On computing distribution $\partial_i \partial_j (1/|x|)$ in three dimensions using regularization $1/\sqrt{|x|^2 + \varepsilon^2}$.

Question

Motivated by questions and answers in here and here, I would like to understand how to interpret and prove the following identity in a sense of distributions, where $x \in \mathbb{R}^3$ (assuming it is true; if not -- is there an obvious way how to change it so that it is true?).

$$ \lim \limits_{\varepsilon \to 0^+} \frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) = \mathrm{P.V.} \left( \frac{3x_i x_j-|x|^2 \delta_{ij}}{|x|^5} \right) - \frac{4 \pi}{3}\delta_{ij}\delta(x) $$

I understand previous equality in a sense that for each sufficiently nice (for example, smooth and with compact support) $f: \mathbb{R}^3 \to \mathbb{R}$ we have the following identity.

$$ \lim \limits_{\varepsilon \to 0^+} \int_{\mathbb{R}^3} \frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) f(x) \mathrm{d}^3x = \lim \limits_{R \to 0^+} \int_{\mathbb{R}^3 - B_R(0)} \left( \frac{3x_i x_j-|x|^2 \delta_{ij}}{|x|^5} \right) f(x) \mathrm{d}^3x - \frac{4 \pi}{3} \delta_{ij} f(0)$$

My main confusion is with proving existence of certain integrals and evaluating limits of integrals. I now provide some background for my calculations and then list of questions that I am confused about. I am okay with assuming dominated convergence / monotone convergence (or related) theorems.

For each $\varepsilon > 0$, function with values $1/\sqrt{|x|^2 + \varepsilon^2}$ is smooth, so one can compute the following. $$ \frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) = \frac{3x_i x_j-(|x|^2 + \varepsilon^2) \delta_{ij}}{(|x|^2 + \varepsilon^2)^{5/2}} $$

So, I understand why for each fixed $\varepsilon > 0$, $\int_{\mathbb{R}^3}\frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) f(x) \mathrm{d}^3x$ exists as it is an integral of smooth (so, continuous) function over a compact interval.

Question 1: However, how do we argue that limit $\lim \limits_{\varepsilon \to 0^+} \int_{\mathbb{R}^3} \frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) f(x) \mathrm{d}^3x$ exists?

I understand that for each $\varepsilon >0$ and $R > 0$ we have the following well-defined equality.

$$ \int_{\mathbb{R}^3} \frac{\partial^2 }{\partial x_i \partial x_j} \left( \frac{1}{\sqrt{|x|^2 + \varepsilon^2}} \right) f(x) \mathrm{d}^3x = \int_{\mathbb{R}^3 - B_R(0)} \left( \frac{3x_i x_j-(|x|^2 + \varepsilon^2) \delta_{ij}}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x + \int_{B_R(0)} \left( \frac{3x_i x_j}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x -\delta_{ij} \int_{B_R(0)} \left( \frac{1}{(|x|^2 + \varepsilon^2)^{3/2}} \right) f(x) \mathrm{d}^3x$$

Question 2: Intuitively, it seems that we first consider limit $\varepsilon \to 0^+$ and then we consider limit $R \to 0^+$. This is motivated by the next equality, which I believe I can get from dominated convergence theorem. Is this the correct strategy?

$$ \lim \limits_{\varepsilon \to 0^+} \int_{\mathbb{R}^3 - B_R(0)} \left( \frac{3x_i x_j-(|x|^2 + \varepsilon^2) \delta_{ij}}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x = \int_{\mathbb{R}^3 - B_R(0)} \left( \frac{3x_i x_j-|x|^2 \delta_{ij}}{|x|^5} \right) f(x) \mathrm{d}^3x$$

Question 3: In this case, how to show that the limit of previous integral as $R \to 0^+$ exists?

Now, I believe for the remaining terms (assuming that the strategy is to take $\varepsilon \to 0^+$ first and then $R \to 0^+$) I would like to show the following equalities.

$$ \lim \limits_{R \to 0^+} \lim \limits_{\varepsilon \to 0^+} \int_{B_R(0)} \left( \frac{3x_i x_j}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x = 0 \quad (*) \quad \lim \limits_{R \to 0^+} \lim \limits_{\varepsilon \to 0^+} \int_{B_R(0)} \left( \frac{1}{(|x|^2 + \varepsilon^2)^{3/2}} \right) f(x) \mathrm{d}^3x = \frac{4\pi}{3} f(0) \quad (*)$$

Let me consider my ideas for the first of these results as my confusion is the same for the second result. First, the usual bound with $|f|_{\infty}$ does not seem to work, as I get the following. Here, I first use triangle inequality, then use that for all $i$, $|x_i| \leq |x|$ and for all $x \in \mathbb{R}^3$, $|f(x)| \leq |f|_{\infty}$, by the assumption on $f$.

$$ \left| \int_{B_R(0)} \left( \frac{3x_i x_j}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x \right| \leq \int_{B_R(0)} \left( \frac{3 |x_i| |x_j|}{(|x|^2 + \varepsilon^2)^{5/2}} \right) |f(x)| \mathrm{d}^3x \leq |f|_{\infty} \int_{B_R(0)} \left( \frac{3 |x|^2}{(|x|^2 + \varepsilon^2)^{5/2}} \right) \mathrm{d}^3x$$

However, I believe that by going to spherical coordinates, previously established bound diverges as $\varepsilon \to 0^+$. So, I am forced to try another bound. Motivated by answer here, I agree that by symmetry one can show the following.

$$ \int_{B_R(0)} \left( \frac{3x_i x_j}{(|x|^2 + \varepsilon^2)^{5/2}} \right) f(x) \mathrm{d}^3x = \int_{B_R(0)} \left( \frac{3x_i x_j}{(|x|^2 + \varepsilon^2)^{5/2}} \right) \left( f(x) - f(0) \right) \mathrm{d}^3x $$

However, even though I understand that now we can use from continuity of $f$ that $\lim_{x \to 0} (f(x) - f(0)) = 0$, it is not clear to me how that helps with the bound.

Question 4: How to rigorously show that equalities in $(*)$ hold?

Answer 1

Let $T(x)=\frac{1}{|x|}$ for $x\in\Bbb{R}^3\setminus\{0\}$, and let $T_{\epsilon}(x)=\frac{1}{(|x|^2+\epsilon^2)^{1/2}}$ be the regularization. Also, let $f\in\mathcal{D}(\Bbb{R}^3)$ (smooth, compactly supported). Note that $T$ and $T_{\epsilon}$ are locally Lebesgue integrable, hence define distributions via integration. Also, $T_{\epsilon}\to T$ in the topology of distributions $\mathcal{D}'(\Bbb{R}^3)$ (which is the weak* topology, meaning $\langle T_{\epsilon},\phi \rangle:=\int_{\Bbb{R}^3} T_{\epsilon}\phi\,dx\to \int_{\Bbb{R}^3} T\phi\,dx=\langle T,\phi\rangle$ for all $\phi\in \mathcal{D}(\Bbb{R}^3)$) due to dominated convergence. As a result, we have that $\partial_i\partial_jT_{\epsilon}\to \partial_i\partial_j T$ in the topology of distributions as well (unwind the definitions to see why this is true).

1. My previous paragraph already answers why the limit in question (1) exists, but here's a more direct answer (actually it's the same argument, just without the "fancy" distributional vocabulary). Since $f$ is smooth and compactly supported, we can integrate by parts twice (and due to compact support of $f$, we can throw away the boundary terms) to get $\int_{\Bbb{R}^3}\frac{\partial^2T_{\epsilon}}{\partial x_i\partial x_j}(x)f(x)\,dx=\int_{\Bbb{R}^3}T_{\epsilon}(x)\frac{\partial^2f}{\partial x_i\partial x_j}(x)\,dx$. Now, $T_{\epsilon}f\to Tf$ pointwise except at the origin (i.e pointwise a.e), and $|T_{\epsilon}f|\leq |Tf|$ a.e, and the latter function is in $L^1$, so by Lebesgue's DCT, the RHS expression has a limit as $\epsilon\to 0^+$, and the limit is as we expect: \begin{align} \lim\limits_{\epsilon\to 0^+}\int_{\Bbb{R}^3}\frac{\partial^2T_{\epsilon}}{\partial x_i\partial x_j}(x)f(x)\,dx = \int_{\Bbb{R}^3}T(x)\frac{\partial^2f}{\partial x_i\partial x_j}(x)\,dx. \end{align}

Now, for the remaining questions it seems like you have a bunch of small issues everywhere, so rather than pinpointing each error (either aritmethic/typo or whatever) let me instead write out how the calculation goes. For each $\epsilon>0$, we have \begin{align} \int_{\Bbb{R}^3}\frac{\partial^2T_{\epsilon}}{\partial x_i\partial x_j}(x)f(x)\,dx&= \int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}(|x|^2+\epsilon^2)}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx\\ &=\int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx- \int_{\Bbb{R}^3}\frac{\epsilon^2\delta_{ij}}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx\tag{$*$}. \end{align}

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Dealing with the First integral in $(*)$

For the first term, we let $R>0$, and split the integral into the intergal over the ball and outside: \begin{align} \int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx&= \int_{B_R(0)}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx + \int_{B_R(0)^c}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx\\ &= \int_{B_R(0)}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}[f(x)-f(0)]\,dx + \int_{B_R(0)^c}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx,\tag{i} \end{align} where we have invoked symmetry, or more explicitly, that $\int_{S^2}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}\,d\sigma(x)=0$, where $d\sigma$ is the surface measure on the unit sphere (and hence the integral over the ball $B_R(0)$ is also zero, and hence $\int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)}f(0)\,dx=0$). Now, the reason for subtracting $f(0)$ is the following: as $\epsilon\to 0^+$, the original integrand $\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}} f(x)$ behaves like $\frac{C|x|^2}{(|x|^2+0^2)^{5/2}}M=\frac{CM}{|x|^3}$, where $M=\sup\limits_{|x|\leq R}|f(x)|$. Now, remembering that in polar coordinates the volume element (for radial functions) is $4\pi r^2\,dr$, this means after converting to polar coordinates, we have an integral like $\int_0^R\frac{1}{r}\,dr=\infty$. Ouch. So, we want to avoid the overall $\frac{1}{r}$ decay in the end. To accomplish this, we need to "upgrade" from a $\frac{1}{|x|^3}$ singularity, to something slightly milder. Now, by subtracting $f(0)$, we now have the term $[f(x)-f(0)]$, which by the mean-value theorem can be bounded by $|f(x)-f(0)|\leq M'|x|$, where $M'=\sup\limits_{|x|\leq R}\|Df(x)\|$ (supremum of the operator norm of the derivative). This extra power of $|x|$ now gives us a $\frac{1}{|x|^2}$ singularity, which is perfect since after changing to polar coordinates it will cancel with the $4\pi r^2\,dr$ term to yield a finite integral. So, long story short, $x\mapsto \frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}[f(x)-f(0)]$ is integrable over $B_R(0)$, and is dominated by $x\mapsto \left|\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}[f(x)-f(0)]\right|$, which we just argued is integrable over $B_R(0)$ (since it is bounded by $\frac{\text{const}}{|x|^2}$).

Hence, we can use dominated convergence on the first integral in (i) to claim the limit as $\epsilon\to 0^+$ exists, and equals what we expect. FOr the second term in (i), we can definitely use dominated convergence since we're integrating away from the origin. Therefore, by this double use of dominated convergence, we have \begin{align} \lim_{\epsilon\to 0^+}\int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx&= \int_{B_R(0)}\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}[f(x)-f(0)]\,dx+ \int_{B_R(0)^c}\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}f(x)\,dx. \end{align} Finally, since $R>0$ was arbitrary, we can take the limit $R\to 0^+$ (which exists since the LHS doesn't even depend on $R$). Now, note in particular that for the first integral here, over $B_R(0)$, we have an $L^1(B_1(0))$ function, so as $R\to 0^+$, the dominated convergence theorem tells us the limit is $0$. Therefore, \begin{align} \lim_{\epsilon\to 0^+}\int_{\Bbb{R}^3}\frac{3x_ix_j-\delta_{ij}|x|^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx&=0+\lim_{R\to 0^+}\int_{B_R(0)^c}\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}f(x)\,dx\\ &=\left\langle \text{P.V.}\left(\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}\right),f\right\rangle \end{align}

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Dealing with the second integral in $(*)$

Here, we have \begin{align} \int_{\Bbb{R}^3}\frac{\epsilon^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx&= \int_{\Bbb{R}^3}\frac{1}{\epsilon^3}\frac{1}{(1+|x/\epsilon|^2)^{5/2}}f(x)\,dx. \end{align} Now, define $\zeta(x)=\frac{1}{(1+|x|^2)^{5/2}}$, and $\zeta_{\epsilon}(x)=\frac{1}{\epsilon^3}\zeta\left(\frac{x}{\epsilon}\right)$. So the integral above is $\int_{\Bbb{R}^3}\zeta_{\epsilon}(x)f(x)\,dx$. As I described in Dirac's $\delta$ distribution smooth approximation, this equals $\left(\int_{\Bbb{R}^3}\zeta(x)\,dx\right)f(0)$. In other words, $\zeta_{\epsilon}\to \left(\int_{\Bbb{R}^3}\zeta(x)\,dx\right)\delta_0$ in the topology of distributions $\mathcal{D}'(\Bbb{R}^3)$. I leave it as a 'simple' calculus exercise for you to change to polar coordinates and prove that $\int_{\Bbb{R}^3}\zeta(x)\,dx=\frac{4\pi}{3}$. Hence, we have \begin{align} \lim_{\epsilon\to 0^+}\int_{\Bbb{R}^3}\frac{\epsilon^2}{(|x|^2+\epsilon^2)^{5/2}}f(x)\,dx&= \frac{4\pi}{3}f(0). \end{align}

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Therefore, putting everything together, we have that in the sense of distributions, \begin{align} \frac{\partial^2}{\partial x_i\partial x_j}\left(\frac{1}{|x|}\right)&=\text{P.V.}\left(\frac{3x_ix_j-\delta_{ij}|x|^2}{|x|^5}\right)-\frac{4\pi}{3}\delta_{ij}\delta_0. \end{align}