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I was dealing with a problem where a delta approximating function is given. Apart from the fact that this delta approximating functions has the form $$g_\epsilon(x) = \epsilon^{-3}g(\epsilon^{-1}x)\tag{1}$$ and that $g\in L^1(\mathbb{R}^3)\cap L^2(\mathbb{R}^3)$ with $\int g(x)\,\mathrm{d}^3x = 1$, no specific form of the $g_\epsilon$ is given. What i then was trying to evaluate is the integral $$\int g_\epsilon(x)f(x)\,\mathrm{d}^3x\tag{2}$$ for which i would imagine that $$\lim_{\epsilon\to 0}\int g_\epsilon(x)f(x)\,\mathrm{d}^3x = f(0)\tag{3}$$ but I'm not quite sure what to do with the integral $(2)$ which is what interests me.

What i can say about $(2)$? And, if there's one, what's an explicit form of that integral?

Maybe a bit of context could help. I'm trying to study the resolvent of an hamiltonian which contains a term $$\mu_\epsilon(g_\epsilon, \cdot)g_\epsilon\qquad \mu_\epsilon\in \mathbb{R}$$ and to do so i'm trying to find the action of the hamiltonian in Fourier space by computing the integral $$\mu_\epsilon\int(g_\epsilon, f)g_\epsilon(x)e^{-ikx}\,\mathrm{d}^3x = \int g_\epsilon(x) e^{-ikx}\,\mathrm{d}^3x\int \overline{g_\epsilon}(y)f(y)\,\mathrm{d}^3 y$$ where the second integral is what I'm trying to understand.

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  • $\begingroup$ You might want to replace each $\mathrm{d}x$ with $\mathrm{d}^3x$. $\endgroup$
    – J.G.
    Jul 6, 2020 at 17:14
  • $\begingroup$ @J.G. Done! I thought it could be understood $\endgroup$
    – Quiver
    Jul 6, 2020 at 17:15
  • $\begingroup$ It is understood. It should be understood. $\endgroup$
    – Medo
    Jul 6, 2020 at 17:16
  • $\begingroup$ This $\mathrm d^3x$ is rarely used in math papers (and this male sense since it correspond to an order $1$ difference, and I will not even mention differential forms ...). why this advice? :0 It was clearly stated that $g\in L^1(\mathbb{R}^3)$, so everything was clear $\endgroup$
    – LL 3.14
    Jul 6, 2020 at 17:50

2 Answers 2

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If $f\in C^{\infty}_{c}(\mathbb{R}^{3})$ (infinitely differentiable with compact support), then your guess is correct. Explanation: Change variables $y=x/\epsilon$, then $dx=\epsilon^{3}dy$ (notice that $x,y\in \mathbb{R}^{3}$). Now

$$\int_{\mathbb{R}^{3}} \frac{1}{\epsilon^3}g(\frac{x}{\epsilon})f(x)dx=\int_{\mathbb{R}^{3}}g(y) f(\epsilon y) dy\rightarrow \int_{\mathbb{R}^{3}}g(y) f(0) dy$$ by the dominated convergence theorem (notice that $fg\in L^{1}(\mathbb{R}^{3})$ and that $f$ is continuous at $0$). Finally recall that $\int_{\mathbb{R}^{3}}g(y)dy=1$.

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  • $\begingroup$ But is there something I can say about the integral without taking the limit? Like some approximate form which in the limit takes the desired value? $\endgroup$
    – Quiver
    Jul 6, 2020 at 17:29
  • $\begingroup$ For a general function $g$, you will certainly not find a simpler expression ... $\endgroup$
    – LL 3.14
    Jul 6, 2020 at 17:54
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The integral in (2) is $\int_{\Bbb R^3}g(x)f(\epsilon y)\mathrm{d}^3y$. For sufficiently nice $f$ continuous at $0$ with finite $f(0)$, we can move a $\lim_{\epsilon\to0^+}$ operator inside the integral. This gives $\int_{\Bbb R^3}g(x)f(0)\mathrm{d}^3y=f(0)\int_{\Bbb R^3}g(x)\mathrm{d}^3y=f(0)$.

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  • $\begingroup$ Same question done to Medo: is there something I can say about the integral without taking the limit? Like some approximate form which in the limit takes the desired value? $\endgroup$
    – Quiver
    Jul 6, 2020 at 17:29
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    $\begingroup$ @DavideMorgante You could use $f(\epsilon y)\approx f(0)+\epsilon yf^\prime(0)$ to approximate the integral as $f(0)+\epsilon f^\prime(0)\color{red}{\int_{\Bbb R}yg(y)dy}$. However, if the red integral vanishes, as happens e.g. if $g$ is even (which common with nascent delta functions), we'd need to go to the next order, getting $f(0)+\tfrac12\epsilon^2f^{\prime\prime}(0)\int_{\Bbb R}y^2g(y)dy$. $\endgroup$
    – J.G.
    Jul 6, 2020 at 17:47
  • $\begingroup$ Ok, that could be something! Thank you, l'll try with this. $\endgroup$
    – Quiver
    Jul 6, 2020 at 17:48

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