymbol{x}_0$, and I would think that we can prove that by using the fact that $\int_{D\setminus B(\boldsymbol{x},\varepsilon_n)}\int_{D\setminus B(\boldsymbol{x}_0,\varepsilon_n)}$ $=\int_{(D\cap B(\boldsymbol{x}_0,\varepsilon_n))\setminus B(\boldsymbol{x},\varepsilon_n)}-\int_{(D\cap B(\boldsymbol{x},\varepsilon_n))\setminus B(\boldsymbol{x}_0,\varepsilon_n)}$ we can write, for $\boldsymbol{x}\in B(\boldsymbol{x}_0,\delta)$, $\delta\le\varepsilon_n/2$, $$\left\|\int_{D\setminus B(\boldsymbol{x},\varepsilon_n)}\frac{k\rho(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}(\boldsymbol{x}-\boldsymbol{y})\text{d}y_1\text{d}y_2\text{d}y_3-\int_{D\setminus B(\boldsymbol{x}_0,\varepsilon_n)}\frac{k\rho(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}(\boldsymbol{x}-\boldsymbol{y})\text{d}y_1\text{d}y_2\text{d}y_3\right\|$$ $$=\Bigg\| \int_\limits{(D\cap B(\boldsymbol{x}_0,\varepsilon_n))\setminus B(\boldsymbol{x},\varepsilon_n)}^{} \frac{k\rho(\boldsymbol{y})(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}\text{d}y_1\text{d}y_2\text{d}y_3-\int_\limits{(D\cap B(\boldsymbol{x},\varepsilon_n))\setminus B(\boldsymbol{x}_0,\varepsilon_n)}^{}\frac{k\rho(\boldsymbol{y})(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^{3}}\text{d}y_1\text{d}y_2\text{d}y_3 \Bigg\|$$$$\le\int_\limits{(D\cap B(\boldsymbol{x}_0,\varepsilon_n))\setminus B(\boldsymbol{x},\varepsilon_n)}^{} \frac{k\sup|\rho|}{\varepsilon_n^2}\text{d}y_1\text{d}y_2\text{d}y_3+\int_\limits{(D\cap B(\boldsymbol{x},\varepsilon_n))\setminus B(\boldsymbol{x}_0,\varepsilon_n)}^{}\frac{k\sup|\rho|}{(\varepsilon_n-\delta)^2}\text{d}y_1\text{d}y_2\text{d}y_3 $$$$\le k\sup|\rho|\left(\frac{\mu((D\cap B(\boldsymbol{x}_0,\varepsilon_n))\setminus B(\boldsymbol{x},\varepsilon_n))}{\varepsilon_n^2}+\frac{\mu((D\cap B(\boldsymbol{x},\varepsilon_n))\setminus B(\boldsymbol{x}_0,\varepsilon_n))}{(\varepsilon_n -\delta)^2}\right)$$where both volumes $\mu$ approaches 0 as $\delta\to 0$. Which completes my proof.

Answer 1

Second answer (no measure theory): Let $f$ be a nice bounded function on $D$ (like $k\rho$). For suitable $g,$ define $Tg(x) = \int_D f(y)g(x-y)\,dy, x \in \mathbb {R}^3.$ One suitable $g$ is any $g\in C_c$ (where $C_c$ is the set of continuous functions on $\mathbb {R}^3$ with compact support). Other examples are $g(y) = y_k/|y|^3.$ These are the only examples we need worry about.

Claim 1: If $g \in C_c,$ then $Tg$ is continuous on $\mathbb {R}^3.$ This follows straightforwardly from the uniform continuity of $g$ on $\mathbb {R}^3,$ the boundedness of $f,$ and the boundedness of $D.$

Now each $g(y)=y_k/|y|^3$ is not in $C_c,$ but we'll take care of that. For small $\epsilon > 0$ and large $R> 0,$ define $\phi_{\epsilon,R}$ on $[0,\infty)$ in the following way: Join the points $(0,0),(\epsilon,0), (2\epsilon, 1), (R,1), (R+\epsilon,0)$ with straight line segments. That is the graph of $\phi_{\epsilon,R}$ on $[0, R+\epsilon].$ Define $\phi_{\epsilon,R} = 0$ on $(R+\epsilon,\infty).$ (It's good to draw a picture.) Note that each $g(y)\phi_{\epsilon,R}(|y|)\in C_c.$

Claim 2: Let $g(y) = y_k/|y|^3,$ and suppose $R>0.$ Then

$$\int_{B(0,R)} |g(y)-\phi_{\epsilon,R}(|y|)g(y)|\, dy < 12\pi\epsilon.$$

Proof: This comes right out of integrating in polar coordinates in $\mathbb {R}^3.$

So here's how we finish. Let $g$ be as in claim 2. We have $D \subset B(0,r)$ for some $r.$ Fix $a.$ Let $R = 1+|a|+r.$ Set $h(y) = g(y)\phi_{\epsilon,R}(|y|).$ Then for $x \in B(a,1),$

$$|Tg(x) - Tg(a)| \le |Tg(x) - Th(x)| + |Th(x) - Th(a)| + |Th(a) - Tg(a)|.$$

Now recall $|f|$ is bounded by some $M$ on $D.$ So

$$|Tg(x) - Th(x)|\le M\int_D |g(x-y) - h(x-y)|\,dy = M\int_{-D+x} |g(y) - h(y)|\,dy$$ $$ \le M\int_{B(0,R)}|g-h| \le M\cdot 12\pi\epsilon,$$

where we have used the fact that $-D+x\subset B(0,R).$ Same for $|Th(a) - Tg(a)|.$ Thus

$$\limsup_{x\to a}|Tg(x) - Tg(a)| \le M\cdot12\pi\epsilon + 0 + M\cdot 12\pi\epsilon.$$

We have used claim 1 to get that $0$ in the middle. Since $\epsilon$ is arbitrary, the $\limsup$ is $0,$ giving us the result.

Answer 2

I looked at your proof a bit and it seems to me you have the right ideas. But a little measure theory can make life easier.

Well known result: If $f\in L^\infty(\mathbb {R}^3), g \in L^1(\mathbb {R}^3),$ then

$$\tag 1 f\ast g (x) = \int_{\mathbb {R}^3} f(y)g(x-y)\,dy$$

is continuous on $\mathbb {R}^3.$ (And in fact $f\ast g (x)\to 0$ as $|x|\to \infty.$)

How would this apply to your situation? I'm thinking of $f=k\rho$ in $D,$ $f=0$ elsewhere, and $g(y) = y_k/|y|^3,k=1,2,3.$ There's a problem however: $g\not \in L^1(\mathbb {R}^3)$ because of the behavior of $g(y)$ for $|y|$ large.

Let's dispense with this problem. Let $g(y) = y_1/|y|^3$ for example. We have $D\subset B(0,R)$ for some $R>0.$ Fix $a;$ we will prove the desired continuity for $x \in B(a,1).$ Verify that if $x\in B(a,1),y\in D,$ then $|x-y| \le |a| + 1 + R.$ If we define $h = g\chi_{B(0,|a| + 1 + R)},$ then $h(x-y) = g(x-y)$ for $x\in B(a,1),y\in D.$ This tells us

$$\int_{\mathbb {R}^3} f(y)g(x-y)\,dy = \int_{D} f(y)g(x-y)\,dy =\int_{D} f(y)h(x-y)\,dy = \int_{\mathbb {R}^3} f(y)h(x-y)\,dy$$

for $x \in B(a,1).$ Because $h \in L^1(\mathbb {R}^3),$ $(1)$ shows the last integral is continuous everywhere, hence the second integral is continuous for $x\in B(a,1)$as promised.

So how do you get $(1)?$ You prove it first for continuous $g$ with compact support; here it follows from the uniform continuity of $g.$ Then you use the fact that if $g\in L^1(\mathbb {R^3})$ and $\epsilon>0,$ then there exists a continuous $h$ with compact support such that

$$\tag 2 \int_{\mathbb {R}^3} |g(y) - h(y)|\,dy < \epsilon.$$

So you approximate $g$ by such an $h$ and take an appropriate $\limsup$ to get $(1).$

If I've used some measure theory results you haven't seen, note that in your problem you have only the functions $g(y) = y_k/|y|^3$ to worry about, so approximating as in $(2)$ is pretty simple. I think you could fashion a proof using these ideas, probably avoiding measure theory altogether.